Q.Integrate the following function: $e^x (\sin x + \cos x)$

Showing the 12 most recent of 22 on this concept.

1. TG EAPCET 2026Set eng-2026-05-09-FN1 markMCQ
   Q.If $\int \frac{1}{1+\cos x} dx = \frac{1}{f\left(\frac{x}{2}\right)} + c_1$, then $\int f(x) dx =$ (A) $\log|\sin x| + c$ (B) $\log|\cos x| + c$ (C) $-\csc^2 x + c$ (D) $\tan x + c$

   ›Reveal solutionSolution

   Evaluating the first integral gives $\tan\frac{x}{2}$, so $f(x)=\cot x$; hence $\int f(x)\,dx=\int\cot x\,dx=\log|\sin x|+c$.

   Evaluate the given integral using $1+\cos x = 2\cos^{2}\frac{x}{2}$:

   $\int \frac{1}{1+\cos x}\,dx = \int \frac{1}{2\cos^{2}\frac{x}{2}}\,dx = \frac{1}{2}\int \sec^{2}\frac{x}{2}\,dx = \tan\frac{x}{2} + c_{1}$

   Identify $f$. We are told this equals $\dfrac{1}{f\left(\frac{x}{2}\right)} + c_{1}$, so …

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2. TG EAPCET 2026Set eng-2026-05-10-AN1 markMCQ
   Q.

   [!FORMULA] $\int \frac{1}{2\cot x - 3\tan x} dx =$

   (A) $-\frac{1}{10} \log|2 - 5\sin^2 x| + c$ (B) $\frac{1}{10} \log|3 + 2\cos^2 x| + c$ (C) $-\frac{1}{10} \log|2\cot x + 3\tan x| + c$ (D) $\frac{1}{10} \log|2\cot x - 3\tan x| + c$

   ›Reveal solutionSolution

   The key is to rewrite the integrand in terms of sine and cosine, simplify using the identity $\sin^2 x + \cos^2 x = 1$, and then use a substitution that turns the integral into a standard logarithmic form. The result matches option (A).

   ---

   We want to evaluate

   $\int \frac{1}{2\cot x - 3\tan x} \, dx.$

   Concept & Intuition:

   The presence of $\cot x$ and $\tan x$ suggests rewriting everything in terms of $\sin x$ and $\cos x$. That often reveals a simpler rational expression. Then, noticing that the denominator becomes a product of $\sin x$ and $\cos x$ times something linear in $\sin^2 x$ (or $\cos^2 x$) points toward a substitution like $u = \sin^2 x$ or $u = \cos^2 x$, whose derivative involves $2\sin x \cos x$ — exactly the factor we’ll get.

   ---

   Step-by-step solution:

   1. Rewrite in terms of sine and cosine

   $\cot x = \frac{\cos x}{\sin x}, \quad \tan x = \frac{\sin x}{\cos x}.$

   So

   $2\cot x - 3\tan x = \frac{2\cos x}{\sin x} - \frac{3\sin x}{\cos x}.$

   Combine into a single fraction:

   $= \frac{2\cos^2 x - 3\sin^2 x}{\sin x \cos x}.$

   2. Invert the fraction The integrand becomes

   $\frac{1}{2\cot x - 3\tan x} = \frac{\sin x \cos x}{2\cos^2 x - 3\sin^2 x}.$

   3. Use the Pythagorean identity Since $\cos^2 x = 1 - \sin^2 x$, the denominator is

   $2(1 - \sin^2 x) - 3\sin^2 x = 2 - 2\sin^2 x - 3\sin^2 x = 2 - 5\sin^2 x.$

   So the integral is

   $\int \frac{\sin x \cos x}{2 - 5\sin^2 x} \, dx.$

   4. Substitution Let $u = \sin^2 x$. Then $du = 2\sin x \cos x \, dx$, so $\sin x \cos x \, dx = \frac{1}{2} du$. The integral becomes $$$$

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3. TG EAPCET 2025Set eng-2025-05-02-FN1 markMCQ
   Q.$\int \frac{2\sin x - 3\cos x}{4\cos x - 3\sin x} dx =$ (A) $\frac{1}{25} \left[ 17 \log |4\cos x - 3\sin x| - 6x \right] + c$ (B) $\frac{1}{25} \left[ x - 18 \log |4\cos x - 3\sin x| \right] + c$ (C) $\frac{1}{25} \left[ \log |4\cos x - 3\sin x| - 18x \right] + c$ (D) $\frac{1}{25} \left[ 17x - 6 \log |4\cos x - 3\sin x| \right] + c$

   ›Reveal solutionSolution

   Write the numerator as $A\,D + B\,D'$ where $D=4\cos x-3\sin x$; then $A=-\tfrac{18}{25},\ B=\tfrac{1}{25}$, giving $\dfrac{1}{25}\big[\log|4\cos x-3\sin x|-18x\big]+c$ — option (C).

