Q.Integrate the following functions w.r.t. $x$:

Showing the 12 most recent of 51 on this concept.

1. AP EAPCET 2025Set eng-2025-05-22-FN1 markMCQ
   Q.$\int\frac{x^4 - 1}{x^2\sqrt{x^4 + x^2 + 1}}dx =$ (A) $\frac{2\sqrt{x^4+x^2+1}}{x} + c$ (B) $\frac{\sqrt{x^4+x^2+1}}{x} + c$ (C) $\frac{\sqrt{x^4+x^2+1}}{2x} + c$ (D) $\frac{4\sqrt{x^4+x^2+1}}{x} + c$

   ›Reveal solutionSolution

   Differentiating the candidate $\frac{\sqrt{x^4+x^2+1}}{x}$ reproduces the given integrand exactly, confirming it as the antiderivative.

   Concept and Intuition

   When an integrand looks like it could come from a quotient rule (a square root over a power of $x$), it is often faster to differentiate a plausible candidate of that shape and check, rather than search for a substitution from scratch.

   Step-by-Step Solution

   1. Try $g(x)=\dfrac{\sqrt{x^4+x^2+1}}{x}=\dfrac{N}{x}$ where $N=\sqrt{x^4+x^2+1}$.
   2. $N'=\dfrac{4x^3+2x}{2\sqrt{x^4+x^2+1}}=\dfrac{x(2x^2+1)}{N}$.
   3. Quotient rule: $g'(x)=\dfrac{N'x-N}{x^2}=\dfrac{\frac{x^2(2x^2+1)}{N}-N}{x^2}=\dfrac{x^2(2x^2+1)-N^2}{Nx^2}$.
   4. $N^2=x^4+x^2+1$, so the numerator is $x^2(2x^2+1)-(x^4+x^2+1)=2x^4+x^2-x^4-x^2-1=x^4-1$.
   5. So $g'(x)=\dfrac{x^4-1}{x^2\sqrt{x^4+x^2+1}}$ — exactly the given integrand.
   6. Hence $\displaystyle\int\frac{x^4-1}{x^2\sqrt{x^4+x^2+1}}\,dx=\frac{\sqrt{x^4+x^2+1}}{x}+c$.

   Common Mistakes

   • Attempting a substitution like $t=x-1/x$ or $t=x+1/x$ and getting tangled in cross terms instead of recognising the quotient-rule shape.
   • Dropping the $x^2$ in the denominator when differentiating $N/x$ (quotient rule, not just $N'/x$).

   ✓Final answer

   The correct option is (B) — $\frac{\sqrt{x^4+x^2+1}}{x} + c$.

   ANSWER: B

2. AP EAPCET 2021Set eng-2021-10-05-FN1 markMCQ
   Q.$\displaystyle\int \dfrac{\sin x}{\sin 3x}\, dx =$ (A) $\dfrac{1}{2\sqrt{3}} \log\left|\dfrac{\sqrt{3} + \tan x}{\sqrt{3} - \tan x}\right| + c$ (B) $\dfrac{1}{2\sqrt{3}} \log\left|\dfrac{\sqrt{3} - \tan x}{\sqrt{3} + \tan x}\right| + c$ (C) $\dfrac{1}{4\sqrt{3}} \log\left|\dfrac{\sqrt{3} + \tan x}{\sqrt{3} - \tan x}\right| + c$ (D) $\dfrac{1}{4\sqrt{3}} \log\left|\dfrac{\sqrt{3} - \tan x}{\sqrt{3} + \tan x}\right| + c$

   ›Reveal solutionSolution

   Use the triple-angle identity to cancel $\sin x$, rewrite in terms of $\cos2x$, then apply the Weierstrass-type substitution $t=\tan x$ to reduce to a standard rational integral. The answer is (A).

   Concept and Intuition

   $\sin3x$ factors as $\sin x(3-4\sin^2x)$, so $\sin x$ cancels immediately with the numerator, turning a trigonometric-looking integral into a much simpler one in $\sin^2x$ (hence in $\cos2x$), which is a textbook target for the $t=\tan x$ substitution.

