Q.Evaluate the integral using substitution $\int_{0}^{1}\sin^{-1}\left(\frac{2x}{1+x^2}\right)dx$

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 2026Set eng-2026-05-14-FN1 markMCQ
   Q.$\displaystyle\int \dfrac{\sin^2 x \cos^2 x}{\cos^6 x + \sin^6 x}\,dx =$ (A) $\dfrac{1}{2}\mathrm{Tan}^{-1}(\tan^2 x) + c$ (B) $\dfrac{1}{3}\mathrm{Tan}^{-1}(\tan^2 x) + c$ (C) $\dfrac{1}{3}\mathrm{Tan}^{-1}(\tan^3 x) + c$ (D) $\mathrm{Tan}^{-1}(\tan^3 x) + c$

   ›Reveal solutionSolution

   Dividing through by $\cos^6x$ to introduce $\tan x$, then a double substitution ($t=\tan x$, then $u=t^3$), reduces the integral to a standard arctangent form, giving $\dfrac13\mathrm{Tan}^{-1}(\tan^3x)+c$.

   Concept and Intuition

   When an integrand has $\sin$ and $\cos$ appearing only in even powers that can be grouped, dividing everything by the highest power of $\cos x$ converts the whole expression into a rational function purely of $\tan x$ — a very common trick that opens the door to the substitution $t=\tan x$. Here, since the denominator naturally forms $1+\tan^6x$, a further substitution $u=t^3$ turns it into the standard $\int\frac{du}{1+u^2}$ arctangent integral.

   Step-by-Step Solution

   1. Divide numerator and denominator of $\dfrac{\sin^2x\cos^2x}{\cos^6x+\sin^6x}$ by $\cos^6x$: numerator becomes $\dfrac{\sin^2x\cos^2x}{\cos^6x} = \dfrac{\sin^2x}{\cos^4x} = \tan^2x\sec^2x$; denominator becomes $1 + \tan^6x$.
   2. The integral is now $\displaystyle\int \dfrac{\tan^2x\sec^2x}{1+\tan^6x}\,dx$.
   3. Substitute $t=\tan x$, so $dt = \sec^2x\,dx$. The integral becomes $\displaystyle\int\dfrac{t^2}{1+t^6}\,dt$.
   4. Substitute $u = t^3$, so $du = 3t^2\,dt$, i.e. $t^2\,dt = \dfrac{du}{3}$. The integral becomes $\dfrac13\displaystyle\int\dfrac{du}{1+u^2}$.
   5. This is the standard arctangent integral: $\dfrac13\mathrm{Tan}^{-1}(u) + c$.
   6. Substitute back $u=t^3=\tan^3x$: the answer is $\dfrac13\mathrm{Tan}^{-1}(\tan^3x)+c$.

   Common Mistakes

   • Mis-simplifying $\dfrac{\sin^2x\cos^2x}{\cos^6x}$ (arithmetic slip in exponents), which changes the power of $\tan x$ obtained.
   • Forgetting the second substitution ($u=t^3$) and trying to directly integrate $\int\frac{t^2}{1+t^6}dt$ as though it were already a standard arctan form.

   ✓Final answer

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

   ANSWER: C

3. 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

4. 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

5. 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

6. AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQ
   Q.If $\displaystyle\int \dfrac{1}{\sin^3 x+\cos^3 x}\,dx = A\log\left|\dfrac{\sqrt{2}+t}{\sqrt{2}-t}\right|+B\,\mathrm{Tan}^{-1}(t)+c$, then $\left(\dfrac{B}{A},t\right)=$ (A) $(2\sqrt{2},\sin x+\cos x)$ (B) $\left(\dfrac{\sqrt{2}}{9},\sin x+\cos x\right)$ (C) $\left(\dfrac{\sqrt{2}}{9},\sin x-\cos x\right)$ (D) $(2\sqrt{2},\sin x-\cos x)$

   ›Reveal solutionSolution

   Factoring the sum of cubes and substituting $t=\sin x-\cos x$ turns the trigonometric integral into a clean rational-function integral in $t$, giving $B/A=2\sqrt2$ with $t=\sin x-\cos x$.

