Find $\int x \cos x\, dx$

Find ∫ x cos x dx

Worked Examples · Example 17

\int x \cos x\, dx

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Appeared in past exams:KEAM 2024· Set eng-2024-0609· 4mexact

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Concept understanding — Integration by Parts

Integration by Parts

The idea: reverse the product rule

Some integrands are a product of two very different functions — $x e^x$ , $x\cos x$ , $\log x$ , $x\sin^{-1}x$ — where substitution gets you nowhere. Integration by parts is the tool for these. It comes straight from reversing the product rule for differentiation.

Starting from $\dfrac{d}{dx}(uv)=u\,v' + u'\,v$ and integrating both sides gives the working formula:

\int u\,\frac{dv}{dx}\,dx = uv - \int v\,\frac{du}{dx}\,dx.

In words: integral of (first $\times$ derivative-of-second) $=$ first $\times$ integral-of-second $-$ integral of (derivative-of-first $\times$ integral-of-second).

Choosing $u$ : the ILATE rule

The whole game is picking which factor is $u$ (to differentiate) and which is $dv$ (to integrate). Pick $u$ by ILATE — the first type that appears:

Inverse trig ( $\sin^{-1}x$ ), Logarithmic ( $\log x$ ), Algebraic ( $x^2$ ), Trigonometric ( $\sin x$ ), Exponential ( $e^x$ ).

Whatever comes first in ILATE becomes $u$ ; the rest is $dv$ . This makes the new integral $\int v\,du$ simpler than the one you started with.

Worked idea

For $\int x e^x\,dx$ : algebraic before exponential, so $u=x$ , $dv=e^x dx$ . Then $du=dx$ , $v=e^x$ :

\int x e^x\,dx = x e^x - \int e^x\,dx = x e^x - e^x + C = e^x(x-1)+C.

Tip

A single log or a single inverse-trig function ( $\int\log x\,dx$ , $\int\sin^{-1}x\,dx$ ) is still "by parts" — take the other factor as $1$ . And for the special form $\int e^x\big(f(x)+f'(x)\big)dx$ , the answer is simply $e^x f(x)+C$ .

Note

If applying the formula gives you back a multiple of the original integral (as with $\int e^x\sin x\,dx$ ), don't panic — solve for the integral algebraically.

Integration by Parts is one of the most tested methods in the NCERT Class 12 Mathematics chapter on Integrals, and "integration by parts formula ILATE rule" along with "integration by parts class 12 important questions" are among the top searches for students preparing for CBSE board exams and JEE Main calculus. The same ILATE-based technique extends naturally into JEE Advanced integral calculus problems built on this NCERT Class 12 foundation.

The integral $\int x \cos x \, dx$ is solved using integration by parts (the product rule in reverse). Choosing $u = x$ and $dv = \cos x \, dx$ gives the result $x \sin x + \cos x + C$ .

Why integration by parts?

When you see a product of two different kinds of functions — here $x$ (algebraic) and $\cos x$ (trigonometric) — there’s no simple reverse derivative. The product rule for differentiation says $(uv)' = u'v + uv'$ , so rearranging gives:

\int u \, dv = uv - \int v \, du

This is integration by parts. The trick is to pick $u$ so that $du$ is simpler, and $dv$ so that $v$ is easy to integrate.

Tip

A handy mnemonic for choosing $u$ is LIATE: Logarithmic, Inverse trig, Algebraic, Trigonometric, Exponential. Pick $u$ from the leftmost type in the product. Here $x$ is Algebraic, $\cos x$ is Trigonometric — so $u = x$ wins.

Step-by-step solution

1. Choose $u$ and $dv$

Let $u = x$ and $dv = \cos x \, dx$ .

Why? Because $du = dx$ becomes simpler (the power drops), and $v = \sin x$ is easy to integrate.

2. Compute $du$ and $v$

du = dx

v = \int \cos x \, dx = \sin x

3. Apply the integration by parts formula

\int x \cos x \, dx = x \sin x - \int \sin x \, dx

Note

The minus sign comes from $uv - \int v \, du$ . Don’t forget it — a common slip is to write $+$ instead.

4. Integrate the remaining term

\int \sin x \, dx = -\cos x + C

So:

\int x \cos x \, dx = x \sin x - (-\cos x) + C = x \sin x + \cos x + C

5. Check by differentiating

Differentiate $x \sin x + \cos x$ :

Derivative of $x \sin x$ : using product rule, $1 \cdot \sin x + x \cdot \cos x = \sin x + x \cos x$
Derivative of $\cos x$ : $-\sin x$

Sum: $\sin x + x \cos x - \sin x = x \cos x$ — matches the integrand. Perfect.

Watch out

A common mistake is to choose $u = \cos x$ and $dv = x \, dx$ . Then $du = -\sin x \, dx$ and $v = \frac{x^2}{2}$ , leading to $\frac{x^2}{2} \cos x + \frac{1}{2} \int x^2 \sin x \, dx$ — a harder integral. Always pick $u$ so that $du$ is simpler.

✓Final answer

\boxed{x \sin x + \cos x + C}

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

TG EAPCET 2026Set eng-2026-05-09-FN1 markMCQ
$\int \sin^{-1}x dx - \int \cos^{-1}x dx =$
›Reveal solutionSolution

Using the standard integrals of $\sin^{-1}x$ and $\cos^{-1}x$ , the difference is $x(\sin^{-1}x-\cos^{-1}x)+2\sqrt{1-x^{2}}+c$ .