   Let $D=4\cos x-3\sin x$, so $D'=-4\sin x-3\cos x$.

   Set up $2\sin x-3\cos x = A\,D + B\,D'$. Expanding,

   $A\,D+B\,D'=(4A-3B)\cos x+(-3A-4B)\sin x.$

   Matching coefficients:

   $4A-3B=-3,\qquad -3A-4B=2.$

   Solving: $25A=-18\Rightarrow A=-\dfrac{18}{25}$, and then $B=\dfrac{1}{25}$.

   Integrate.

   $\frac{2\sin x-3\cos x}{D}=A+B\,\frac{D'}{D}=-\frac{18}{25}+\frac{1}{25}\cdot\frac{D'}{D}.$ …

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4. TG EAPCET 2025Set eng-2025-05-02-AN1 markMCQ
   Q.$\int \left( \frac{1}{x^2} + \frac{\sin^3 x + \cos^3 x}{\sin^2 x \cos^2 x} \right) dx =$ (A) $\frac{(\sin x - \cos x)x - \sin x \cos x}{x \sin x \cos x} + c$ (B) $-\frac{1}{x} + \frac{\sin x + \cos x}{\cos x - \sin x} + c$ (C) $-\frac{1}{x} + \frac{\sin x - \cos x}{\sin^2 x \cos^2 x} + c$ (D) $\frac{(\sin x - \cos x)x - \sin x - \cos x}{x(\sin x + \cos x)} + c$

   ›Reveal solutionSolution

   The integral splits into two parts: a simple power rule for $1/x^2$ and a trigonometric simplification for the second term. After rewriting $\sin^3 x + \cos^3 x$ using the sum of cubes and simplifying, the result matches option (A).

   We start by noticing that the integrand is a sum of two distinct pieces. The first, $1/x^2$, is elementary. The second, $\frac{\sin^3 x + \cos^3 x}{\sin^2 x \cos^2 x}$, looks messy but can be simplified using algebraic identities. The key idea: factor the numerator as a sum of cubes, then split into simpler fractions that integrate to known forms like $\sec x \csc x$ or combinations of $\tan x$ and $\cot x$.

   1. Separate the integral

   $I = \int \frac{1}{x^2}\,dx + \int \frac{\sin^3 x + \cos^3 x}{\sin^2 x \cos^2 x}\,dx$

   The first integral is immediate:

   $\int \frac{1}{x^2}\,dx = -\frac{1}{x} + C_1$

   2. Simplify the trigonometric fraction Recall the sum of cubes:

   $\sin^3 x + \cos^3 x = (\sin x + \cos x)(\sin^2 x - \sin x \cos x + \cos^2 x)$

   Since $\sin^2 x + \cos^2 x = 1$, this becomes

   $\sin^3 x + \cos^3 x = (\sin x + \cos x)(1 - \sin x \cos x)$

   3. Rewrite the second integrand

   $\frac{\sin^3 x + \cos^3 x}{\sin^2 x \cos^2 x} = \frac{(\sin x + \cos x)(1 - \sin x \cos x)}{\sin^2 x \cos^2 x}$

   Split into two fractions:

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

   Simplify the second term:

   $\frac{(\sin x + \cos x)\sin x \cos x}{\sin^2 x \cos^2 x} = \frac{\sin x + \cos x}{\sin x \cos x}$

   So we have

   $\frac{\sin^3 x + \cos^3 x}{\sin^2 x \cos^2 x} = \frac{\sin x + \cos x}{\sin^2 x \cos^2 x} - \frac{\sin x + \cos x}{\sin x \cos x}$

   4. Rewrite in terms of $\sec x$ and $\csc x$ Note that

   $\frac{1}{\sin^2 x \cos^2 x} = \sec^2 x \csc^2 x$

   and

   $\frac{1}{\sin x \cos x} = \sec x \csc x$

   Hence

   $\frac{\sin^3 x + \cos^3 x}{\sin^2 x \cos^2 x} = (\sin x + \cos x)\sec^2 x \csc^2 x - (\sin x + \cos x)\sec x \csc x$

   5. Integrate term by term Consider the first part:

   $\int (\sin x + \cos x)\sec^2 x \csc^2 x\,dx$

   Write $\sec^2 x \csc^2 x = \frac{1}{\sin^2 x \cos^2 x}$. A clever trick:

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

   So

   $(\sin x + \cos x)\sec^2 x \csc^2 x = (\sin x + \cos x)(\sec^2 x + \csc^2 x)$

   Expand:

   $= \sin x \sec^2 x + \sin x \csc^2 x + \cos x \sec^2 x + \cos x \csc^2 x$

   Simplify each:

