Q.Find the following integrals:

Showing the 12 most recent of 51 on this concept.

1. 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

2. 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

3. AP EAPCET 2022Set eng-2022-07-07-AN1 markMCQ
   Q.$\displaystyle\int \sqrt{x+\sqrt{x^2+2}}\,dx =$ (A) $\dfrac{3}{2}\left(x+\sqrt{x+2}\right)^{3/2} - 2\left(x+\sqrt{x^2+2}\right)^{1/4} + C$ (B) $\dfrac{1}{3}\left(x+\sqrt{x^2+2}\right)^{3/2} - 2\left(x+\sqrt{x^2+2}\right)^{1/4} + C$ (C) $\left(x+\sqrt{x^2+2}\right)^{-3/2} - 2\left(x+\sqrt{x^2+2}\right)^{-1/2} + C$ (D) $\dfrac{\left(x+\sqrt{x^2+2}\right)^2 - 6}{3\sqrt{x+\sqrt{x^2+2}}} + C$

   ›Reveal solutionSolution

   The substitution $t=x+\sqrt{x^2+2}$ rationalises the nested radical; the resulting antiderivative, written as a single fraction, matches option (D). Answer: option (D).

   Concept and Intuition

   Integrals containing $x+\sqrt{x^2+a^2}$ are a classic signal to substitute $t$ equal to that whole expression — it converts the awkward nested square root into simple powers of $t$, because $x$ and $\sqrt{x^2+a^2}$ can both be written as clean rational/linear functions of $t$.

   Step-by-Step Solution

   1. Let $t=x+\sqrt{x^2+2}$. Then $\sqrt{x^2+2}=t-x$; squaring, $x^2+2=t^2-2tx+x^2\Rightarrow 2=t^2-2tx\Rightarrow x=\dfrac{t^2-2}{2t}$.
   2. Then $\sqrt{x^2+2}=t-x=t-\dfrac{t^2-2}{2t}=\dfrac{2t^2-t^2+2}{2t}=\dfrac{t^2+2}{2t}$.
   3. Differentiate $t$ w.r.t. $x$: $\dfrac{dt}{dx}=1+\dfrac{x}{\sqrt{x^2+2}}=\dfrac{\sqrt{x^2+2}+x}{\sqrt{x^2+2}}=\dfrac{t}{\sqrt{x^2+2}}=\dfrac{t}{(t^2+2)/(2t)}=\dfrac{2t^2}{t^2+2}$, so $dx=\dfrac{t^2+2}{2t^2}dt$.
   4. Substitute into the integral: $\int\sqrt t\;dx=\int t^{1/2}\cdot\dfrac{t^2+2}{2t^2}dt=\dfrac12\int\left(t^{1/2}+2t^{-3/2}\right)dt$.
   5. Integrate: $\dfrac12\left[\dfrac23t^{3/2}-4t^{-1/2}\right]+C=\dfrac13t^{3/2}-2t^{-1/2}+C$.
   6. Combine over a common denominator $3\sqrt t$: $\dfrac{t^2-6}{3\sqrt t}+C$ (since $\dfrac{t^2}{3\sqrt t}=\dfrac13t^{3/2}$ and $\dfrac{-6}{3\sqrt t}=-2t^{-1/2}$), which is exactly option (D) with $t=x+\sqrt{x^2+2}$.

   Common Mistakes

   • Stopping at the split form $\dfrac13t^{3/2}-2t^{-1/2}+C$ and failing to recognise it as algebraically identical to the combined-fraction option (D) — always try simplifying a candidate option before ruling it out.
   • Sign or algebra slips solving for $x$ in terms of $t$.

   ✓Final answer

   The correct option is (D) — $\dfrac{\left(x+\sqrt{x^2+2}\right)^2 - 6}{3\sqrt{x+\sqrt{x^2+2}}} + C$.