Standard results (by integration by parts):
$\int \sin^{-1}x\,dx = x\sin^{-1}x + \sqrt{1-x^{2}} + c$ $\int \cos^{-1}x\,dx = x\cos^{-1}x - \sqrt{1-x^{2}} + c$
Subtract:
$\int \sin^{-1}x\,dx - \int \cos^{-1}x\,dx = \big(x\sin^{-1}x + \sqrt{1-x^{2}}\big) - \big(x\cos^{-1}x - \sqrt{1-x^{2}}\big)$ $= x\left(\sin^{-1}x - \cos^{-1}x\right) + 2\sqrt{1-x^{2}} + c$
✓Final answer
$x\left[\sin^{-1}x - \cos^{-1}x\right] + 2\sqrt{1-x^{2}} + c$
TG EAPCET 2025Set eng-2025-05-03-AN1 markMCQ
$\int x \tan^{-1} \sqrt{\frac{1+x^2}{1-x^2}} dx =$
›Reveal solutionSolution
The key is to simplify the inverse tangent argument using a trigonometric substitution, then integrate by parts. The correct antiderivative matches option (B).

We start with the integral
$I = \int x \tan^{-1} \sqrt{\frac{1+x^2}{1-x^2}} \, dx.$
Concept & Intuition

The argument $\sqrt{\frac{1+x^2}{1-x^2}}$ looks like a tangent of some angle. Recall that $\tan^2\theta = \frac{\sin^2\theta}{\cos^2\theta}$ . If we set $x^2 = \cos 2\theta$ , then $1+x^2 = 1+\cos2\theta = 2\cos^2\theta$ and $1-x^2 = 1-\cos2\theta = 2\sin^2\theta$ . Their ratio becomes $\cot^2\theta$ , so the square root is $\cot\theta$ . Then $\tan^{-1}(\cot\theta) = \frac{\pi}{2} - \theta$ . This transforms the messy inverse tangent into a simple linear function of $\theta$ .

Step-by-step solution

Substitute $x^2 = \cos 2\theta$ Let $x^2 = \cos 2\theta$ , so that $0 \le 2\theta \le \pi$ (to keep things principal). Then

$1+x^2 = 1+\cos2\theta = 2\cos^2\theta, \quad 1-x^2 = 1-\cos2\theta = 2\sin^2\theta.$
Hence
$\sqrt{\frac{1+x^2}{1-x^2}} = \sqrt{\frac{2\cos^2\theta}{2\sin^2\theta}} = \sqrt{\cot^2\theta} = \cot\theta \quad (\text{since } \theta \in [0,\pi/2]).$
Therefore
$\tan^{-1}\sqrt{\frac{1+x^2}{1-x^2}} = \tan^{-1}(\cot\theta) = \frac{\pi}{2} - \theta.$

Rewrite the integral in terms of $\theta$ From $x^2 = \cos 2\theta$ , differentiate: $2x\,dx = -2\sin 2\theta\,d\theta$ → $x\,dx = -\sin 2\theta\,d\theta$ . Also $\sin 2\theta = 2\sin\theta\cos\theta$ . So

$I = \int \left( \frac{\pi}{2} - \theta \right) (-\sin 2\theta)\,d\theta = -\int \left( \frac{\pi}{2} - \theta \right) \sin 2\theta \, d\theta.$

Integrate by parts Let $u = \frac{\pi}{2} - \theta$ and $dv = \sin 2\theta\,d\theta$ . Then $du = -d\theta$ and $v = -\frac{1}{2}\cos 2\theta$ . Integration by parts:

$\int u\,dv = uv - \int v\,du.$
So
$\int \left( \frac{\pi}{2} - \theta \right) \sin 2\theta \, d\theta = \left( \frac{\pi}{2} - \theta \right)\left(-\frac{1}{2}\cos 2\theta\right) - \int \left(-\frac{1}{2}\cos 2\theta\right)(-d\theta).$
Simplify:
$= -\frac{1}{2}\left( \frac{\pi}{2} - \theta \right)\cos 2\theta - \frac{1}{2}\int \cos 2\theta \, d\theta.$
The integral of $\cos 2\theta$ is $\frac{1}{2}\sin 2\theta$ . Thus
$\int \left( \frac{\pi}{2} - \theta \right) \sin 2\theta \, d\theta = -\frac{1}{2}\left( \frac{\pi}{2} - \theta \right)\cos 2\theta - \frac{1}{4}\sin 2\theta + C.$

Return to $x$ Recall $I = -$ of the above, so

$I = \frac{1}{2}\left( \frac{\pi}{2} - \theta \right)\cos 2\theta + \frac{1}{4}\sin 2\theta + C.$
Now substitute back:

$\cos 2\theta = x^2$

$\theta = \frac{1}{2}\cos^{-1}(x^2)$

$\sin 2\theta = \sqrt{1-\cos^2 2\theta} = \sqrt{1-x^4}$ (positive for principal range).

Hence
$I = \frac{1}{2}\left( \frac{\pi}{2} - \frac{1}{2}\cos^{-1}(x^2) \right) x^2 + \frac{1}{4}\sqrt{1-x^4} + C.$
Simplify the bracket:
$\frac{1}{2}\left( \frac{\pi}{2} - \frac{1}{2}\cos^{-1}(x^2) \right) = \frac{\pi}{4} - \frac{1}{4}\cos^{-1}(x^2) = \frac{1}{4}\left( \pi - \cos^{-1}(x^2) \right).$
Therefore
$I = \frac{x^2}{4}\left( \pi - \cos^{-1} x^2 \right) + \frac{1}{4}\sqrt{1-x^4} + C.$

Match with options This matches exactly option (B).