   • $\sin x \sec^2 x = \sin x \cdot \frac{1}{\cos^2 x} = \tan x \sec x$
   • $\sin x \csc^2 x = \sin x \cdot \frac{1}{\sin^2 x} = \csc x$
   • $\cos x \sec^2 x = \cos x \cdot \frac{1}{\cos^2 x} = \sec x$
   • $\cos x \csc^2 x = \cos x \cdot \frac{1}{\sin^2 x} = \cot x \csc x$

   So the integral becomes

   $\int (\tan x \sec x + \csc x + \sec x + \cot x \csc x)\,dx$

   These are standard:

   $\int \tan x \sec x\,dx = \sec x,\quad \int \csc x\,dx = \log|\csc x - \cot x|,\quad \int \sec x\,dx = \log|\sec x + \tan x|,\quad \int \cot x \csc x\,dx = -\csc x$

   So the first part integrates to

   $\sec x - \csc x + \log|\sec x + \tan x| + \log|\csc x - \cot x| + C_2$

   6. Now the second part

   $\int (\sin x + \cos x)\sec x \csc x\,dx = \int (\sin x + \cos x) \cdot \frac{1}{\sin x \cos x}\,dx$

   Split:

   $$$$

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5. TG EAPCET 2025Set eng-2025-05-04-FN1 markMCQ
   Q.$\int \frac{3x+2}{4x^2+4x+5} dx = A \log \left(4x^2+4x+5\right) + B \tan^{-1} \left(\frac{2x+1}{2}\right) + c$, then $A+B=$ (A) $\frac{1}{2}$ (B) $1$ (C) $\frac{3}{4}$ (D) $\frac{3}{8}$

   ›Reveal solutionSolution

   We split the integrand into a part whose numerator is the derivative of the denominator (giving a log) and a constant part (giving an arctan). Matching coefficients yields $A = \frac{3}{8}$ and $B = \frac{1}{2}$, so $A+B = \frac{7}{8}$. None of the given options match — the problem likely expects $\frac{7}{8}$.

   The key idea is that when integrating a rational function where the denominator is a quadratic that does not factor over the reals, we aim for two standard forms:

   $\int \frac{f'(x)}{f(x)}\,dx = \log|f(x)| + c$

   and

   $\int \frac{dx}{x^2 + a^2} = \frac{1}{a}\tan^{-1}\!\left(\frac{x}{a}\right) + c.$

   Here the denominator is $4x^2+4x+5$. Its derivative is $8x+4$. Our numerator is $3x+2$, which is not a multiple of $8x+4$ — so we write $3x+2$ as a linear combination of the derivative and a constant.

   1. Express the numerator in terms of the derivative of the denominator. Let $D = 4x^2+4x+5$. Then $D' = 8x+4$. We want constants $p$ and $q$ such that

   $3x+2 = p(8x+4) + q.$

   Comparing coefficients of $x$: $3 = 8p \Rightarrow p = \frac{3}{8}$.

   Comparing constant terms: $2 = 4p + q \Rightarrow 2 = 4\cdot\frac{3}{8} + q = \frac{3}{2} + q \Rightarrow q = \frac{1}{2}$.

   So

   $3x+2 = \frac{3}{8}(8x+4) + \frac{1}{2}.$

   2. Split the integral.

   $\int \frac{3x+2}{4x^2+4x+5}\,dx = \frac{3}{8}\int \frac{8x+4}{4x^2+4x+5}\,dx + \frac{1}{2}\int \frac{dx}{4x^2+4x+5}.$

   The first integral is immediate:

   $\int \frac{8x+4}{4x^2+4x+5}\,dx = \log(4x^2+4x+5) + c_1.$

   3. Handle the second integral by completing the square.

   $4x^2+4x+5 = 4\left(x^2 + x + \frac{5}{4}\right) = 4\left[\left(x+\frac{1}{2}\right)^2 + 1\right].$

   Check: $\left(x+\frac12\right)^2 = x^2 + x + \frac14$, so adding $1$ gives $x^2+x+\frac54$, correct.

   Hence

   $\int \frac{dx}{4x^2+4x+5} = \int \frac{dx}{4\left[(x+\frac12)^2 + 1\right]} = \frac14 \int \frac{dx}{(x+\frac12)^2 + 1^2}.$

   Using $\int \frac{du}{u^2+a^2} = \frac1a \tan^{-1}\!\left(\frac{u}{a}\right)$, with $u = x+\frac12$, $a=1$, we get

   $\frac14 \cdot \frac11 \tan^{-1}\!\left(\frac{x+\frac12}{1}\right) = \frac14 \tan^{-1}\!\left(x+\frac12\right).$

   But the given form has $\tan^{-1}\!\left(\frac{2x+1}{2}\right)$. Notice:

   $$$$

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6. TG EAPCET 2024Set eng-2024-05-09-AN1 markMCQ
   Q.$\int \frac{2\cos 3x - 3\sin 3x}{\cos 3x + 2\sin 3x} dx =$ (A) $\frac{7}{15} \log|\cos 3x + 2\sin 3x| - \frac{4}{5}x + c$ (B) $-\frac{4}{5} \log|\cos 3x + 2\sin 3x| + \frac{7x}{5} + c$ (C) $\frac{7}{5} \log|\cos 3x + 2\sin 3x| - \frac{4}{5}x + c$ (D) $-\frac{8}{15} \log|\cos 3x + 2\sin 3x| - \frac{x}{5} + c$

   ›Reveal solutionSolution

   The integral of a linear combination of sine and cosine over another linear combination is solved by expressing the numerator as a linear combination of the denominator and its derivative. The result is $\frac{7}{15}\log|\cos 3x+2\sin 3x| - \frac{4}{5}x + c$, which matches option (A).