   Step-by-Step Solution

   1. $\sin3x = 3\sin x - 4\sin^3x = \sin x(3-4\sin^2x)$.
   2. $\dfrac{\sin x}{\sin3x} = \dfrac{1}{3-4\sin^2x}$.
   3. Using $\sin^2x=\dfrac{1-\cos2x}2$: $4\sin^2x = 2-2\cos2x$, so $3-4\sin^2x = 1+2\cos2x$.
   4. Integral becomes $\displaystyle\int\dfrac{dx}{1+2\cos2x}$.
   5. Substitute $t=\tan x$, $dx=\dfrac{dt}{1+t^2}$, $\cos2x=\dfrac{1-t^2}{1+t^2}$: $1+2\cos2x = \dfrac{(1+t^2)+2(1-t^2)}{1+t^2} = \dfrac{3-t^2}{1+t^2}$.
   6. Integral $= \displaystyle\int \dfrac{1+t^2}{3-t^2}\cdot\dfrac{dt}{1+t^2} = \int\dfrac{dt}{3-t^2}$.
   7. Standard form $\displaystyle\int\dfrac{dt}{a^2-t^2}=\dfrac1{2a}\log\left|\dfrac{a+t}{a-t}\right|+c$ with $a=\sqrt3$: $= \dfrac1{2\sqrt3}\log\left|\dfrac{\sqrt3+\tan x}{\sqrt3-\tan x}\right|+c$.

   Common Mistakes

   • Forgetting the triple-angle identity and instead trying product-to-sum formulas, which lead to a much messier route.
   • Sign or coefficient slip converting $\sin^2x$ to $\cos2x$ terms.

   ✓Final answer

   The correct option is (A) — $\dfrac{1}{2\sqrt{3}} \log\left|\dfrac{\sqrt{3} + \tan x}{\sqrt{3} - \tan x}\right| + c$.

   ANSWER: A

3. AP EAPCET 2026Set eng-2026-05-13-AN1 markMCQ
   Q.$\displaystyle\int \dfrac{dx}{x^2(x^4+1)^{3/4}}=$ (A) $\left(1+\dfrac{1}{x^4}\right)^{3/4}+c$ (B) $\left(1+\dfrac{1}{x^6}\right)^{1/2}+c$ (C) $-\left(1+\dfrac{1}{x^4}\right)^{-1/4}+c$ (D) $-\left(1+\dfrac{1}{x^4}\right)^{1/4}+c$

   ›Reveal solutionSolution

   Pulling $x^4$ out from under the radical and substituting $t=1+x^{-4}$ reduces the integral to a simple power rule, giving $-(1+1/x^4)^{1/4}+c$.

   Concept and Intuition

   When the integrand has $(x^n+1)^p$, it often helps to factor out the highest power of $x$ from inside the bracket so that a substitution like $t = 1+x^{-n}$ produces a clean differential matching the rest of the integrand.

   Step-by-Step Solution

   1. Write $(x^4+1)^{3/4} = \left(x^4\left(1+\dfrac{1}{x^4}\right)\right)^{3/4} = x^3\left(1+\dfrac{1}{x^4}\right)^{3/4}$.
   2. So $\dfrac{1}{x^2(x^4+1)^{3/4}} = \dfrac{1}{x^2\cdot x^3\left(1+\frac{1}{x^4}\right)^{3/4}} = \dfrac{1}{x^5\left(1+\frac{1}{x^4}\right)^{3/4}} = x^{-5}\left(1+x^{-4}\right)^{-3/4}$.
   3. Let $t = 1+x^{-4}$. Then $dt = -4x^{-5}dx$, so $x^{-5}dx = -\dfrac{dt}{4}$.
   4. The integral becomes $\displaystyle\int t^{-3/4}\left(-\dfrac{dt}{4}\right) = -\dfrac14\cdot\dfrac{t^{1/4}}{1/4}+c = -t^{1/4}+c$.
   5. Substituting back: $-\left(1+\dfrac{1}{x^4}\right)^{1/4}+c$.

   Common Mistakes

   • Forgetting the negative sign that comes from $dt=-4x^{-5}dx$.
   • Not factoring $x^4$ out correctly before substituting, leading to a mismatched power.