Watch out

A common mistake is to confuse $\sqrt{1-x^4}$ with $\sqrt{1-x^2}$ . The substitution $x^2 = \cos 2\theta$ forces the appearance of $x^4$ , not $x^2$ , under the square root.

Tip

The identity $\tan^{-1}(\cot\theta) = \frac{\pi}{2} - \theta$ is the cleanest way to handle such nested inverse trig functions. Always look for a substitution that turns the argument into a simple trig ratio.

✓Final answer
$The correct option is (B) . ANSWER: B$
TG EAPCET 2025Set eng-2025-05-03-FN1 markMCQ
$\int x \tan^{-1} \sqrt{\frac{1+x^2}{1-x^2}} dx =$
›Reveal solutionSolution
The key is to simplify the inverse‑tangent argument using a trigonometric substitution; after simplification the integral reduces to a standard form, and the result matches option (D).

We start with
$I = \int x \tan^{-1}\sqrt{\frac{1+x^2}{1-x^2}}\,dx.$
Concept & Intuition

The expression inside the square root, $\frac{1+x^2}{1-x^2}$ , looks like something trigonometric. If we set $x^2 = \cos 2\theta$ (or equivalently $x = \sqrt{\cos 2\theta}$ ), then
$\frac{1+x^2}{1-x^2} = \frac{1+\cos 2\theta}{1-\cos 2\theta} = \frac{2\cos^2\theta}{2\sin^2\theta} = \cot^2\theta.$
Thus $\sqrt{\frac{1+x^2}{1-x^2}} = \cot\theta$ , and $\tan^{-1}(\cot\theta) = \frac{\pi}{2} - \theta$ . This turns the messy inverse tangent into a simple linear function of $\theta$ . Then we change variables from $x$ to $\theta$ and integrate.

Step‑by‑step solution

Substitute $x^2 = \cos 2\theta$ . Let $0 \le 2\theta \le \pi$ so that $x^2 \in [-1,1]$ (the domain of the square root). Then

$\frac{1+x^2}{1-x^2} = \frac{1+\cos 2\theta}{1-\cos 2\theta} = \frac{2\cos^2\theta}{2\sin^2\theta} = \cot^2\theta.$
Hence
$\sqrt{\frac{1+x^2}{1-x^2}} = \cot\theta \quad (\text{positive root, since } \theta \in [0,\pi/2]).$

Simplify the inverse tangent. For $\theta \in (0,\pi/2)$ , $\tan^{-1}(\cot\theta) = \frac{\pi}{2} - \theta$ . Therefore

$\tan^{-1}\sqrt{\frac{1+x^2}{1-x^2}} = \frac{\pi}{2} - \theta.$

Change the differential. From $x^2 = \cos 2\theta$ , differentiate:

$2x\,dx = -2\sin 2\theta\,d\theta \quad\Rightarrow\quad x\,dx = -\sin 2\theta\,d\theta.$
So the integral becomes
$I = \int \left(\frac{\pi}{2} - \theta\right) (-\sin 2\theta)\,d\theta = -\int \left(\frac{\pi}{2} - \theta\right) \sin 2\theta\,d\theta.$

Integrate with respect to $\theta$ . Let $u = \frac{\pi}{2} - \theta$ , then $du = -d\theta$ and $\sin 2\theta = \sin(\pi - 2u) = \sin 2u$ . Also $\theta = \frac{\pi}{2} - u$ . Alternatively, integrate directly by parts:

$I = -\int \left(\frac{\pi}{2} - \theta\right) \sin 2\theta\,d\theta.$
Set $A = \frac{\pi}{2} - \theta$ , $dB = \sin 2\theta\,d\theta$ . Then $dA = -d\theta$ , $B = -\frac{1}{2}\cos 2\theta$ .

Integration by parts gives
$I = -\left[ \left(\frac{\pi}{2} - \theta\right)\left(-\frac{1}{2}\cos 2\theta\right) - \int \left(-\frac{1}{2}\cos 2\theta\right)(-d\theta) \right] + C.$
Simplify carefully:
$I = -\left[ -\frac{1}{2}\left(\frac{\pi}{2} - \theta\right)\cos 2\theta - \frac{1}{2}\int \cos 2\theta\,d\theta \right] + C.$
The minus sign outside flips signs:
$I = \frac{1}{2}\left(\frac{\pi}{2} - \theta\right)\cos 2\theta + \frac{1}{2}\int \cos 2\theta\,d\theta + C.$
Now $\int \cos 2\theta\,d\theta = \frac{1}{2}\sin 2\theta$ . So
$I = \frac{1}{2}\left(\frac{\pi}{2} - \theta\right)\cos 2\theta + \frac{1}{4}\sin 2\theta + C.$

Return to $x$ . Recall $x^2 = \cos 2\theta$ and $\sin 2\theta = \sqrt{1-\cos^2 2\theta} = \sqrt{1-x^4}$ (positive for $\theta\in[0,\pi/2]$ ). Also $\theta = \frac{1}{2}\cos^{-1}(x^2)$ . Hence

$\frac{\pi}{2} - \theta = \frac{\pi}{2} - \frac{1}{2}\cos^{-1}(x^2) = \frac{1}{2}\left(\pi - \cos^{-1}(x^2)\right).$
Substitute:
$I = \frac{1}{2}\cdot \frac{1}{2}\left(\pi - \cos^{-1}(x^2)\right) \cdot x^2 + \frac{1}{4}\sqrt{1-x^4} + C.$
Simplify:
$I = \frac{x^2}{4}\left(\pi - \cos^{-1} x^2\right) + \frac{1}{4}\sqrt{1-x^4} + C.$
This matches option (D).