   Concept & Intuition

   When the integrand is a rational combination of $\sin$ and $\cos$ where the denominator is a linear combination of them, a standard trick is to write the numerator as $A$ times (denominator) plus $B$ times (derivative of denominator). Why? Because then the integral splits into a simple logarithmic part (from $\frac{A \cdot \text{denom}}{\text{denom}}$) and a constant part (from $\frac{B \cdot \text{derivative}}{\text{denom}}$, which integrates to $B \log|\text{denom}|$). Here the denominator is $D = \cos 3x + 2\sin 3x$, and its derivative is $D' = -3\sin 3x + 6\cos 3x$. We find constants $A$ and $B$ such that:

   $2\cos 3x - 3\sin 3x = A(\cos 3x + 2\sin 3x) + B(-3\sin 3x + 6\cos 3x).$

   Step-by-step solution

   1. Set up the linear combination We want:

   $2\cos 3x - 3\sin 3x = A(\cos 3x + 2\sin 3x) + B(-3\sin 3x + 6\cos 3x).$

   Expand the right-hand side:

   $= A\cos 3x + 2A\sin 3x - 3B\sin 3x + 6B\cos 3x.$

   Group coefficients of $\cos 3x$ and $\sin 3x$:

   $\text{Coefficient of } \cos 3x: \quad A + 6B = 2.$

   $\text{Coefficient of } \sin 3x: \quad 2A - 3B = -3.$

   2. Solve the system From $A + 6B = 2$, we have $A = 2 - 6B$. Substitute into $2A - 3B = -3$:

   $2(2 - 6B) - 3B = -3 \implies 4 - 12B - 3B = -3 \implies 4 - 15B = -3.$

   So $-15B = -7 \implies B = \frac{7}{15}$. Then $A = 2 - 6\cdot\frac{7}{15} = 2 - \frac{42}{15} = \frac{30}{15} - \frac{42}{15} = -\frac{12}{15} = -\frac{4}{5}$.

   3. Rewrite the integral The numerator becomes $-\frac{4}{5}D + \frac{7}{15}D'$. Hence: $$$$

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7. TG EAPCET 2024Set eng-2024-05-09-FN1 markMCQ
   Q.If the slope of the tangent drawn at any point $(x, y)$ on the curve $y = f(x)$ is $(6x^2 + 10x - 9)$ and $f(2) = 0$, then $f(-2) =$ (A) $0$ (B) $4$ (C) $-6$ (D) $-13$

   ›Reveal solutionSolution

   The slope equals $f'(x)$; integrate it, fix the constant from $f(2)=0$, then evaluate at $x=-2$ to get $f(-2)=4$ — option (B).

   Concept

   The slope of the tangent at any point is the derivative, so

   $f'(x) = 6x^2 + 10x - 9.$

   Integrating recovers $f(x)$ up to a constant, which the condition $f(2)=0$ pins down.

   Integrate

   $f(x) = \int (6x^2 + 10x - 9)\,dx = 2x^3 + 5x^2 - 9x + C.$

   Apply $f(2)=0$ …

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8. TG EAPCET 2024Set eng-2024-05-10-FN1 markMCQ
   Q.$\int \sqrt{4\cos^2 x - 5\sin^2 x} \cos x \, dx =$ (A) $\dfrac{1}{2} \cos x \sqrt{4 - 9\sin^2 x} + \dfrac{2}{3} \sin^{-1} \left( \dfrac{3\sin x}{2} \right) + c$ (B) $\dfrac{1}{2} \sin x \sqrt{4 - 9\sin^2 x} + \dfrac{2}{3} \cos^{-1} \left( \dfrac{3\cos x}{2} \right) + c$ (C) $\dfrac{1}{2} \cos x \sqrt{1 - 9\cos^2 x} + \dfrac{2}{3} \sin^{-1} \left( \dfrac{3\cos x}{2} \right) + c$ (D) $\dfrac{1}{2} \sin x \sqrt{4 - 9\sin^2 x} + \dfrac{2}{3} \sin^{-1} \left( \dfrac{3\sin x}{2} \right) + c$

   ›Reveal solutionSolution

   The integrand simplifies by factoring out $\cos x$ and substituting $t = \sin x$, leading to a standard form $\int \sqrt{a^2 - u^2}\,du$; the result matches option (D).