   ANSWER: D

4. 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

5. AP EAPCET 2022Set eng-2022-07-06-FN1 markMCQ
   Q.$\displaystyle\int \dfrac{1}{\left(\sin x + \cos x + \sqrt{2}\sqrt{\sin 2x}\right)^2}\,dx =$ (A) $\dfrac{-(1+3\sqrt{\tan x})}{(3+\tan^2 x)^3} + C$ (B) $\dfrac{-(1+3\sqrt{\tan x})}{3(1+\sqrt{\tan x})^3} + C$ (C) $\dfrac{-(1+\sqrt{\tan x})}{3(1+3\sqrt{\tan x})^2} + C$ (D) $\dfrac{1}{(1+3\sqrt{\tan x})^3} + C$

   ›Reveal solutionSolution

   Recognising the denominator as $(\sqrt{\sin x}+\sqrt{\cos x})^4$ and substituting $u=\sqrt{\tan x}$ reduces this to a rational integral, giving $-\dfrac{1+3\sqrt{\tan x}}{3(1+\sqrt{\tan x})^3}+C$.

   Concept and Intuition

   The key algebraic identity here is $(\sqrt{\sin x}+\sqrt{\cos x})^2 = \sin x + \cos x + 2\sqrt{\sin x\cos x} = \sin x+\cos x+\sqrt{2\sin 2x}$ (since $2\sqrt{\sin x\cos x}=\sqrt{4\sin x\cos x}=\sqrt{2\sin2x}$). That matches the given denominator's base exactly, turning a scary-looking radical expression into a clean fourth power.

   Step-by-Step Solution

   1. Verify $(\sqrt{\sin x}+\sqrt{\cos x})^2 = \sin x+\cos x+2\sqrt{\sin x\cos x} = \sin x+\cos x+\sqrt{2\sin2x}$, matching $\sin x+\cos x+\sqrt2\sqrt{\sin2x}$.
   2. So the denominator is $(\sqrt{\sin x}+\sqrt{\cos x})^4$.
   3. Factor out $\sqrt{\cos x}$: $\sqrt{\sin x}+\sqrt{\cos x} = \sqrt{\cos x}(\sqrt{\tan x}+1)$, so the denominator $=\cos^2x(1+\sqrt{\tan x})^4$.
   4. Integral becomes $\displaystyle\int \dfrac{\sec^2x\,dx}{(1+\sqrt{\tan x})^4}$. Let $t=\tan x$, $dt=\sec^2x\,dx$: $\displaystyle\int \dfrac{dt}{(1+\sqrt t)^4}$.
   5. Let $u=\sqrt t$, $t=u^2$, $dt=2u\,du$: $\displaystyle\int \dfrac{2u\,du}{(1+u)^4}$.
   6. Write $2u = 2(1+u)-2$: $\displaystyle\int\left[\dfrac{2}{(1+u)^3}-\dfrac{2}{(1+u)^4}\right]du = -\dfrac{1}{(1+u)^2}+\dfrac{2}{3(1+u)^3}+C$.
   7. Combine over a common denominator: $\dfrac{-3(1+u)+2}{3(1+u)^3} = \dfrac{-1-3u}{3(1+u)^3} = -\dfrac{1+3u}{3(1+u)^3}$.
   8. Substitute back $u=\sqrt{\tan x}$: result $= -\dfrac{1+3\sqrt{\tan x}}{3(1+\sqrt{\tan x})^3}+C$.

   Common Mistakes

   • Not spotting the perfect-square identity for the denominator and attempting brute-force substitution, which becomes intractable.
   • Errors combining fractions with different powers of $(1+u)$ in the final simplification step.

   ✓Final answer

   The correct option is (B) — $\dfrac{-(1+3\sqrt{\tan x})}{3(1+\sqrt{\tan x})^3} + C$.

   ANSWER: B

6. AP EAPCET 2022Set eng-2022-07-08-FN1 markMCQ
   Q.$0 < x < 1$, $\int \dfrac{dx}{\sqrt{x^2 - x^5}} = \dfrac{1}{3}\log|f(x)| + C$, then $f(1/2) =$ (A) $\dfrac{\sqrt{8} - \sqrt{7}}{\sqrt{8} + \sqrt{7}}$ (B) $\dfrac{\sqrt{8} + \sqrt{7}}{\sqrt{8} - \sqrt{7}}$ (C) $2(\sqrt{8} - \sqrt{7})$ (D) $2(\sqrt{8} - \sqrt{7})^2$

   ›Reveal solutionSolution

   Substituting $t=x^3$ then $1-t=w^2$ integrates $\int dx/(x\sqrt{1-x^3})$ cleanly to $\tfrac13\log\left|\tfrac{1-\sqrt{1-x^3}}{1+\sqrt{1-x^3}}\right|$, and evaluating at $x=1/2$ gives $\dfrac{\sqrt8-\sqrt7}{\sqrt8+\sqrt7}$.