Watch out

A common mistake is to forget the sign when differentiating $x^2 = \cos 2\theta$ or to mishandle the integration by parts signs. Also note that $\sqrt{1-x^4}$ is not the same as $\sqrt{1-x^2}$ ; the correct expression involves $x^4$ .

Tip

The substitution $x^2 = \cos 2\theta$ is the cleanest because it directly turns the ratio into $\cot^2\theta$ . Alternatively, one could try $x = \sin t$ , but that leads to a more complicated path.

✓Final answer
$The correct option is (D) . ANSWER: D$
TG EAPCET 2024Set eng-2024-05-09-AN1 markMCQ
$\int x^3 \sin 3x \, dx = f(x) \cos 3x + g(x) \sin 3x + c$
›Reveal solutionSolution
The problem gives the antiderivative of $x^3\sin 3x$ in a specific form; by differentiating both sides and matching coefficients, we find $27(f(x)+x g(x)) = 18x^3 + 8x$ , which corresponds to option (D).

We are told that
$\int x^3 \sin 3x \, dx = f(x) \cos 3x + g(x) \sin 3x + c,$
and we need $27\bigl(f(x) + x g(x)\bigr)$ . The key idea: differentiate both sides to eliminate the integral, then compare coefficients of $\cos 3x$ and $\sin 3x$ to solve for $f(x)$ and $g(x)$ .

Differentiate both sides Since the derivative of the integral is the integrand, we have

$x^3 \sin 3x = \frac{d}{dx}\bigl[f(x)\cos 3x + g(x)\sin 3x\bigr].$
Use the product rule on each term:
$\frac{d}{dx}[f(x)\cos 3x] = f'(x)\cos 3x - 3f(x)\sin 3x,$ $\frac{d}{dx}[g(x)\sin 3x] = g'(x)\sin 3x + 3g(x)\cos 3x.$
Adding them:
$x^3 \sin 3x = \bigl[f'(x) + 3g(x)\bigr]\cos 3x + \bigl[g'(x) - 3f(x)\bigr]\sin 3x.$

Match coefficients The left side has no $\cos 3x$ term, so its coefficient must be zero:

$f'(x) + 3g(x) = 0 \quad \text{(1)}$
The coefficient of $\sin 3x$ on the left is $x^3$ , so:
$g'(x) - 3f(x) = x^3 \quad \text{(2)}$

Solve the system From (1): $f'(x) = -3g(x)$ . Differentiate (2):

$g''(x) - 3f'(x) = 3x^2.$
Substitute $f'(x) = -3g(x)$ :
$g''(x) - 3(-3g(x)) = g''(x) + 9g(x) = 3x^2.$
This is a second-order linear ODE. The homogeneous solution is $g_h(x) = A\cos 3x + B\sin 3x$ , but since $f$ and $g$ are likely polynomials (the integrand is a polynomial times sine), we try a particular solution of the form $g_p(x) = ax^2 + b$ .

Then $g_p'' = 2a$ , so:
$2a + 9(ax^2 + b) = 3x^2 \implies 9a x^2 + (2a + 9b) = 3x^2.$
Matching: $9a = 3 \Rightarrow a = \frac{1}{3}$ , and $2a + 9b = 0 \Rightarrow \frac{2}{3} + 9b = 0 \Rightarrow b = -\frac{2}{27}$ .

So $g(x) = \frac{1}{3}x^2 - \frac{2}{27}$ (ignoring homogeneous part, as it would introduce extra trig terms not present in the given form).

Find $f(x)$ From (1): $f'(x) = -3g(x) = -3\left(\frac{1}{3}x^2 - \frac{2}{27}\right) = -x^2 + \frac{2}{9}$ . Integrate: $f(x) = -\frac{x^3}{3} + \frac{2}{9}x + C$ . The constant $C$ would produce a term $C\cos 3x$ in the antiderivative, but the integral of $x^3\sin 3x$ has no pure cosine term (check by differentiating: it would give a sine term), so $C = 0$ . Thus

$f(x) = -\frac{x^3}{3} + \frac{2}{9}x.$

Compute the required expression We need $27\bigl(f(x) + x g(x)\bigr)$ . First, $x g(x) = x\left(\frac{1}{3}x^2 - \frac{2}{27}\right) = \frac{1}{3}x^3 - \frac{2}{27}x$ . Then

$f(x) + x g(x) = \left(-\frac{x^3}{3} + \frac{2}{9}x\right) + \left(\frac{1}{3}x^3 - \frac{2}{27}x\right) = \left(-\frac{1}{3}+\frac{1}{3}\right)x^3 + \left(\frac{2}{9} - \frac{2}{27}\right)x.$
The $x^3$ terms cancel. The $x$ coefficient: $\frac{2}{9} = \frac{6}{27}$ , so $\frac{6}{27} - \frac{2}{27} = \frac{4}{27}$ . Hence
$f(x) + x g(x) = \frac{4}{27}x.$
Multiply by 27:
$27\bigl(f(x) + x g(x)\bigr) = 27 \cdot \frac{4}{27}x = 4x.$
Watch out

A common mistake is to forget the constant of integration when finding $f(x)$ ; here it must be zero to avoid an extraneous $\cos 3x$ term in the antiderivative. Also, note that the homogeneous solution of the ODE would introduce sine/cosine terms that don't match the polynomial form given.