   ✓Final answer

   The correct option is (D) — $-\left(1+\dfrac{1}{x^4}\right)^{1/4}+c$.

   ANSWER: D

4. AP EAPCET 2021Set eng-2021-08-25-AN1 markMCQ
   Q.$\int \cos\sqrt{x}\, dx =$ (A) $2\sqrt{x}\sin\sqrt{x} + 2\cos\sqrt{x} + c$ (B) $2\sqrt{x}\sin\sqrt{x} + 2\sin\sqrt{x} + c$ (C) $2\sqrt{x}\sin\sqrt{x} - 2\cos\sqrt{x} + c$ (D) $\sqrt{x}\cos\sqrt{x} - 2\sin\sqrt{x} + c$

   ›Reveal solutionSolution

   Substitute $t=\sqrt x$ to turn the integral into a standard integration-by-parts problem. Answer: $2\sqrt x\sin\sqrt x+2\cos\sqrt x+c$.

   Concept and Intuition

   Whenever you see $\sqrt x$ trapped inside a trig or exponential function, substituting $t=\sqrt x$ converts it into a polynomial-times-trig integral solvable by parts.

   Step-by-Step Solution

   1. Let $t=\sqrt x \Rightarrow x=t^2,\ dx=2t\,dt$.
   2. $\int\cos\sqrt x\,dx = \int\cos t\cdot 2t\,dt = 2\int t\cos t\,dt$.
   3. Integrate by parts: $\int t\cos t\,dt = t\sin t - \int\sin t\,dt = t\sin t+\cos t$.
   4. So the integral $=2(t\sin t+\cos t)+c = 2t\sin t+2\cos t+c$.
   5. Substitute back $t=\sqrt x$: $=2\sqrt x\sin\sqrt x+2\cos\sqrt x+c$.

   Common Mistakes

   • Forgetting the factor of $2t$ from $dx=2t\,dt$ when substituting.

   ✓Final answer

   The correct option is (A) — $2\sqrt x\sin\sqrt x+2\cos\sqrt x+c$.

   ANSWER: A

5. AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
   Q.$\displaystyle\int \dfrac{\cos^3 x}{(1 + \sin x)^4}\, dx =$ (A) $-\dfrac{\cos^4 x}{5(1 + \sin x)^5} + c$ (B) $\dfrac{\cos^4 x}{5(1 + \sin x)^5} + c$ (C) $\dfrac{\cos^4 x}{4(1 + \sin x)^4} + c$ (D) $-\dfrac{\cos^4 x}{4(1 + \sin x)^4} + c$

   ›Reveal solutionSolution

   Factor $\cos^3x$ using $\cos^2x=(1-\sin x)(1+\sin x)$ and substitute $t=\sin x$; the resulting antiderivative can equivalently be written in the $\cos^4x/(1+\sin x)^4$ form given in the options (they differ only by an added constant). Answer: $-\dfrac{\cos^4x}{4(1+\sin x)^4}+c$.

   Concept and Intuition

   Integrals of $\cos^{\text{odd}}x$ over powers of $(1+\sin x)$ are handled by peeling off one factor of $\cos x$ to pair with $dx$ (making $d(\sin x)$) and expressing the remaining even power of $\cos x$ in terms of $\sin x$. Since the MCQ options are phrased in terms of $\cos^4x$ rather than $\sin x$ directly, it is often faster (and safer against sign traps) to guess-and-check an antiderivative of that shape by differentiating a general form $A\cos^4x(1+\sin x)^{-n}$ and matching powers/coefficients — this is exactly how the printed option is confirmed.