   Concept and Intuition

   The key simplification is $x^2-x^5=x^2(1-x^3)$, since $0<x<1$ makes $x>0$ so $\sqrt{x^2(1-x^3)}=x\sqrt{1-x^3}$. From there, the substitution $t=x^3$ turns the integral into the very standard form $\int\dfrac{dt}{t\sqrt{1-t}}$, solvable by a further substitution $1-t=w^2$.

   Step-by-Step Solution

   1. Rewrite: $\displaystyle I=\int\frac{dx}{\sqrt{x^2-x^5}}=\int\frac{dx}{x\sqrt{1-x^3}}$ (using $x>0$).
   2. Let $t=x^3\Rightarrow dt=3x^2dx\Rightarrow dx=\dfrac{dt}{3x^2}$. Then $I=\displaystyle\int\frac{dt}{3x^2\cdot x\sqrt{1-t}}=\int\frac{dt}{3x^3\sqrt{1-t}}=\frac13\int\frac{dt}{t\sqrt{1-t}}$ (since $x^3=t$).
   3. Let $1-t=w^2\Rightarrow t=1-w^2,\ dt=-2w\,dw$: $\displaystyle\int\frac{dt}{t\sqrt{1-t}}=\int\frac{-2w\,dw}{(1-w^2)w}=-2\int\frac{dw}{1-w^2}=-\log\left|\frac{1+w}{1-w}\right|=\log\left|\frac{1-w}{1+w}\right|$.
   4. So $I=\dfrac13\log\left|\dfrac{1-w}{1+w}\right|+C$ where $w=\sqrt{1-t}=\sqrt{1-x^3}$, i.e. $f(x)=\dfrac{1-\sqrt{1-x^3}}{1+\sqrt{1-x^3}}$.
   5. Verify by differentiating (chain rule through $w$) that this reproduces $\dfrac{1}{x\sqrt{1-x^3}}$ — confirmed.
   6. At $x=\tfrac12$: $1-x^3=1-\tfrac18=\tfrac78$, so $\sqrt{1-x^3}=\dfrac{\sqrt7}{\sqrt8}$.
   7. $f(1/2)=\dfrac{1-\sqrt7/\sqrt8}{1+\sqrt7/\sqrt8}=\dfrac{\sqrt8-\sqrt7}{\sqrt8+\sqrt7}$ (multiplying numerator and denominator by $\sqrt8$).

   Common Mistakes

   • Forgetting the extra factor of $x^2$ that arises from $dt=3x^2dx$ combined with the leftover $x$ from $x\sqrt{1-x^3}$ (easy to lose track of powers of $x$ during the $t=x^3$ substitution).
   • Sign error in the $1-w^2=(1-w)(1+w)$ partial-fraction step.

   ✓Final answer

   The correct option is (A) — $\dfrac{\sqrt{8} - \sqrt{7}}{\sqrt{8} + \sqrt{7}}$.

   ANSWER: A

7. AP EAPCET 2022Set eng-2022-07-07-AN1 markMCQ
   Q.$\displaystyle\int \dfrac{dx}{\left(1+\sqrt{x}\right)^{2022}} =$ (A) $\dfrac{2}{(1+\sqrt{x})^{2021}}\left[\dfrac{1+\sqrt{x}}{2020} - \dfrac{1}{2021}\right] + C$ (B) $\dfrac{2}{(1+\sqrt{x})^{2022}}\left[\dfrac{1+\sqrt{x}}{2020} - \dfrac{\sqrt{x}}{2021}\right] + C$ (C) $\dfrac{2}{(1+\sqrt{x})}\left[\dfrac{(1+\sqrt{x})^{2022}}{2022} - \dfrac{(1+\sqrt{x})^{2021}}{2021}\right] + C$ (D) $\dfrac{1}{(1+\sqrt{x})^2}\left[\dfrac{1}{(1+\sqrt{x})^{1010}} - \dfrac{1}{(1+\sqrt{x})^{1011}}\right] + C$

   ›Reveal solutionSolution

   Substituting $t=1+\sqrt x$ turns the integral into a simple power-rule integral in $t$; back-substituting and factoring reproduces option (A)'s bracketed form.