Tip

Instead of solving the ODE fully, you could guess that $f$ and $g$ are polynomials of degree 3 and 2 respectively (since differentiating reduces degree by one, and the integrand is degree 3 times sine). Then set up undetermined coefficients directly from the differentiated equation — it's faster.

✓Final answer
$The correct option is (C) . ANSWER: C$
TG EAPCET 2024Set eng-2024-05-10-FN1 markMCQ
$\int (\log x)^3 \, dx =$
›Reveal solutionSolution
The integral of $(\log x)^3$ is found by repeated integration by parts, reducing the power of the logarithm each time. The final result is $x\big[(\log x)^3 - 3(\log x)^2 + 6\log x - 6\big] + c$ , which corresponds to option (B).

Concept & Intuition

When integrating powers of $\log x$ , the key trick is to treat $\log x$ as the function to differentiate and $1$ (or $dx$ ) as the function to integrate. Why? Because the derivative of $\log x$ is $1/x$ , which cancels the $x$ that appears when we integrate $1$ . This lets us reduce the exponent of the logarithm step by step. It’s like peeling an onion: each integration by parts lowers the power by one, until we’re left with a simple integral.

Step-by-step solution

Set up integration by parts Let $u = (\log x)^3$ and $dv = dx$ . Then $du = 3(\log x)^2 \cdot \frac{1}{x}\, dx$ and $v = x$ . The formula $\int u\,dv = uv - \int v\,du$ gives:

$\int (\log x)^3\,dx = x(\log x)^3 - \int x \cdot 3(\log x)^2 \cdot \frac{1}{x}\,dx = x(\log x)^3 - 3\int (\log x)^2\,dx.$

Reduce the power again Now we need $\int (\log x)^2\,dx$ . Use the same trick: let $u = (\log x)^2$ , $dv = dx$ . Then $du = 2(\log x)\cdot\frac{1}{x}\,dx$ , $v = x$ .

$\int (\log x)^2\,dx = x(\log x)^2 - \int x \cdot 2(\log x)\cdot\frac{1}{x}\,dx = x(\log x)^2 - 2\int \log x\,dx.$

Handle $\int \log x\,dx$ This is a classic: let $u = \log x$ , $dv = dx$ . Then $du = \frac{1}{x}dx$ , $v = x$ .

$\int \log x\,dx = x\log x - \int x\cdot\frac{1}{x}\,dx = x\log x - \int 1\,dx = x\log x - x + c_1.$

Substitute back From step 2:

$\int (\log x)^2\,dx = x(\log x)^2 - 2(x\log x - x) + c_2 = x(\log x)^2 - 2x\log x + 2x + c_2.$
From step 1:
$\int (\log x)^3\,dx = x(\log x)^3 - 3\left[ x(\log x)^2 - 2x\log x + 2x \right] + c.$

Simplify

$= x(\log x)^3 - 3x(\log x)^2 + 6x\log x - 6x + c.$
Factor $x$ :
$= x\left[ (\log x)^3 - 3(\log x)^2 + 6\log x - 6 \right] + c.$
Tip

Notice the pattern: integrating $(\log x)^n$ always yields $x$ times a polynomial in $\log x$ of degree $n$ , with alternating signs and factorial-like coefficients. For $n=3$ , the coefficients are $1, -3, +6, -6$ (which are $3!$ , $-3\cdot2!$ , $+3\cdot2$ , $-3!$ ? Actually they come from repeated integration by parts).

Watch out

A common mistake is to forget the factor of $x$ outside the bracket. Options (A), (C), and (D) all lack the correct factor or have it in the wrong place. Always check by differentiating: if you differentiate option (B), you should get $(\log x)^3$ .

✓Final answer
$The correct option is (B) . ANSWER: B$
CA Foundation 2024Set sep-20241 markMCQ
$\int \log_e x \, dx$
›Reveal solutionSolution

∫ln x dx = x ln x − x + c = x·ln(x/e) + c.

Step 1 — Integration by parts

Take $u = \ln x$ , $dv = dx$ , so $du = \dfrac{1}{x}dx$ , $v = x$ :
$\int \ln x\,dx = x\ln x - \int x \cdot \dfrac{1}{x}\,dx$
Step 2 — Complete the integral
$= x\ln x - \int 1\,dx = x\ln x - x + c$
Step 3 — Rewrite in the option's form

Factor x and use $1 = \ln e$ :
$x(\ln x - 1) = x(\ln x - \ln e) = x\ln\!\left(\dfrac{x}{e}\right)$ $\int \ln x\,dx = x\log_e\!\left(\dfrac{x}{e}\right) + c$
Why the other options are wrong: (A) x·ln(ex) = x(ln x + 1) has the wrong sign; (C) x·ln(e/x) reverses the ratio; (D) drops the leading x factor. Only (B) matches x ln x − x.

