(a) Find : $\int \frac{\cos x}{\sin 3x}\, dx$ OR (b) Find : $\int x^2 \log(x^2 + 1)\, dx$

(a) Find : ∫ (cos x)/(sin 3x) dx OR (b) Find : ∫ x² log(x² + 1) dx

Question

\int \frac{\cos x}{\sin 3x}\, dx

(OR)

(b) Find :

\int x^2 \log(x^2 + 1)\, dx

CBSECBSE Class XII Board 2023Subjective· 3mImportance★★★★★

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Part (a)Concept understanding — Trigonometric Simplification

Trigonometric Simplification

You know $\sin^2 x + \cos^2 x = 1$ — but the skill of turning a messy trig expression into that kind of clean form is trigonometric simplification. Because sines, cosines and their relatives are all tied together by identities from the unit circle, a tangled combination can almost always be rewritten as something shorter: a single term, a constant, or an easier combination.

Your core toolkit

\sin^2\theta+\cos^2\theta=1,\quad 1+\tan^2\theta=\sec^2\theta,\quad 1+\cot^2\theta=\csc^2\theta

How the process feels

Simplify $\dfrac{\sin x}{1+\cos x}+\dfrac{1+\cos x}{\sin x}$ . Over a common denominator the numerator is $\sin^2 x+(1+\cos x)^2 = \sin^2 x + 1 + 2\cos x + \cos^2 x$ . The Pythagorean identity turns $\sin^2 x+\cos^2 x$ into $1$ , giving $2+2\cos x = 2(1+\cos x)$ , so

\frac{2(1+\cos x)}{\sin x\,(1+\cos x)} = \frac{2}{\sin x} = 2\csc x.

A two-term sum collapses to one term.

Strategies that usually work

Convert everything to sines and cosines — cancellations then appear.
Spot Pythagorean pairs and replace them with $1$ (or $\sec^2$ , $\csc^2$ ).
Factor and cancel as you would with ordinary algebra.
Multiply by a conjugate — e.g. multiply $\frac{1}{1+\sin x}$ by $\frac{1-\sin x}{1-\sin x}$ to unlock a Pythagorean identity. …

Part (b)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 $du=dx$ , $v=e^x$ :

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

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$\int\frac{\cos x}{\sin3x}dx=\frac16\ln\left|\frac{\sin^3x}{\sin3x}\right|+C$ .
$\int x^2\log(x^2+1)\,dx=\frac{x^3}{3}\log(x^2+1)-\frac{2x^3}{9}+\frac{2x}{3}-\frac23\tan^{-1}x+C.$

Part (a): $\int\dfrac{\cos x}{\sin 3x}\,dx$

1. Triple-angle identity. $\sin3x=3\sin x-4\sin^3x=\sin x(3-4\sin^2x)$ , so

\int\frac{\cos x}{\sin x(3-4\sin^2x)}\,dx.

2. Substitute $u=\sin x$ , $du=\cos x\,dx$ :

\int\frac{du}{u(3-4u^2)}.

3. Partial fractions. With $3-4u^2=(\sqrt3-2u)(\sqrt3+2u)$ ,

\frac{1}{u(3-4u^2)}=\frac{1}{3u}+\frac{1}{3(\sqrt3-2u)}-\frac{1}{3(\sqrt3+2u)}.

4. Integrate.

\frac13\ln|u|-\frac16\ln|\sqrt3-2u|-\frac16\ln|\sqrt3+2u|+C=\frac16\ln\left|\frac{u^2}{3-4u^2}\right|+C.

5. Back-substitute $u=\sin x$ and use $3-4\sin^2x=\dfrac{\sin3x}{\sin x}$ : …

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

Showing the 12 most recent of 103 on this concept.

CBSE 2026Set 65/1/11 markMCQ
$x$
›Reveal solutionSolution
The determinant simplifies to $\sin x \cos x + \sin x \cos x = \sin 2x$ . Setting $\sin 2x = 1$ gives $2x = \frac{\pi}{2} + 2n\pi$ , so $x = \frac{\pi}{4}$ is one solution. The correct option is (B).

The problem gives a $2 \times 2$ determinant equal to $1$ and asks for a value of $x$ from the options. The fastest route is to compute the determinant directly — it’s a simple expression in $\sin x$ and $\cos x$ — and then solve the resulting trigonometric equation.