   Step-by-Step Solution

   1. $\cos^3x=\cos x\cdot\cos^2x=\cos x(1-\sin^2x)=\cos x(1-\sin x)(1+\sin x)$.
   2. Integrand $=\dfrac{\cos x(1-\sin x)(1+\sin x)}{(1+\sin x)^4}=\dfrac{\cos x(1-\sin x)}{(1+\sin x)^3}$.
   3. Let $t=\sin x$, $dt=\cos x\,dx$: $I=\displaystyle\int\frac{1-t}{(1+t)^3}dt$. Writing $1-t=2-(1+t)$: $I=\displaystyle\int\left(\frac{2}{(1+t)^3}-\frac{1}{(1+t)^2}\right)dt=-\frac{1}{(1+t)^2}+\frac{1}{1+t}+c=\frac{t}{(1+t)^2}+c$.
   4. So $I=\dfrac{\sin x}{(1+\sin x)^2}+c$ is one valid closed form.
   5. To match the option's shape, try $F(x)=A\dfrac{\cos^4x}{(1+\sin x)^n}$ and differentiate: using $\cos^2x=(1-\sin x)(1+\sin x)$, one finds $F'(x)=\cos^3x/(1+\sin x)^4$ exactly when $n=4$ and $A=-\dfrac14$ — i.e. $F(x)=-\dfrac{\cos^4x}{4(1+\sin x)^4}$ is also a valid antiderivative (differs from step 4's form only by the constant $\tfrac14$, confirmed by evaluating both at, say, $x=0$ and $x=\pi/2$).

   Common Mistakes

   • Sign error: option (C) has the same magnitude but wrong sign — differentiating (C) gives $+\cos^3x/(1+\sin x)^4$, not matching.
   • Using the wrong power $n=5$ (options A/B) instead of the correct $n=4$.

   ✓Final answer

   The correct option is (D) — $-\dfrac{\cos^4 x}{4(1 + \sin x)^4} + c$.

   ANSWER: D

6. AP EAPCET 2024Set eng-2024-05-18-FN1 markMCQ
   Q.$\displaystyle\int \dfrac{1}{x^5\sqrt[5]{x^5+1}}\,dx=$ (A) $\dfrac{4}{\sqrt[5]{x^5+1}}+c$ (B) $4x^4(x^5+1)^{4/5}+c$ (C) $-\dfrac{(x^5+1)^{4/5}}{4x^4}+c$ (D) $-\dfrac{(x^5+1)^{4/5}}{4x^5}+c$

   ›Reveal solutionSolution

   Rewriting the integrand to expose $1+x^{-5}$ as the natural substitution variable solves this cleanly; the answer is $-\dfrac{(x^5+1)^{4/5}}{4x^4}+c$.

   Concept and Intuition

   When an integral mixes a power of $x$ with a root of a polynomial in $x$, factoring out the highest power of $x$ from inside the root often converts the expression into a function of $1/x$ (or $x^{-5}$ here), whose derivative is already present elsewhere in the integrand — a clean substitution.

   Step-by-Step Solution

   1. $(x^5+1)^{-1/5}=\left(x^5\left(1+x^{-5}\right)\right)^{-1/5}=x^{-1}(1+x^{-5})^{-1/5}$.
   2. So the integrand $x^{-5}(x^5+1)^{-1/5}=x^{-6}(1+x^{-5})^{-1/5}$.
   3. Let $t=1+x^{-5}$, so $dt=-5x^{-6}dx\Rightarrow x^{-6}dx=-\dfrac{dt}{5}$.
   4. Integral $=\displaystyle\int t^{-1/5}\left(-\dfrac{dt}{5}\right)=-\dfrac15\cdot\dfrac{t^{4/5}}{4/5}+c=-\dfrac14t^{4/5}+c$.
   5. Substitute back: $-\dfrac14(1+x^{-5})^{4/5}+c=-\dfrac14\left(\dfrac{x^5+1}{x^5}\right)^{4/5}+c=-\dfrac{(x^5+1)^{4/5}}{4x^4}+c$.

   Common Mistakes

   • Trying $u=x^5+1$ directly, which does not match the $x^{-5}$ factor present and leads to a messier, non-matching form.
   • Sign or exponent slip converting $x^5\cdot x^{-4}$ powers back after substitution.

   ✓Final answer

   The correct option is (C) — $-\dfrac{(x^5+1)^{4/5}}{4x^4}+c$.