Watch out

The result is $x\ln x - x$ (minus x), so the ratio inside the log is x/e, NOT e/x or ex. A sign slip lands you on option (A) or (C).

Tip

Memorise ∫ln x dx = x(ln x − 1) + c; then just express it as x·ln(x/e) to match ICAI's answer form.

✓Final answer
$x \log_e \left(\dfrac{x}{e}\right) + c$
TG EAPCET 2023Set eng-2023-05-12-AN1 markMCQ
$\int x^4 (\log x)^3 \, dx = x^5 \left[ A (\log x)^3 + B (\log x)^2 + C \log x + D \right] + k$
›Reveal solutionSolution
The integral $\int x^4 (\log x)^3 \, dx$ is solved by repeated integration by parts, yielding coefficients $A = \frac{1}{5}$ , $B = -\frac{3}{25}$ , $C = \frac{6}{125}$ , $D = -\frac{6}{625}$ . Then $A+B+C+5D = \frac{16}{125}$ , so the correct option is (D).

Concept & Intuition

When integrating a product of a polynomial and a logarithm, integration by parts is the natural tool. Here $x^4$ is easy to integrate and $(\log x)^3$ becomes simpler when differentiated. Repeating the process three times reduces the power of $\log x$ to zero, leaving a pure polynomial integral. The final expression matches the given form, so we can read off $A, B, C, D$ by comparing coefficients.

Step-by-step solution

Set up integration by parts Let $u = (\log x)^3$ and $dv = x^4\,dx$ . Then

$du = 3(\log x)^2 \cdot \frac{1}{x}\,dx, \quad v = \frac{x^5}{5}.$
Integration by parts gives
$\int x^4 (\log x)^3\,dx = \frac{x^5}{5}(\log x)^3 - \int \frac{x^5}{5} \cdot 3(\log x)^2 \cdot \frac{1}{x}\,dx = \frac{x^5}{5}(\log x)^3 - \frac{3}{5} \int x^4 (\log x)^2\,dx.$

Second integration by parts Now handle $\int x^4 (\log x)^2\,dx$ . Let $u = (\log x)^2$ , $dv = x^4\,dx$ . Then

$du = 2(\log x)\cdot\frac{1}{x}\,dx, \quad v = \frac{x^5}{5}.$
So
$\int x^4 (\log x)^2\,dx = \frac{x^5}{5}(\log x)^2 - \frac{2}{5}\int x^4 \log x\,dx.$
Substituting back:
$\int x^4 (\log x)^3\,dx = \frac{x^5}{5}(\log x)^3 - \frac{3}{5}\left[ \frac{x^5}{5}(\log x)^2 - \frac{2}{5}\int x^4 \log x\,dx \right]$ $= \frac{x^5}{5}(\log x)^3 - \frac{3x^5}{25}(\log x)^2 + \frac{6}{25}\int x^4 \log x\,dx.$

Third integration by parts For $\int x^4 \log x\,dx$ , let $u = \log x$ , $dv = x^4\,dx$ . Then

$du = \frac{1}{x}\,dx, \quad v = \frac{x^5}{5}.$
Hence
$\int x^4 \log x\,dx = \frac{x^5}{5}\log x - \int \frac{x^5}{5}\cdot\frac{1}{x}\,dx = \frac{x^5}{5}\log x - \frac{1}{5}\int x^4\,dx = \frac{x^5}{5}\log x - \frac{x^5}{25}.$
Substitute this into the previous expression:
$\int x^4 (\log x)^3\,dx = \frac{x^5}{5}(\log x)^3 - \frac{3x^5}{25}(\log x)^2 + \frac{6}{25}\left( \frac{x^5}{5}\log x - \frac{x^5}{25} \right)$ $= \frac{x^5}{5}(\log x)^3 - \frac{3x^5}{25}(\log x)^2 + \frac{6x^5}{125}\log x - \frac{6x^5}{625}.$

Identify coefficients The given form is

$x^5\left[ A (\log x)^3 + B (\log x)^2 + C \log x + D \right] + k.$
Matching term by term:
$A = \frac{1}{5},\quad B = -\frac{3}{25},\quad C = \frac{6}{125},\quad D = -\frac{6}{625}.$

Compute $A + B + C + 5D$

$A + B + C = \frac{1}{5} - \frac{3}{25} + \frac{6}{125} = \frac{25}{125} - \frac{15}{125} + \frac{6}{125} = \frac{16}{125}.$ $5D = 5 \cdot \left(-\frac{6}{625}\right) = -\frac{30}{625} = -\frac{6}{125}.$
Therefore
$A + B + C + 5D = \frac{16}{125} - \frac{6}{125} = \frac{10}{125} = \frac{2}{25}.$
Watch out

A common mistake is to forget the factor of $5$ in $5D$ and simply add $D$ directly. Always check the expression carefully.

Tip

Notice that $A+B+C+5D$ simplifies nicely to $\frac{2}{25}$ , which is option (A). But wait — we must double-check: the problem asks for $A+B+C+5D$ , not $A+B+C+D$ . Our calculation gives $\frac{2}{25}$ , which is option (A). However, let's verify the arithmetic once more:

$A = \frac{1}{5} = \frac{25}{125}$ , $B = -\frac{3}{25} = -\frac{15}{125}$ , $C = \frac{6}{125}$ , so $A+B+C = \frac{25-15+6}{125} = \frac{16}{125}$ .

$5D = 5 \cdot \left(-\frac{6}{625}\right) = -\frac{30}{625} = -\frac{6}{125}$ .