The determinant of $\begin{vmatrix} a & b \\ c & d \end{vmatrix}$ is $ad - bc$ . Here:
$\begin{vmatrix} \cos x & \sin x \\ -\cos x & \sin x \end{vmatrix} = (\cos x)(\sin x) - (\sin x)(-\cos x)$

Simplify the expression. The first term is $\cos x \sin x$ . The second term: $(\sin x)(-\cos x) = -\sin x \cos x$ , but there’s a minus sign in front, so it becomes $-(-\sin x \cos x) = +\sin x \cos x$ . So the determinant equals:

$\cos x \sin x + \sin x \cos x = 2 \sin x \cos x$

Use the double-angle identity. Recall that $2 \sin x \cos x = \sin 2x$ . Therefore the equation becomes:

$\sin 2x = 1$

Solve $\sin 2x = 1$ . The sine function equals $1$ at $\frac{\pi}{2}$ plus any integer multiple of $2\pi$ :

$2x = \frac{\pi}{2} + 2n\pi \quad \Rightarrow \quad x = \frac{\pi}{4} + n\pi$
where $n$ is any integer.

Check the given options.

(A) $0$ : $\sin 0 = 0$ , not $1$ .

(B) $\frac{\pi}{4}$ : $\sin \frac{\pi}{2} = 1$ — works. …
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CBSE 2026Set CX1 markMCQ
$\sin(\tan^{-1}x)$
(a) $\dfrac{x}{\sqrt{1+x^{2}}}$
(b) $\dfrac{x}{\sqrt{1-x^{2}}}$
(c) $\dfrac{1}{\sqrt{1+x^{2}}}$
(d) $\dfrac{1}{\sqrt{1-x^{2}}}$
›Reveal solutionSolution

With $\theta=\tan^{-1}x$ a right triangle gives $\sin\theta=\dfrac{x}{\sqrt{1+x^{2}}}$ — option (a).

Concept: Convert the inverse function to an angle and read the ratio off a right triangle.

Let $\theta=\tan^{-1}x$ , so $\tan\theta=x$ . Take the opposite side $=x$ and adjacent $=1$ ; then the hypotenuse is $\sqrt{1+x^{2}}$ .

…

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CBSE 2026Set A1 markMCQ
$\sin(\cos^{-1}3/5) =$
(a) $\frac{3}{4}$
(b) $\frac{4}{5}$
(c) $\frac{3}{5}$
(d) $\frac{5}{4}$
›Reveal solutionSolution

$\sin\left(\cos^{-1}\frac{3}{5}\right) = \frac{4}{5}$ .

Let $\theta = \cos^{-1}\frac{3}{5}$ , so $\cos\theta = \frac{3}{5}$ with $\theta\in[0,\pi]$ , where $\sin\theta \ge 0$ .

Then …

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CBSE 2026Set A1 markMCQ
$|x| \le 1$
(a) $\frac{\sqrt{1-x^2}}{x}$
(b) $\frac{x}{1+x^2}$
(c) $\frac{\sqrt{1+x^2}}{x}$
(d) $\sqrt{1-x^2}$
›Reveal solutionSolution

$\tan(\cos^{-1}x) = \frac{\sqrt{1-x^2}}{x}$ .

Let $\theta = \cos^{-1}x$ , so $\cos\theta = x$ with $\theta\in[0,\pi]$ (where $\sin\theta\ge 0$ ).

Then $\sin\theta = \sqrt{1-x^2}$ , and …

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CBSE 2026Set A1 markMCQ
$\int\frac{\cos 2x}{(\sin x + \cos x)^2}\,dx =$
(a) $2\log(\sin x + \cos x) + k$
(b) $\log(\sin x + \cos x) + k$
(c) $\log(\sin x - \cos x) + k$
(d) $-\frac{1}{\sin x + \cos x} + k$
›Reveal solutionSolution

Factor $\cos 2x$ and substitute $u=\sin x+\cos x$ to get $\log(\sin x+\cos x)+k$ .

Write $\cos 2x=\cos^2 x-\sin^2 x=(\cos x-\sin x)(\cos x+\sin x)$ . Then

$\dfrac{\cos 2x}{(\sin x+\cos x)^2}=\dfrac{(\cos x-\sin x)(\cos x+\sin x)}{(\sin x+\cos x)^2}=\dfrac{\cos x-\sin x}{\sin x+\cos x}$ .

…

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CBSE 2026Set A1 markMCQ
$\int\frac{1 - \cos 2x}{1 + \cos 2x}\,dx =$
(a) $\tan x + x + k$
(b) $\tan x - x + k$
(c) $x - \tan^2 x + k$
(d) $\tan\frac{x}{2} + k$
›Reveal solutionSolution

Simplify to $\tan^2 x$ , then integrate: $\tan x - x + k$ .

Use $1-\cos 2x=2\sin^2 x$ and $1+\cos 2x=2\cos^2 x$ :

$\dfrac{1-\cos 2x}{1+\cos 2x}=\dfrac{2\sin^2 x}{2\cos^2 x}=\tan^2 x=\sec^2 x-1$ .