   Concept and Intuition

   $\sin^3x+\cos^3x$ factors as a sum of cubes: $(\sin x+\cos x)(1-\sin x\cos x)$. Both $\sin x+\cos x$ and $1-\sin x\cos x$ can be written purely in terms of $u=\sin x-\cos x$, because $(\sin x+\cos x)^2+(\sin x-\cos x)^2=2$ and $1-\sin x\cos x=\tfrac{1+(\sin x-\cos x)^2}{2}$. Crucially, $\dfrac{du}{dx}=\cos x+\sin x$, which is exactly the factor left over after using the second identity — so the whole integral collapses into a rational function of $u$ alone.

   Step-by-Step Solution

   1. Factor: $\sin^3x+\cos^3x=(\sin x+\cos x)(1-\sin x\cos x)$.
   2. Let $u=\sin x-\cos x$. Then $u^2=1-2\sin x\cos x$, so $1-\sin x\cos x=\dfrac{1+u^2}{2}$.
   3. Also $(\sin x+\cos x)^2=1+2\sin x\cos x=2-u^2$, so $\sin x+\cos x=\sqrt{2-u^2}$ (taking the appropriate branch), and $\dfrac{du}{dx}=\cos x+\sin x=\sqrt{2-u^2}$.
   4. So $\sin^3x+\cos^3x=\sqrt{2-u^2}\cdot\dfrac{1+u^2}{2}$, and

   $I=\int\frac{dx}{\sin^3x+\cos^3x}=\int\frac{2\,dx}{\sqrt{2-u^2}(1+u^2)}=\int\frac{2}{\sqrt{2-u^2}(1+u^2)}\cdot\frac{du}{\sqrt{2-u^2}}=\int\frac{2\,du}{(2-u^2)(1+u^2)}.$

   5. Partial fractions (by symmetry, only even terms survive): $\dfrac{2}{(2-u^2)(1+u^2)}=\dfrac{2/3}{2-u^2}+\dfrac{2/3}{1+u^2}$.
   6. Integrate: $\displaystyle\int\frac{du}{2-u^2}=\frac{1}{2\sqrt2}\log\left|\frac{\sqrt2+u}{\sqrt2-u}\right|$ and $\int\frac{du}{1+u^2}=\mathrm{Tan}^{-1}u$.
   7. So $I=\dfrac{1}{3\sqrt2}\log\left|\dfrac{\sqrt2+u}{\sqrt2-u}\right|+\dfrac{2}{3}\mathrm{Tan}^{-1}u+c$, matching the given form with $A=\dfrac{1}{3\sqrt2}$, $B=\dfrac{2}{3}$, $t=u=\sin x-\cos x$.
   8. Hence $\dfrac{B}{A}=\dfrac{2/3}{1/(3\sqrt2)}=2\sqrt2$.

   Common Mistakes

   • Trying $t=\sin x+\cos x$ directly — it doesn't decouple the $\sqrt{2-t^2}$ factor from the rest as cleanly as $t=\sin x-\cos x$ does here.
   • Sign slip when computing $(\sin x+\cos x)^2$ vs $(\sin x-\cos x)^2$ and which one equals $1\pm2\sin x\cos x$.

   ✓Final answer

   The correct option is (D) — $(2\sqrt{2},\sin x-\cos x)$.

   ANSWER: D

7. AP EAPCET 2025Set eng-2025-05-21-FN1 markMCQ
   Q.If $\displaystyle\int \dfrac{dx}{\sin^3 x+\cos^3 x} = A\log\left|\dfrac{\sqrt{2}+t}{\sqrt{2}-t}\right|+B\mathrm{Tan}^{-1}(t)+c$, then $\left(\dfrac{B}{A},t\right) =$ (A) $(3\sqrt{2},\sin x-\cos x)$ (B) $(2\sqrt{2},\sin x-\cos x)$ (C) $\left(\dfrac{\sqrt{2}}{3},\sin x-\cos x\right)$ (D) $\left(\dfrac{3}{\sqrt{2}},\sin x+\cos x\right)$

   ›Reveal solutionSolution

   The standard $\sin^3x+\cos^3x$ integral, solved via the substitution $t=\sin x-\cos x$; matching to the given $A\log|\cdots|+B\tan^{-1}t$ form gives $B/A=2\sqrt2$ with $t=\sin x-\cos x$.

   Concept and Intuition

   $\sin^3x+\cos^3x$ factors as a sum of cubes, $(\sin x+\cos x)(\sin^2x-\sin x\cos x+\cos^2x)=(\sin x+\cos x)(1-\sin x\cos x)$. Both remaining factors can be expressed in terms of $t=\sin x-\cos x$ (since $t^2=1-2\sin x\cos x$ links $\sin x\cos x$ to $t$, and $dt=(\sin x+\cos x)\,dx$ conveniently cancels the other factor).