   ANSWER: C

7. AP EAPCET 2024Set eng-2024-05-18-FN1 markMCQ
   Q.$\displaystyle\int(\tan^7x+\tan x)\,dx=$ (A) $\dfrac{\tan^2 x}{12}(2\tan^4 x - 3\tan^2 x+6)+c$ (B) $\dfrac{\tan^2 x}{6}-\dfrac{\tan^5 x}{4}+\dfrac{\tan^4 x}{2}+c$ (C) $\dfrac{\tan^2 x}{6}(\tan^4 x+3\tan^2 x+4)+c$ (D) $\dfrac{\tan x}{12}(\tan^4 x-3\tan^2 x+6)+c$

   ›Reveal solutionSolution

   Factoring $\tan^7x+\tan x$ using the sum-of-like-terms identity $t^7+t=t(t^6+1)=t(t^2+1)(t^4-t^2+1)$ exposes the $\sec^2x$ needed for a clean $t=\tan x$ substitution. Answer: $\dfrac{\tan^2x}{12}(2\tan^4x-3\tan^2x+6)+c$.

   Concept and Intuition

   Whenever an integrand is built purely from powers of $\tan x$ together with an explicit or hidden $\sec^2x$, substituting $t=\tan x$ turns it into a polynomial integral — the key is recognizing that $t^6+1=(t^2+1)(t^4-t^2+1)$ supplies exactly the $\sec^2x=1+\tan^2x$ factor needed.

   Step-by-Step Solution

   1. $\tan^7x+\tan x=\tan x(\tan^6x+1)=\tan x(\tan^2x+1)(\tan^4x-\tan^2x+1)=\tan x\sec^2x(\tan^4x-\tan^2x+1)$.
   2. Let $t=\tan x$, $dt=\sec^2x\,dx$. Integral becomes $\displaystyle\int t(t^4-t^2+1)dt=\int(t^5-t^3+t)dt$.
   3. $=\dfrac{t^6}{6}-\dfrac{t^4}{4}+\dfrac{t^2}{2}+c=\dfrac{\tan^6x}{6}-\dfrac{\tan^4x}{4}+\dfrac{\tan^2x}{2}+c$.
   4. Factor out $\dfrac{\tan^2x}{12}$: $\dfrac{\tan^2x}{12}\left(2\tan^4x-3\tan^2x+6\right)+c$ (check: $\dfrac{2}{12}=\dfrac16$, $\dfrac{3}{12}=\dfrac14$, $\dfrac{6}{12}=\dfrac12$ — all match).

   Common Mistakes

   • Missing the factorization $t^6+1=(t^2+1)(t^4-t^2+1)$ and instead trying to expand $\tan^7x$ directly, which is much harder.
   • Sign error in the middle term when factoring $t^4-t^2+1$ vs $t^4+t^2+1$.

   ✓Final answer

   The correct option is (A) — $\dfrac{\tan^2 x}{12}(2\tan^4 x - 3\tan^2 x+6)+c$.

   ANSWER: A

8. AP EAPCET 2025Set eng-2025-05-27-FN1 markMCQ
   Q.$\displaystyle\int \dfrac{dx}{(1 + \sqrt{x})\sqrt{x - x^2}} =$ (A) $-2\sqrt{\dfrac{1 + \sqrt{x}}{1 - \sqrt{x}}} + c$ (B) $-\sqrt{\dfrac{1 - \sqrt{x}}{1 + \sqrt{x}}} + c$ (C) $-2\sqrt{\dfrac{1 - \sqrt{x}}{1 + \sqrt{x}}} + c$ (D) $2\sqrt{\dfrac{1 + \sqrt{x}}{1 - \sqrt{x}}} + c$

   ›Reveal solutionSolution

   Substituting $t=\sqrt x$ turns the surd-heavy integrand into $\int \frac{2\,dt}{(1+t)^{3/2}(1-t)^{1/2}}$, whose antiderivative is exactly $-2\sqrt{\frac{1-t}{1+t}}$.

   Concept and Intuition

   When an integrand mixes $\sqrt x$ and $\sqrt{x-x^2}=\sqrt{x}\sqrt{1-x}$, substituting $t=\sqrt x$ clears every square root of $x$ at once, converting the whole thing into a rational-power integral in $t$ that matches the derivative of $\sqrt{\frac{1-t}{1+t}}$ — a standard "recognise the derivative" pattern worth memorising for CET-style problems.