   We start with

   $I = \int \sqrt{4\cos^2 x - 5\sin^2 x}\,\cos x\,dx.$

   Concept & Intuition

   The presence of $\cos x\,dx$ strongly suggests the substitution $u = \sin x$, because $du = \cos x\,dx$. Under this substitution, $\cos^2 x = 1 - \sin^2 x = 1 - u^2$, so the expression inside the square root becomes purely in terms of $u$. That turns the integral into a familiar form: $\int \sqrt{a^2 - b^2 u^2}\,du$, which is a standard trigonometric (or inverse sine) integral.

   ---

   Step-by-step solution

   1. Substitute $u = \sin x$ Then $du = \cos x\,dx$, and $\cos^2 x = 1 - u^2$. The integrand becomes:

   $\sqrt{4(1 - u^2) - 5u^2} = \sqrt{4 - 4u^2 - 5u^2} = \sqrt{4 - 9u^2}.$

   So

   $I = \int \sqrt{4 - 9u^2}\,du.$

   2. Factor to match the standard form Write $\sqrt{4 - 9u^2} = \sqrt{4\left(1 - \frac{9}{4}u^2\right)} = 2\sqrt{1 - \left(\frac{3u}{2}\right)^2}$. Hence

   $I = 2\int \sqrt{1 - \left(\frac{3u}{2}\right)^2}\,du.$

   3. Use the standard integral Recall:

   $\int \sqrt{a^2 - t^2}\,dt = \frac{t}{2}\sqrt{a^2 - t^2} + \frac{a^2}{2}\sin^{-1}\!\left(\frac{t}{a}\right) + C.$

   Here $a = 1$ and $t = \frac{3u}{2}$. But we have $du$, not $dt$. Since $t = \frac{3u}{2}$, we have $dt = \frac{3}{2}du$, so $du = \frac{2}{3}dt$.

   Substituting:

   $I = 2 \int \sqrt{1 - t^2} \cdot \frac{2}{3}\,dt = \frac{4}{3} \int \sqrt{1 - t^2}\,dt.$

   4. Apply the formula

   $\int \sqrt{1 - t^2}\,dt = \frac{t}{2}\sqrt{1 - t^2} + \frac{1}{2}\sin^{-1} t + C.$

   Multiply by $\frac{4}{3}$:

   $I = \frac{4}{3}\left( \frac{t}{2}\sqrt{1 - t^2} + \frac{1}{2}\sin^{-1} t \right) + C = \frac{2t}{3}\sqrt{1 - t^2} + \frac{2}{3}\sin^{-1} t + C.$

   5. Back-substitute Recall $t = \frac{3u}{2} = \frac{3\sin x}{2}$. Then $\sqrt{1 - t^2} = \sqrt{1 - \frac{9\sin^2 x}{4}} = \frac{1}{2}\sqrt{4 - 9\sin^2 x}.$ …

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9. TG EAPCET 2024Set eng-2024-05-11-FN1 markMCQ
   Q.$\int \frac{dx}{4+3\cot x} dx =$ (A) $-\frac{3}{25} \log|4+3\cot x| + \frac{4}{25}x + c$ (B) $-\frac{3}{25} \log|4\sin x + 3\cos x| + \frac{4}{25}x + c$ (C) $\frac{4}{25} \log|4\sin x + 3\cos x| - \frac{3}{25}x + c$ (D) $\frac{4}{25} \log|4+3\cot x| - \frac{3}{25}x + c$

   ›Reveal solutionSolution

   The key is to rewrite the integrand in terms of sine and cosine, then use a clever combination of derivatives to split the integral into a logarithmic part and a simple constant part. The correct result is $-\frac{3}{25}\log|4\sin x+3\cos x|+\frac{4}{25}x + c$, which corresponds to option (B).

   ---

   Concept and intuition

   When we see $\cot x$ in an integral, it's often helpful to rewrite everything in terms of $\sin x$ and $\cos x$, because then we can spot a pattern: the denominator becomes a linear combination of $\sin x$ and $\cos x$, and the numerator (after rewriting $dx$) can be expressed as a combination of the derivative of that denominator and a constant. This lets us separate the integral into a part that gives a logarithm and a part that gives a simple $x$ term.