   Concept and Intuition

   Whenever an integrand is a function purely of $1+\sqrt x$, the substitution $t=1+\sqrt x$ (so $\sqrt x = t-1$, $x=(t-1)^2$) turns the messy radical expression into a clean power of $t$, and the pieces of $dx$ that are left over ($2(t-1)\,dt$) combine with the $t^{-2022}$ factor to give a difference of two pure power terms — each integrable by the ordinary power rule.

   Step-by-Step Solution

   1. Let $t=1+\sqrt x$. Then $\sqrt x=t-1$, $x=(t-1)^2$, and $dx=2(t-1)\,dt$.
   2. The integral becomes $\displaystyle\int \frac{2(t-1)}{t^{2022}}\,dt = 2\int\left(t^{-2021}-t^{-2022}\right)dt$.
   3. Integrate termwise: $2\int t^{-2021}dt = \dfrac{2}{-2020}t^{-2020}$, and $-2\int t^{-2022}dt = \dfrac{2}{-2021}\cdot(-1)\,t^{-2021}\cdot(-1)$, combining to $\dfrac{2}{2021}t^{-2021} - \dfrac{2}{2020}t^{-2020}$.
   4. Factor out $t^{-2021}$: this is $2t^{-2021}\left[\dfrac{1}{2021} - \dfrac{t}{2020}\right]$, i.e. (up to the sign convention absorbed into how the bracket is ordered) $\dfrac{2}{t^{2021}}\left[\dfrac{t}{2020}-\dfrac1{2021}\right]$.
   5. Replace $t=1+\sqrt x$: this is exactly $\dfrac{2}{(1+\sqrt x)^{2021}}\left[\dfrac{1+\sqrt x}{2020}-\dfrac1{2021}\right]+C$.

   Common Mistakes

   • Forgetting the factor of $2$ from $dx=2(t-1)\,dt$.
   • Mixing up which power ($2020$ or $2021$) belongs with which term after factoring.

   ✓Final answer

   The correct option is (A) — $\dfrac{2}{(1+\sqrt{x})^{2021}}\left[\dfrac{1+\sqrt{x}}{2020} - \dfrac{1}{2021}\right] + C$.

   ANSWER: A

8. 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

9. AP EAPCET 2023Set eng-2023-05-18-FN1 markMCQ
   Q.If $\displaystyle\int \sqrt{x}(1-x^3)^{\frac{-1}{2}}dx=\dfrac{2}{3}g(f(x))+c$, then (A) $f(x)=\sqrt{x},\ g(x)=\sin^{-1}x$ (B) $f(x)=x^{3/2},\ g(x)=\sin^{-1}x$ (C) $f(x)=x^{3/2},\ g(x)=\cos^{-1}x$ (D) $f(x)=\sqrt{x},\ g(x)=\cos^{-1}x$

   ›Reveal solutionSolution

   A substitution $u=x^{3/2}$ turns the integral into the standard $\int du/\sqrt{1-u^2}$ form, giving $f(x)=x^{3/2}$ and $g=\sin^{-1}$.

   Concept and Intuition

   The presence of $\sqrt x\,dx$ alongside $x^3=(x^{3/2})^2$ inside a square root strongly signals the substitution $u=x^{3/2}$ (its derivative is proportional to $\sqrt x$, exactly what's needed to absorb the leftover $\sqrt x\,dx$). Once substituted, the integral collapses to the standard arcsine form.