   Step-by-Step Solution

   1. Write $\sqrt{x-x^2}=\sqrt{x(1-x)}=\sqrt x\sqrt{1-x}$, so the integral is

   $I=\int\frac{dx}{(1+\sqrt x)\sqrt x\sqrt{1-x}}.$

   2. Let $t=\sqrt x$, so $x=t^2$, $dx=2t\,dt$:

   $I=\int\frac{2t\,dt}{(1+t)\cdot t\cdot\sqrt{1-t^2}}=\int\frac{2\,dt}{(1+t)\sqrt{(1-t)(1+t)}}=\int\frac{2\,dt}{(1+t)^{3/2}(1-t)^{1/2}}.$

   3. Let $y=\sqrt{\dfrac{1-t}{1+t}}$. Differentiating $y^2=\dfrac{1-t}{1+t}$:

   $2y\,y'=\frac{-(1+t)-(1-t)}{(1+t)^2}=\frac{-2}{(1+t)^2}\ \Rightarrow\ y'=\frac{-1}{y(1+t)^2}=\frac{-1}{(1+t)^{3/2}(1-t)^{1/2}}.$

   4. So $I=2\int y'\,dt\cdot(-1)^{-1}$, i.e. $\dfrac{d}{dt}\left(-2y\right)=\dfrac{2}{(1+t)^{3/2}(1-t)^{1/2}}$, matching the integrand exactly.
   5. Hence $I=-2y+c=-2\sqrt{\dfrac{1-t}{1+t}}+c=-2\sqrt{\dfrac{1-\sqrt x}{1+\sqrt x}}+c$.

   Common Mistakes

   • Flipping the ratio inside the square root (getting $\sqrt{\frac{1+t}{1-t}}$ instead of $\sqrt{\frac{1-t}{1+t}}$) — check by differentiating your guess before committing.
   • Losing the negative sign in front.

   ✓Final answer

   The correct option is (C) — $-2\sqrt{\dfrac{1-\sqrt x}{1+\sqrt x}}+c$.

   ANSWER: C

9. AP EAPCET 2024Set eng-2024-05-20-AN1 markMCQ
   Q.$\displaystyle\int \dfrac{dx}{\sqrt{\sin^3 x \cos(x-\alpha)}} =$ (A) $\dfrac{1}{\cos\alpha}\sqrt{\cot x+\tan\alpha}+c$ (B) $\dfrac{1}{\sqrt{\cos\alpha}}\sqrt{\cot x-\tan\alpha}+c$ (C) $\dfrac{-1}{\sqrt{\sin\alpha}}\sqrt{\cot x+\tan\alpha}+c$ (D) $\dfrac{-2}{\sqrt{\cos\alpha}}\sqrt{\cot x+\tan\alpha}+c$

   ›Reveal solutionSolution

   Expanding $\cos(x-\alpha)$ and substituting $u=\cot x$ reduces this to a simple square-root integral; final answer is (D).

   Concept and Intuition

   The key move is expanding $\cos(x-\alpha)$ and factoring out $\sin^4x$ from inside the square root so that a substitution $u=\cot x$ (whose differential is $-\csc^2x\,dx$, conveniently matching what's left outside) linearizes the whole thing.