Sum: $\frac{16}{125} - \frac{6}{125} = \frac{10}{125} = \frac{2}{25}$ . Yes, that is correct.

✓Final answer
$The correct option is (A) . ANSWER: A$
TG EAPCET 2023Set eng-2023-05-12-FN1 markMCQ
$\int_{1/2}^{2} \left| \log_{10} x \right| dx =$
›Reveal solutionSolution
The integral of the absolute value of a logarithm splits at the point where the argument equals 1 (here $x=1$ ). Evaluating the two pieces gives $\frac{3}{2}\log_{10}e$ , which simplifies to $\frac12\log_{10}\left(\frac{8}{e}\right)$ , matching option (B).

The key idea: $\left|\log_{10} x\right|$ changes sign at $x=1$ because $\log_{10}1=0$ . For $x<1$ , the logarithm is negative, so the absolute value flips its sign; for $x>1$ , it stays positive. So we split the integral at $x=1$ and integrate each piece separately, remembering that $\log_{10} x = \frac{\ln x}{\ln 10}$ .

Find where the integrand changes behavior

$\log_{10} x = 0 \implies x=1$ .

On $[\frac12,1]$ , $\log_{10} x \le 0$ , so $|\log_{10} x| = -\log_{10} x$ .

On $[1,2]$ , $\log_{10} x \ge 0$ , so $|\log_{10} x| = \log_{10} x$ .

Write the integral as a sum

$I = \int_{1/2}^{1} (-\log_{10} x)\,dx + \int_{1}^{2} \log_{10} x\,dx.$

Convert to natural logs for easy integration $\log_{10} x = \frac{\ln x}{\ln 10}$ , so

$I = \frac{1}{\ln 10}\left( -\int_{1/2}^{1} \ln x\,dx + \int_{1}^{2} \ln x\,dx \right).$

Integrate $\ln x$ Recall $\int \ln x\,dx = x\ln x - x + C$ .

For the first integral:

$\int_{1/2}^{1} \ln x\,dx = \left[ x\ln x - x \right]_{1/2}^{1} = (1\cdot0 - 1) - \left( \frac12 \ln\frac12 - \frac12 \right) = -1 - \left( \frac12 (-\ln 2) - \frac12 \right) = -1 + \frac12 \ln 2 + \frac12 = \frac12 \ln 2 - \frac12.$

For the second integral:

$\int_{1}^{2} \ln x\,dx = \left[ x\ln x - x \right]_{1}^{2} = (2\ln 2 - 2) - (0 - 1) = 2\ln 2 - 2 + 1 = 2\ln 2 - 1.$

Combine the pieces

$I = \frac{1}{\ln 10}\left( -\left( \frac12 \ln 2 - \frac12 \right) + (2\ln 2 - 1) \right) = \frac{1}{\ln 10}\left( -\frac12 \ln 2 + \frac12 + 2\ln 2 - 1 \right) = \frac{1}{\ln 10}\left( \frac32 \ln 2 - \frac12 \right).$

Simplify using logarithm properties Factor $\frac12$ :

$I = \frac{1}{\ln 10} \cdot \frac12 \left( 3\ln 2 - 1 \right) = \frac12 \cdot \frac{3\ln 2 - 1}{\ln 10}.$
Write $3\ln 2 = \ln(2^3) = \ln 8$ , and $1 = \ln e$ , so
$3\ln 2 - 1 = \ln 8 - \ln e = \ln\left(\frac{8}{e}\right).$
Thus
$I = \frac12 \cdot \frac{\ln(8/e)}{\ln 10} = \frac12 \log_{10}\left(\frac{8}{e}\right).$
Tip

Notice that $\frac{1}{\ln 10}$ converts a natural log to a base‑10 log: $\frac{\ln a}{\ln 10} = \log_{10} a$ . This shortcut avoids rewriting the final expression.

Watch out

A common mistake is forgetting the absolute value changes sign at $x=1$ , not at $x=0$ or $x=10$ . Also, $\log_{10} x$ is negative for $x<1$ , so the absolute value introduces a minus sign on that interval.

✓Final answer
$The correct option is (B) . ANSWER: B$
TG EAPCET 2023Set eng-2023-05-14-FN1 markMCQ
$\int (\log x)^3 \, x^5 \, dx = \frac{x^6}{A} \left[ B(\log x)^3 + C(\log x)^2 + D(\log x) - 1 \right] + k$
›Reveal solutionSolution
Integrating $\int x^5(\log x)^3\,dx$ by parts three times and matching the result to the given template gives $A=216,\ B=36,\ C=-18,\ D=6$ , so $A-(B+C+D) = 216-24 = 192$ — option (C).

Concept and Intuition

When you see a product of a polynomial and a power of $\log x$ , integration by parts is the natural tool: each application reduces the power of $\log x$ by one, until only a pure polynomial integral remains. The given form is that result, factored so its coefficients are integers — we just need to integrate carefully and match coefficients.

Step-by-step solution

Set up the integral

Let $I = \int x^5 (\log x)^3\,dx$ . Take $u=(\log x)^3,\ dv=x^5dx$ , so $du = 3(\log x)^2\cdot\frac1x\,dx$ , $v=\frac{x^6}{6}$ .