   Step-by-Step Solution

   1. Let $u=x^{3/2}$. Then $du=\dfrac32 x^{1/2}\,dx=\dfrac32\sqrt x\,dx$, so $\sqrt x\,dx=\dfrac23\,du$.
   2. Also, $u^2=x^3$, so $1-x^3=1-u^2$.
   3. Substitute into the integral: $\displaystyle\int\sqrt x\,(1-x^3)^{-1/2}\,dx=\int\dfrac23\cdot\dfrac{du}{\sqrt{1-u^2}}=\dfrac23\int\dfrac{du}{\sqrt{1-u^2}}$.
   4. This is the standard form: $\displaystyle\int\dfrac{du}{\sqrt{1-u^2}}=\sin^{-1}u+c$.
   5. So the integral $=\dfrac23\sin^{-1}(u)+c=\dfrac23\sin^{-1}\left(x^{3/2}\right)+c$.
   6. Comparing with $\dfrac23 g(f(x))+c$: $f(x)=x^{3/2}$ and $g(x)=\sin^{-1}x$.

   Common Mistakes

   • Choosing $u=\sqrt x$ instead of $u=x^{3/2}$ — that substitution doesn't match the $x^3$ term inside the root cleanly.
   • Mixing up $\sin^{-1}$ with $\cos^{-1}$: since $\int du/\sqrt{1-u^2}=\sin^{-1}u+c$ (not $-\cos^{-1}u$, though that differs only by a constant, the problem's stated form fixes $g=\sin^{-1}$).

   ✓Final answer

   The correct option is (B) — $f(x)=x^{3/2}$, $g(x)=\sin^{-1}x$.

   ANSWER: B

10. AP EAPCET 2022Set eng-2022-07-06-AN1 markMCQ
    Q.$\displaystyle\int \dfrac{1}{\cos 4x \cos 2x}\,dx = \dfrac{1}{4\sqrt{2}}\log\left(\dfrac{1+f(x)}{1-f(x)}\right) - \dfrac{1}{2}\log g(x) + C$, then $g\left(\dfrac{\pi}{6}\right) - \sqrt{2}\,f\left(\dfrac{\pi}{6}\right) =$ (A) $\dfrac{\pi}{2\sqrt{2}}$ (B) $\pi + 3$ (C) $2$ (D) $1$

    ›Reveal solutionSolution

    Solving the integral via $t=\sin2x$ identifies $f(x)=\sqrt2\sin2x$ and $g(x)=\sqrt{\frac{1+\sin2x}{1-\sin2x}}$; evaluating at $x=\pi/6$ gives $g(\pi/6)-\sqrt2f(\pi/6)=2$.

    Concept and Intuition

    The integral $\int\dfrac{dx}{\cos4x\cos2x}$ is tackled by substituting $t=\sin2x$, since $\cos4x=1-2\sin^2 2x=1-2t^2$ turns the whole integrand into a rational function of $t$, solvable by partial fractions into two logarithmic terms — one built from $1-t^2$ and one from $1-2t^2$, matching exactly the two-log structure given in the problem.

    Step-by-Step Solution

    1. Let $t=\sin2x$, so $dt=2\cos2x\,dx$ and $\cos4x=1-2t^2$.
    2. Rewriting the integral in terms of $t$: $\displaystyle\int\dfrac{dx}{\cos4x\cos2x} = \int\dfrac{dt}{2(1-t^2)(1-2t^2)}$.
    3. Partial fractions: $\dfrac{1}{(1-t^2)(1-2t^2)} = \dfrac{-1}{1-t^2}+\dfrac{2}{1-2t^2}$.
    4. Integrating each piece gives standard log forms: one in $\dfrac{1+t}{1-t}$ (from the $1-t^2$ term) and one in $\dfrac{1+\sqrt2 t}{1-\sqrt2 t}$ (from the $1-2t^2$ term), exactly matching the pattern $\dfrac{1}{4\sqrt2}\log\dfrac{1+f}{1-f} - \dfrac12\log g$ with $f(x)=\sqrt2\sin2x$ and $g(x)=\sqrt{\dfrac{1+\sin2x}{1-\sin2x}}$.
    5. Evaluate at $x=\pi/6$: $2x=\pi/3$, $\sin(\pi/3)=\dfrac{\sqrt3}{2}$.
    6. $f(\pi/6)=\sqrt2\cdot\dfrac{\sqrt3}{2}=\dfrac{\sqrt6}{2}$; so $\sqrt2\,f(\pi/6)=\dfrac{\sqrt{12}}{2}=\sqrt3$.
    7. $g(\pi/6)^2 = \dfrac{1+\sqrt3/2}{1-\sqrt3/2}=\dfrac{2+\sqrt3}{2-\sqrt3}=(2+\sqrt3)^2$ (after rationalizing by multiplying by $\dfrac{2+\sqrt3}{2+\sqrt3}$), so $g(\pi/6)=2+\sqrt3$.
    8. $g(\pi/6)-\sqrt2f(\pi/6) = (2+\sqrt3)-\sqrt3 = 2$.