   Step-by-Step Solution

   1. $\cos(x-\alpha)=\cos x\cos\alpha+\sin x\sin\alpha$.
   2. $\sin^3x\cos(x-\alpha)=\sin^3x\cos x\cos\alpha+\sin^4x\sin\alpha = \sin^4x\left(\dfrac{\cos x}{\sin x}\cos\alpha+\sin\alpha\right)=\sin^4x(\cot x\cos\alpha+\sin\alpha)$.
   3. So $\sqrt{\sin^3x\cos(x-\alpha)}=\sin^2x\sqrt{\cot x\cos\alpha+\sin\alpha}$ (taking $\sin^2x>0$ outside the root).
   4. The integral is $\displaystyle\int\dfrac{dx}{\sin^2x\sqrt{\cos\alpha\cot x+\sin\alpha}}=\int\dfrac{\csc^2x\,dx}{\sqrt{\cos\alpha\cot x+\sin\alpha}}$.
   5. Let $u=\cot x$, $du=-\csc^2x\,dx$. Integral becomes $-\displaystyle\int\dfrac{du}{\sqrt{\cos\alpha\,u+\sin\alpha}}=-\dfrac{2\sqrt{\cos\alpha\,u+\sin\alpha}}{\cos\alpha}+c$.
   6. Factor $\cos\alpha$ out of the root: $\cos\alpha\,u+\sin\alpha=\cos\alpha(u+\tan\alpha)$, so this is $-\dfrac{2\sqrt{\cos\alpha}\sqrt{u+\tan\alpha}}{\cos\alpha}=-\dfrac2{\sqrt{\cos\alpha}}\sqrt{u+\tan\alpha}+c$.
   7. Substituting back $u=\cot x$: $-\dfrac2{\sqrt{\cos\alpha}}\sqrt{\cot x+\tan\alpha}+c$.

   Common Mistakes

   • Sign error picking $u=\cot x$ vs $u=\tan x$ — the differential $du=-\csc^2x\,dx$ must match the sign of what remains outside the root.
   • Losing the overall factor of $2$ when integrating $u^{-1/2}$.

   ✓Final answer

   The correct option is (D) — $\dfrac{-2}{\sqrt{\cos\alpha}}\sqrt{\cot x+\tan\alpha}+c$.

   ANSWER: D

10. AP EAPCET 2024Set eng-2024-05-22-FN1 markMCQ
    Q.$\displaystyle\int \dfrac{x^3 \tan^{-1} x^4}{1 + x^8}\,dx =$ (A) $\dfrac{(\tan^{-1}(x^4))^2}{8} + c$ (B) $\dfrac{(\tan^{-1}(x^4))^3}{3} + c$ (C) $\dfrac{(\tan^{-1}(x^4))^2}{4} + c$ (D) $\dfrac{(\tan^{-1}(x^4))^2}{2} + c$

    ›Reveal solutionSolution

    A double substitution (first $u=x^4$, then $v=\tan^{-1}u$) turns this into a trivial $\int v\,dv$. Answer: $\dfrac{(\tan^{-1}x^4)^2}{8}+c$.

    Concept and Intuition

    The presence of $x^3\,dx$ alongside $x^4$ inside the arctan and $x^8=(x^4)^2$ in the denominator is a strong signal to substitute $u=x^4$ first. After that, the structure $\tan^{-1}u \cdot \dfrac{du}{1+u^2}$ is exactly of the form "function times its own derivative," solved by a second substitution.

    Step-by-Step Solution

    1. Let $u=x^4$, so $du=4x^3\,dx \Rightarrow x^3\,dx = \dfrac{du}{4}$.
    2. The integral becomes $\displaystyle\int \frac{\tan^{-1}u}{1+u^2}\cdot\frac{du}{4} = \frac14\int \frac{\tan^{-1}u}{1+u^2}\,du$.
    3. Let $v=\tan^{-1}u$, so $dv = \dfrac{du}{1+u^2}$. The integral is $\dfrac14\int v\,dv = \dfrac14\cdot\dfrac{v^2}{2} = \dfrac{v^2}{8}$.
    4. Substituting back: $\dfrac{(\tan^{-1}(x^4))^2}{8}+c$.

    Common Mistakes

    • Forgetting the factor of $\frac14$ introduced by $du=4x^3dx$.
    • Mixing up $\int v\,dv = v^2/2$ with $v^3/3$ (confusing this with $\int v^2 dv$).

    ✓Final answer

    The correct option is (A) — $\dfrac{(\tan^{-1}(x^4))^2}{8} + c$.