First integration by parts

$I = \frac{x^6}{6}(\log x)^3 - \frac12\int x^5(\log x)^2\,dx.$

Second integration by parts For $J=\int x^5(\log x)^2\,dx$ , take $u=(\log x)^2,\ dv=x^5dx$ :

$J = \frac{x^6}{6}(\log x)^2 - \frac13\int x^5\log x\,dx.$
Substituting back,
$I = \frac{x^6}{6}(\log x)^3 - \frac{x^6}{12}(\log x)^2 + \frac16\int x^5\log x\,dx.$

Third integration by parts For $K=\int x^5\log x\,dx$ , take $u=\log x,\ dv=x^5dx$ :

$K = \frac{x^6}{6}\log x - \frac16\int x^5\,dx = \frac{x^6}{6}\log x - \frac{x^6}{36}.$
Substituting into $I$ :
$I = \frac{x^6}{6}(\log x)^3 - \frac{x^6}{12}(\log x)^2 + \frac{x^6}{36}\log x - \frac{x^6}{216} + k.$

Match to the given form Factor out $\dfrac{x^6}{216}$ (since $216/6=36,\ 216/12=18,\ 216/36=6$ ):

$I = \frac{x^6}{216}\Big[36(\log x)^3 - 18(\log x)^2 + 6\log x - 1\Big] + k.$
Comparing to $\dfrac{x^6}{A}\big[B(\log x)^3 + C(\log x)^2 + D(\log x) - 1\big] + k$ :
$A=216,\quad B=36,\quad C=-18,\quad D=6.$

Compute the required expression

$A-(B+C+D) = 216-(36-18+6) = 216-24 = 192.$
Watch out

Each integration by parts introduces a factor of $1/x$ (from differentiating $\log x$ ) that cancels with the $x^5$ , keeping the polynomial degree at $x^6$ throughout. Track the minus sign from the by-parts formula carefully at every step.

Tip

Since differentiating $\log x$ steadily reduces its power (rather than the polynomial part), repeated integration by parts — always peeling off one power of $\log x$ at a time — is more reliable here than the tabular (DI) method.

✓Final answer
$The correct option is (C) . ANSWER: C$
TG EAPCET 2022Set eng-2022-07-19-FN1 markMCQ
$\dfrac{d}{dt}(t \log t - t) = \log t$
›Reveal solutionSolution
The key idea is to evaluate the definite integral using substitution and the given derivative, then exponentiate. The final value is $\frac{4}{e}$ .

The problem gives you a neat hint: $\frac{d}{dt}(t \log t - t) = \log t$ . This tells you that the antiderivative of $\log t$ is $t \log t - t$ . That’s the core tool. You’re asked to find $\exp\left( \int_0^1 2x \log(1+x^2) \, dx \right)$ , so the plan is to transform the integral into a form where you can use that antiderivative.

Notice the $2x$ inside the integral. That’s a strong clue: if you set $t = 1 + x^2$ , then $dt = 2x \, dx$ , which perfectly matches the $2x \, dx$ in the integrand. This substitution will turn the integral into something involving $\log t$ , and then you can apply the given derivative result directly.

Let’s work through it step by step.

Set up the substitution. Let $t = 1 + x^2$ . Then $dt = 2x \, dx$ . When $x = 0$ , $t = 1$ ; when $x = 1$ , $t = 2$ . The integral becomes:

$\int_0^1 2x \log(1 + x^2) \, dx = \int_{t=1}^{2} \log t \, dt$

Evaluate the integral using the given antiderivative. You know $\frac{d}{dt}(t \log t - t) = \log t$ , so:

$\int \log t \, dt = t \log t - t + C$
Therefore:
$\int_1^2 \log t \, dt = \left[ t \log t - t \right]_{1}^{2}$

Plug in the limits. At $t = 2$ : $2 \log 2 - 2$ At $t = 1$ : $1 \cdot \log 1 - 1 = 0 - 1 = -1$ So:

$\int_1^2 \log t \, dt = (2 \log 2 - 2) - (-1) = 2 \log 2 - 2 + 1 = 2 \log 2 - 1$

Simplify the result.

$2 \log 2 = \log(2^2) = \log 4$ , so the integral equals $\log 4 - 1$ .

Exponentiate.

You need $\exp\left( \int_0^1 2x \log(1+x^2) \, dx \right) = e^{\log 4 - 1} = e^{\log 4} \cdot e^{-1} = 4 \cdot \frac{1}{e} = \frac{4}{e}$ .

Watch out

A common mistake is to forget the minus sign when evaluating the lower limit. At $t=1$ , $t \log t - t = 0 - 1 = -1$ , so subtracting $(-1)$ gives $+1$ , not $-1$ . Double-check the arithmetic.

Tip

The substitution $t = 1 + x^2$ is natural because the derivative $2x \, dx$ appears. Always look for such pairings when you see a function and its derivative multiplied together.

✓Final answer
$\boxed{\frac{4}{e}}$

←Previous section · 7.6Integration by Parts →Next question · 7.6Example 18 — Find ∫ log x dx

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Concept understanding — Integration by Parts

Integration by Parts

The idea: reverse the product rule

Choosing $u$ : the ILATE rule

Worked idea

Why integration by parts?

Step-by-step solution

Method: Integration by Parts (Algebraic × Trigonometric)

Steps

Common Mistakes

Step 1 — Integration by parts

Step 2 — Complete the integral

Step 3 — Rewrite in the option's form

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Why integration by parts?

Step-by-step solution

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Q.Find $\int x \cos x\, dx$

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