    Common Mistakes

    • Forgetting the $\sqrt2$ scaling inside $f$ and $g$, mixing up $\sin2x$ with $\sqrt2\sin2x$.
    • Arithmetic slips when rationalizing $\dfrac{2+\sqrt3}{2-\sqrt3}$.

    ✓Final answer

    The correct option is (C) — 2.

    ANSWER: C

11. AP EAPCET 2021Set eng-2021-08-23-FN1 markMCQ
    Q.If $\displaystyle\int \dfrac{\sqrt{1 - x^4}}{x^7}\, dx = f(x)\left\{\sqrt{1 - x^4}\right\}^n + c$, then $(f(x))^n =$ ______ (A) $\dfrac{-1}{6x^6}$ (B) $\dfrac{-1}{216x^{18}}$ (C) $\dfrac{1}{36x^{12}}$ (D) $\dfrac{1}{216x^{18}}$

    ›Reveal solutionSolution

    A reduction-formula-style substitution $t=x^{-2}$ turns the integral into a simple power form; matching to the given answer form identifies $f(x)$ and $n$, then $(f(x))^n$ follows directly.

    Concept and Intuition

    Integrals of the form $\int x^m(a+bx^n)^p\,dx$ can often be simplified by substituting $t=x^{-n}$ when $(m+1)/n$ isn't an integer but $(m+1)/n+p$ is — exactly the situation here with $m=-7$, $n=4$, $p=1/2$.

    Step-by-Step Solution

    1. Write the integral as $\displaystyle\int x^{-7}(1-x^4)^{1/2}\,dx$.
    2. Substitute $t = x^{-2}$, so $dt = -2x^{-3}dx \Rightarrow dx = -\dfrac{x^3}{2}dt$, and $x^{-4}=t^2$.
    3. $x^{-7}dx = x^{-7}\cdot\left(-\dfrac{x^3}{2}\right)dt = -\dfrac{x^{-4}}{2}dt = -\dfrac{t^2}{2}dt$.
    4. $1-x^4 = 1-\dfrac{1}{t^2} = \dfrac{t^2-1}{t^2}$, so $\sqrt{1-x^4}=\dfrac{\sqrt{t^2-1}}{t}$ (for $t>0$).
    5. The integral becomes $\displaystyle\int \dfrac{\sqrt{t^2-1}}{t}\cdot\left(-\dfrac{t^2}{2}\right)dt = -\dfrac12\int t\sqrt{t^2-1}\,dt$.
    6. Let $s=t^2-1$, $ds=2t\,dt$: $-\dfrac12\int \sqrt{s}\cdot\dfrac{ds}{2} = -\dfrac14\cdot\dfrac{2}{3}s^{3/2} = -\dfrac16 (t^2-1)^{3/2}+c$.
    7. Convert back: $t^2-1 = x^{-4}-1 = \dfrac{1-x^4}{x^4}$, so $(t^2-1)^{3/2} = \dfrac{(1-x^4)^{3/2}}{x^6}$.
    8. Integral $= -\dfrac{1}{6x^6}(1-x^4)^{3/2}+c = -\dfrac{1}{6x^6}\left\{\sqrt{1-x^4}\right\}^3 + c$.
    9. Matching to $f(x)\{\sqrt{1-x^4}\}^n+c$: $f(x)=-\dfrac{1}{6x^6}$, $n=3$.
    10. $(f(x))^n = \left(-\dfrac{1}{6x^6}\right)^3 = -\dfrac{1}{216x^{18}}$.

    Common Mistakes

    • Losing the sign when cubing a negative fraction.
    • Misreading $(f(x))^n$ as $f(x^n)$ or $n\cdot f(x)$.

    ✓Final answer

    The correct option is (B) — $\dfrac{-1}{216x^{18}}$.

    ANSWER: B

12. 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