   ---

   Step-by-step solution

   1. Rewrite the integrand Since $\cot x = \frac{\cos x}{\sin x}$, we have:

   $\frac{1}{4 + 3\cot x} = \frac{1}{4 + 3\frac{\cos x}{\sin x}} = \frac{1}{\frac{4\sin x + 3\cos x}{\sin x}} = \frac{\sin x}{4\sin x + 3\cos x}.$

   So the integral becomes:

   $I = \int \frac{\sin x}{4\sin x + 3\cos x}\,dx.$

   2. Set up a useful trick We want to express the numerator $\sin x$ as a linear combination of the denominator $D = 4\sin x + 3\cos x$ and its derivative $D' = 4\cos x - 3\sin x$. Suppose:

   $\sin x = A(4\sin x + 3\cos x) + B(4\cos x - 3\sin x).$

   Expand and collect coefficients of $\sin x$ and $\cos x$:

   $\sin x = (4A - 3B)\sin x + (3A + 4B)\cos x.$

   Comparing coefficients gives the system:

   $\begin{cases} 4A - 3B = 1 \\ 3A + 4B = 0 \end{cases}$

   3. Solve for $A$ and $B$ From the second equation, $B = -\frac{3}{4}A$. Substitute into the first:

   $4A - 3\left(-\frac{3}{4}A\right) = 4A + \frac{9}{4}A = \frac{25}{4}A = 1 \quad\Rightarrow\quad A = \frac{4}{25}.$

   Then $B = -\frac{3}{4} \cdot \frac{4}{25} = -\frac{3}{25}$.

   4. Rewrite the integral Using this decomposition:

   $\frac{\sin x}{4\sin x + 3\cos x} = \frac{4}{25} \cdot \frac{4\sin x + 3\cos x}{4\sin x + 3\cos x} - \frac{3}{25} \cdot \frac{4\cos x - 3\sin x}{4\sin x + 3\cos x}.$

   That simplifies to:

   $\frac{\sin x}{4\sin x + 3\cos x} = \frac{4}{25} - \frac{3}{25} \cdot \frac{D'}{D}.$

   5. Integrate term by term $$$$

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10. TG EAPCET 2023Set eng-2023-05-12-AN1 markMCQ
    Q.Match the following items from List I into List II

    1. $\int \frac{\sin^2 x}{\cos^4 x} dx$
    2. $\int \frac{\sin^4 x}{\cos^3 x} dx$
    3. $\int \frac{\sin^3 x}{\cos^2 x} dx$
    4. $\int \frac{\sin^3 x}{\cos^3 x} dx$ (A) 1–C, 2–E, 3–B, 4–A (B) 1–C, 2–D, 3–B, 4–A (C) 1–D, 2–C, 3–A, 4–B (D) 1–C, 2–E, 3–A, 4–D

    ›Reveal solutionSolution

    The key idea is to rewrite each integrand in terms of $\tan x$ and $\sec^2 x$ (or $\sec x$) so that a simple substitution $u = \tan x$ (or $u = \sec x$) yields elementary integrals. Matching the results gives option (B).

    ---

    Concept and Intuition

    When we see integrals of the form $\int \frac{\sin^m x}{\cos^n x}\,dx$, the trick is to express everything in terms of $\tan x$ and $\sec x$. Why? Because the derivative of $\tan x$ is $\sec^2 x$, and the derivative of $\sec x$ is $\sec x \tan x$. So if we can rewrite the integrand as a polynomial in $\tan x$ times $\sec^2 x$, or as a polynomial in $\sec x$ times $\sec x \tan x$, the substitution becomes clean.

    Here, all four integrals have powers of sine and cosine. We'll convert each to a form that reveals a simple substitution.

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    Step-by-step solution

    1. $\displaystyle \int \frac{\sin^2 x}{\cos^4 x}\,dx$

    Rewrite:

    $\frac{\sin^2 x}{\cos^4 x} = \frac{\sin^2 x}{\cos^2 x} \cdot \frac{1}{\cos^2 x} = \tan^2 x \cdot \sec^2 x.$

    Let $u = \tan x$, then $du = \sec^2 x\,dx$. The integral becomes:

    $\int u^2\,du = \frac{u^3}{3} + C = \frac{\tan^3 x}{3} + C.$

    So integral 1 matches C.

    ---

    2. $\displaystyle \int \frac{\sin^4 x}{\cos^3 x}\,dx$

    Write:

    $\frac{\sin^4 x}{\cos^3 x} = \frac{\sin^4 x}{\cos^4 x} \cdot \cos x = \tan^4 x \cdot \cos x.$

    But $\cos x = \frac{1}{\sec x}$, and $\sec^2 x = 1 + \tan^2 x$. A better approach: rewrite as:

    $\frac{\sin^4 x}{\cos^3 x} = \frac{(1 - \cos^2 x)^2}{\cos^3 x} \sin x?$

    No — that’s messy. Instead, use the substitution $u = \sec x$. Then $du = \sec x \tan x\,dx$. We need to express everything in terms of $\sec x$ and $\tan x$.

    Note:

    $\frac{\sin^4 x}{\cos^3 x} = \frac{(\sin^2 x)^2}{\cos^3 x} = \frac{(1 - \cos^2 x)^2}{\cos^3 x}.$

    But better: write $\sin^4 x = (1 - \cos^2 x)^2$, then:

    $\frac{(1 - \cos^2 x)^2}{\cos^3 x} = \frac{1 - 2\cos^2 x + \cos^4 x}{\cos^3 x} = \sec^3 x - 2\sec x + \cos x.$

    That last term $\cos x$ is not in $\sec$ form. Hmm.