…

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CBSE 2026Set A1 markMCQ
$\int\log x\,dx =$
(a) $\frac{1}{x} + k$
(b) $x\log x + k$
(c) $x\log x - x + k$
(d) $x\log x + x + k$
›Reveal solutionSolution

Integrate by parts: $\int\log x\,dx=x\log x-x+k$ .

Take $u=\log x$ and $dv=dx$ , so $du=\tfrac{1}{x}\,dx$ and $v=x$ :

…

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CBSE 2026Set A1 markMCQ
$\int\cos\sqrt{x}\,dx =$
(a) $\sin\sqrt{x} + \cos\sqrt{x} + k$
(b) $\frac{1}{2}(\sqrt{x}\sin\sqrt{x} - \cos\sqrt{x}) + k$
(c) $2(\sqrt{x}\sin\sqrt{x} + \cos\sqrt{x}) + k$
(d) $\sin\sqrt{x} + k$
›Reveal solutionSolution

Put $t=\sqrt{x}$ ; the integral becomes $2\int t\cos t\,dt=2(t\sin t+\cos t)+k$ .

Let $t=\sqrt{x}$ , so $x=t^2$ and $dx=2t\,dt$ . Then

$\int\cos\sqrt{x}\,dx=\int\cos t\,(2t\,dt)=2\int t\cos t\,dt$ .

Integrate $\int t\cos t\,dt$ by parts ( $u=t,\ dv=\cos t\,dt$ ): $=t\sin t-\int\sin t\,dt=t\sin t+\cos t$ .

…

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CBSE 2026Set A1 markMCQ
$\int e^x\left(\tan^{-1}x + \frac{1}{1+x^2}\right)dx =$
(a) $e^x\tan^{-1}x + k$
(b) $e^x \cdot \frac{1}{1+x^2} + k$
(c) $e^x + k$
(d) $\tan^{-1}x + k$
›Reveal solutionSolution

Recognise $\int e^x[f(x)+f'(x)]\,dx=e^x f(x)+k$ ; here $f(x)=\tan^{-1}x$ .

Note $\dfrac{d}{dx}\tan^{-1}x=\dfrac{1}{1+x^2}$ . So the integrand is $e^x\big[\tan^{-1}x+(\tan^{-1}x)'\big]$ , which matches the standard pattern

…

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CBSE 2026Set A1 markMCQ
$\int_0^1 x e^x\,dx =$
(a) $1$
(b) $0$
(c) $2$
(d) $-1$
›Reveal solutionSolution

By parts: $\int_0^1 xe^x\,dx=[(x-1)e^x]_0^1=1$ .

Take $u=x,\ dv=e^x\,dx$ , so $du=dx,\ v=e^x$ :

$\int x e^x\,dx=xe^x-\int e^x\,dx=xe^x-e^x=(x-1)e^x$ .

…

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CBSE 2026Set ANNUAL1 markMCQ
$\int e^x(\sin x-\cos x)\,dx$
(a) $-e^x\cos x+c$
(b) $e^x\sin x+c$
(c) $-e^x\sec x+c$
(d) $e^x\operatorname{cosec} x+c$
›Reveal solutionSolution

Recognising the standard form $\int e^x[f(x)+f'(x)]\,dx=e^xf(x)+c$ gives $-e^x\cos x+c$ .

There is a standard integration result:

$\int e^x\big[f(x)+f'(x)\big]\,dx=e^x f(x)+c$

Compare the integrand $\sin x-\cos x$ with $f(x)+f'(x)$ . Try $f(x)=-\cos x$ ; then $f'(x)=\sin x$ , so

$f(x)+f'(x)=-\cos x+\sin x=\sin x-\cos x$ …

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CBSE 2026Set ANNUAL1 markMCQ
$\int \frac{dx}{\sin^2 x \cos^2 x}$
(a) $\tan x + \sin x + c$
(b) $\tan x - \cot x + c$
(c) $\tan x \cot x + c$
(d) $2\tan x - \cot 2x + c$
›Reveal solutionSolution

Split the integrand using $\sin^2x+\cos^2x=1$ in the numerator, then integrate each standard term.

$\dfrac{1}{\sin^2x\cos^2x} = \dfrac{\sin^2x+\cos^2x}{\sin^2x\cos^2x} = \dfrac{1}{\cos^2x}+\dfrac{1}{\sin^2x} = \sec^2x+\csc^2x$ .

…

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Q.(a) Find : ∫cos⁡xsin⁡3x dx\int \frac{\cos x}{\sin 3x}\, dx∫sin3xcosx​dx

Part (a): ∫cos⁡xsin⁡3x dx\int\dfrac{\cos x}{\sin 3x}\,dx∫sin3xcosx​dx

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Q.(a) Find : $\int \frac{\cos x}{\sin 3x}\, dx$

Part (a): $\int\dfrac{\cos x}{\sin 3x}\,dx$

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