    ANSWER: A

11. AP EAPCET 2023Set eng-2023-05-15-AN1 markMCQ
    Q.$\displaystyle\int \dfrac{dx}{4+5\cos x} =$ (A) $-\dfrac{1}{3}\log\left|\dfrac{3+\tan\frac{x}{2}}{3-\tan\frac{x}{2}}\right|+C$ (B) $\dfrac{1}{3}\log\left|\dfrac{3+\tan\frac{x}{2}}{3-\tan\frac{x}{2}}\right|+C$ (C) $-\dfrac{1}{9}\log\left|\dfrac{3-\tan\frac{x}{2}}{3+\tan\frac{x}{2}}\right|+C$ (D) $\dfrac{1}{9}\log\left|\dfrac{3-\tan\frac{x}{2}}{3+\tan\frac{x}{2}}\right|+C$

    ›Reveal solutionSolution

    The Weierstrass substitution $t=\tan(x/2)$ turns this into a standard $\int dt/(a^2-t^2)$ integral, giving option (B).

    Concept and Intuition

    For $\int\dfrac{dx}{a+b\cos x}$ the substitution $t=\tan(x/2)$ (so $\cos x=\tfrac{1-t^2}{1+t^2}$, $dx=\tfrac{2\,dt}{1+t^2}$) always converts the integral into a rational function of $t$ alone.

    Step-by-Step Solution

    1. $\cos x=\dfrac{1-t^2}{1+t^2}$, $dx=\dfrac{2\,dt}{1+t^2}$, with $t=\tan(x/2)$.
    2. $4+5\cos x=4+5\cdot\dfrac{1-t^2}{1+t^2}=\dfrac{4(1+t^2)+5(1-t^2)}{1+t^2}=\dfrac{9-t^2}{1+t^2}$.
    3. Integral becomes $\int\dfrac{2\,dt/(1+t^2)}{(9-t^2)/(1+t^2)}=\int\dfrac{2\,dt}{9-t^2}$.
    4. Using $\int\dfrac{dt}{a^2-t^2}=\dfrac{1}{2a}\log\left|\dfrac{a+t}{a-t}\right|+C$ with $a=3$: $\int\dfrac{dt}{9-t^2}=\dfrac16\log\left|\dfrac{3+t}{3-t}\right|+C$.
    5. Multiply by 2: $\dfrac13\log\left|\dfrac{3+t}{3-t}\right|+C$, i.e. $\dfrac13\log\left|\dfrac{3+\tan(x/2)}{3-\tan(x/2)}\right|+C$.

    Common Mistakes

    • Sign slip giving $-\dfrac13\log|\ldots|$ instead of $+\dfrac13$.
    • Using $\dfrac1{2a}$ instead of doubling it correctly after the substitution factor of 2.

    ✓Final answer

    The correct option is (B) — $\dfrac{1}{3}\log\left|\dfrac{3+\tan\frac{x}{2}}{3-\tan\frac{x}{2}}\right|+C$.

    ANSWER: B

12. AP EAPCET 2021Set eng-2021-08-23-AN1 markMCQ
    Q.Find the value of $k$ if $\displaystyle\int \cos^k(x)\sin(x)\,dx = \dfrac{-1}{4}\cos^4(x) + c$ (A) $4$ (B) $3$ (C) $2$ (D) $1$

    ›Reveal solutionSolution

    Differentiate the given antiderivative and match powers of $\cos x$ to find $k=3$.

    Concept and Intuition

    When an integral is given in the form $\int \cos^k(x)\sin(x)\,dx = F(x)+c$, the fastest way to find $k$ is not to integrate but to differentiate the claimed answer $F(x)$ — the derivative must reproduce the original integrand exactly.

    Step-by-Step Solution

    1. Differentiate $F(x) = -\dfrac14\cos^4x$:

    $F'(x) = -\dfrac14\cdot 4\cos^3x\cdot(-\sin x) = \cos^3x\sin x.$

    2. This must equal the integrand $\cos^k x\sin x$.
    3. Comparing powers of $\cos x$: $k=3$.
    4. Sanity check by direct integration: with $u=\cos x$, $du=-\sin x\,dx$, $\int \cos^3x\sin x\,dx = -\int u^3\,du = -\dfrac{u^4}{4}+c = -\dfrac14\cos^4x+c$. Matches.

    Common Mistakes

    • Trying to integrate $\cos^k x\sin x$ for a guessed $k$ instead of differentiating the given answer — differentiating is far faster here.
    • Forgetting the chain-rule factor of $4$ when differentiating $\cos^4x$.

    ✓Final answer

    The correct option is (B) — $3$.

    ANSWER: B