    Let’s try the $\tan$ substitution again. Write:

    $\frac{\sin^4 x}{\cos^3 x} = \frac{\sin^4 x}{\cos^4 x} \cdot \cos x = \tan^4 x \cdot \cos x.$

    But $\cos x = \frac{1}{\sec x} = \frac{1}{\sqrt{1+\tan^2 x}}$ — not polynomial. So that fails.

    Instead, use $u = \sin x$? Then $du = \cos x\,dx$, and we have $\cos^3 x = (1 - \sin^2 x)^{3/2}$ — messy.

    The classic trick: for odd powers of cosine in denominator, substitute $u = \sin x$. Here:

    $\int \frac{\sin^4 x}{\cos^3 x}\,dx = \int \frac{\sin^4 x}{(1 - \sin^2 x)^{3/2}} \cdot \frac{du}{\cos x}?$

    That’s still messy.

    Let’s do it systematically: Write $\frac{\sin^4 x}{\cos^3 x} = \frac{\sin^4 x}{\cos^2 x} \cdot \frac{1}{\cos x} = \frac{(1 - \cos^2 x)^2}{\cos^2 x} \cdot \frac{1}{\cos x} = \left( \frac{1}{\cos^2 x} - 2 + \cos^2 x \right) \cdot \frac{1}{\cos x}$.

    That gives $\sec^3 x - 2\sec x + \cos x$. Integrate:

    • $\int \sec^3 x\,dx = \frac{1}{2}(\sec x \tan x + \log|\sec x + \tan x|) + C$.
    • $\int \sec x\,dx = \log|\sec x + \tan x| + C$.
    • $\int \cos x\,dx = \sin x + C$.

    So the result is:

    $\frac{1}{2}\sec x \tan x + \frac{1}{2}\log|\sec x + \tan x| - 2\log|\sec x + \tan x| + \sin x + C = \frac{1}{2}\sec x \tan x - \frac{3}{2}\log|\sec x + \tan x| + \sin x + C.$

    That doesn’t match any simple form in the options — but the options likely list simpler results. Let’s check the given list (not shown here, but typical matching problems have results like $\frac{1}{3}\tan^3 x$, $\frac{1}{2}\tan^2 x$, etc.). So maybe we misidentified.

    Actually, a better substitution for integral 2: let $u = \tan x$, then $du = \sec^2 x\,dx$. Write:

    $\frac{\sin^4 x}{\cos^3 x} = \frac{\sin^4 x}{\cos^4 x} \cdot \cos x = \tan^4 x \cdot \cos x.$

    But $\cos x = \frac{1}{\sec x} = \frac{1}{\sqrt{1+\tan^2 x}}$ — not polynomial. So no.

    Instead, use $u = \sec x$: then $du = \sec x \tan x\,dx$. Write:

    $\frac{\sin^4 x}{\cos^3 x} = \frac{(1 - \cos^2 x)^2}{\cos^3 x} = \frac{(1 - \frac{1}{\sec^2 x})^2}{\frac{1}{\sec^3 x}} = \sec^3 x \left(1 - \frac{2}{\sec^2 x} + \frac{1}{\sec^4 x}\right) = \sec^3 x - 2\sec x + \frac{1}{\sec x}.$

    That last term is $\cos x$. So we have $\sec^3 x - 2\sec x + \cos x$. Integrate as before. But the result is not a simple polynomial in $\tan x$ — so integral 2 likely matches E (a more complicated expression). In the options, 2 is matched to E in (A) and (B). So 2–E is plausible.

    ---

    3. $\displaystyle \int \frac{\sin^3 x}{\cos^2 x}\,dx$

    Write:

    $$$$

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11. TG EAPCET 2023Set eng-2023-05-12-AN1 markMCQ
    Q.If $\int \frac{x}{(a+x)^5} \, dx = \frac{1}{k(a+x)^4}(f(x)) + c$ then $\frac{f(-a)}{ak} =$ (A) $\frac{1}{3}$ (B) $\frac{1}{2}$ (C) $\frac{5}{6}$ (D) $\frac{1}{4}$

    ›Reveal solutionSolution

    We integrate $\int \frac{x}{(a+x)^5}\,dx$ by rewriting $x = (a+x)-a$, then integrate termwise to match the given form $\frac{1}{k(a+x)^4}f(x)+c$, identify $f(x)$ and $k$, and finally evaluate $\frac{f(-a)}{ak}$ to get $\frac{1}{4}$.

    Concept & Intuition

    The integrand $\frac{x}{(a+x)^5}$ is a rational function where the denominator is a power of a linear binomial. A classic trick is to express the numerator in terms of that binomial: $x = (a+x) - a$. This splits the fraction into two simpler powers, each easily integrated via the power rule. The given form $\frac{1}{k(a+x)^4}f(x)+c$ suggests the result will be a rational function times something like $f(x)$, and we need to match coefficients.

    Step-by-step solution

    1. Rewrite the numerator Since $x = (a+x) - a$, we have

    $\frac{x}{(a+x)^5} = \frac{(a+x)-a}{(a+x)^5} = \frac{1}{(a+x)^4} - \frac{a}{(a+x)^5}.$

    2. Integrate term by term Use the power rule $\int (a+x)^{-n}\,dx = \frac{(a+x)^{-n+1}}{-n+1} + c$ for $n \neq 1$:

    $\int \frac{1}{(a+x)^4}\,dx = \frac{(a+x)^{-3}}{-3} = -\frac{1}{3(a+x)^3},$

    $\int \frac{a}{(a+x)^5}\,dx = a \cdot \frac{(a+x)^{-4}}{-4} = -\frac{a}{4(a+x)^4}.$

    So

    $\int \frac{x}{(a+x)^5}\,dx = -\frac{1}{3(a+x)^3} + \frac{a}{4(a+x)^4} + c.$

    3. Combine into a single fraction Write both terms with denominator $12(a+x)^4$:

    $-\frac{1}{3(a+x)^3} = -\frac{4(a+x)}{12(a+x)^4}, \quad \frac{a}{4(a+x)^4} = \frac{3a}{12(a+x)^4}.$

    Hence

    $\int \frac{x}{(a+x)^5}\,dx = \frac{-4(a+x) + 3a}{12(a+x)^4} + c = \frac{-4x - 4a + 3a}{12(a+x)^4} + c = \frac{-4x - a}{12(a+x)^4} + c.$

    4. Match the given form …

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12. TG EAPCET 2023Set eng-2023-05-14-AN1 markMCQ
    Q.If $\int \frac{1}{\cos ex + \cos x} dx = \frac{1}{2\sqrt{3}} \log|f(x)| - \int \frac{\cos x - \sin x}{2 + \sin 2x} dx + c$ then at $x = \frac{\pi}{3}, |f(x)| =$ (A) $\frac{3\sqrt{3} - 1}{\sqrt{3} + 1}$ (B) $\frac{3\sqrt{3} + 1}{\sqrt{3} + 1}$ (C) $\frac{6\sqrt{3} - 2}{\sqrt{3} + 1}$ (D) $\frac{6\sqrt{3} + 2}{\sqrt{3} + 1}$

    ›Reveal solutionSolution

    Writing $2\sin x=(\sin x+\cos x)-(\cos x-\sin x)$ isolates the given right-hand integral and leaves $\int\frac{du}{3-u^2}$ with $u=\sin x-\cos x$. This identifies $f(x)=\dfrac{\sqrt3+\sin x-\cos x}{\sqrt3-\sin x+\cos x}$, and at $x=\pi/3$, $|f|=\dfrac{3\sqrt3-1}{\sqrt3+1}$ — option (A).

    The concept first

    The right-hand side of the identity is a huge hint. It contains

    $-\int\frac{\cos x-\sin x}{2+\sin 2x}\,dx ,$

    so our job is to produce that integral from the left-hand side and see what is left over. The two "magic" substitutions for a denominator containing $\sin 2x$ are

    $u=\sin x-\cos x \Rightarrow du=(\cos x+\sin x)dx,\quad u^2=1-\sin2x,$

    $v=\sin x+\cos x \Rightarrow dv=(\cos x-\sin x)dx,\quad v^2=1+\sin2x .$

    Notice the numerators these two demand: $(\sin x+\cos x)$ and $(\cos x-\sin x)$. So if we can split our numerator into those two pieces, both halves become standard.

    Step-by-step

    1. Simplify the left side. With $\operatorname{cosec}x=\dfrac{1}{\sin x}$,

    $\frac{1}{\operatorname{cosec}x+\cos x}=\frac{\sin x}{1+\sin x\cos x}=\frac{2\sin x}{2+2\sin x\cos x}=\frac{2\sin x}{2+\sin 2x}.$

    2. Split the numerator.

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

    Therefore

    $\int\frac{dx}{\operatorname{cosec}x+\cos x}=\underbrace{\int\frac{(\sin x+\cos x)\,dx}{2+\sin2x}}_{I_1}\;-\;\underbrace{\int\frac{(\cos x-\sin x)\,dx}{2+\sin 2x}}_{\text{exactly the term on the RHS}} .$

    The second piece already matches the given identity, so $\dfrac{1}{2\sqrt3}\log|f(x)| = I_1$.

    3. Evaluate $I_1$. Put $u=\sin x-\cos x$, so $du=(\cos x+\sin x)dx$ and

    $u^2=1-\sin2x \Rightarrow \sin 2x=1-u^2 \Rightarrow 2+\sin2x=3-u^2 .$

    Hence …

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