   Step-by-Step Solution

   1. Factor: $\sin^3x+\cos^3x = (\sin x+\cos x)(1-\sin x\cos x)$.
   2. Let $t=\sin x-\cos x \Rightarrow dt = (\cos x+\sin x)\,dx$, and $t^2 = 1-2\sin x\cos x \Rightarrow \sin x\cos x = \dfrac{1-t^2}{2}$.
   3. So $1-\sin x\cos x = 1-\dfrac{1-t^2}{2} = \dfrac{1+t^2}{2}$.
   4. Also $(\sin x+\cos x)^2 = 1+2\sin x\cos x = 1+(1-t^2) = 2-t^2$, so $\sin x+\cos x=\sqrt{2-t^2}$.
   5. Rewrite the integral:

   $\int\frac{dx}{(\sin x+\cos x)(1-\sin x\cos x)} = \int\frac{dx}{\sqrt{2-t^2}\cdot\frac{1+t^2}{2}}.$

   Since $dx = \dfrac{dt}{\sin x+\cos x} = \dfrac{dt}{\sqrt{2-t^2}}$:

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

   6. Split using $1=\dfrac{(2-t^2)+(1+t^2)}{3}$:

   $\frac{2}{(2-t^2)(1+t^2)} = \frac23\cdot\frac{(2-t^2)+(1+t^2)}{(2-t^2)(1+t^2)} = \frac23\left[\frac{1}{1+t^2}+\frac{1}{2-t^2}\right].$

   7. Integrate each piece: $\displaystyle\int\frac{dt}{1+t^2}=\tan^{-1}t$; $\displaystyle\int\frac{dt}{2-t^2}=\frac{1}{2\sqrt2}\log\left|\frac{\sqrt2+t}{\sqrt2-t}\right|$ (standard form $\int\frac{dx}{a^2-x^2}=\frac{1}{2a}\log\left|\frac{a+x}{a-x}\right|$ with $a=\sqrt2$).
   8. So the integral is

   $\frac23\tan^{-1}t + \frac23\cdot\frac{1}{2\sqrt2}\log\left|\frac{\sqrt2+t}{\sqrt2-t}\right|+c = \frac{1}{3\sqrt2}\log\left|\frac{\sqrt2+t}{\sqrt2-t}\right| + \frac23\tan^{-1}t + c.$

   9. Matching to $A\log\left|\frac{\sqrt2+t}{\sqrt2-t}\right|+B\tan^{-1}t+c$: $A=\dfrac{1}{3\sqrt2}$, $B=\dfrac23$.
   10. $\dfrac{B}{A} = \dfrac{2/3}{1/(3\sqrt2)} = \dfrac23\times3\sqrt2 = 2\sqrt2$, and $t=\sin x-\cos x$.

   Common Mistakes

   • Using $t=\sin x+\cos x$ instead of $\sin x-\cos x$ (the sign matters — it's the difference that makes $dt$ match the other factor cleanly here).
   • Errors in the standard log-form constant $\dfrac{1}{2a}$ with $a=\sqrt2$, or forgetting the extra $\frac23$ scaling from the partial-fraction split.

   ✓Final answer

   The correct option is (B) — $(2\sqrt2,\ \sin x-\cos x)$.

   ANSWER: B

8. AP EAPCET 2025Set eng-2025-05-22-AN1 markMCQ
   Q.If $\displaystyle\int \dfrac{\cos^3 x}{\sin^2 x + \sin^4 x}dx = c - \mathrm{cosec}\,x - f(x)$, then $f\left(\dfrac{\pi}{2}\right)=$ (A) 1 (B) 0 (C) $\dfrac{\pi}{2}$ (D) $\pi$

   ›Reveal solutionSolution

   Substituting $s=\sin x$ and partial-fractioning gives $-\mathrm{cosec}\,x-2\mathrm{Tan}^{-1}(\sin x)$, so $f(x)=2\mathrm{Tan}^{-1}(\sin x)$ and $f(\pi/2)=\pi/2$.

   Concept and Intuition

   Writing $\cos^3x\,dx = \cos^2x\cdot\cos x\,dx=(1-\sin^2x)\,d(\sin x)$ converts a trig integral into an algebraic one in $s=\sin x$, which is then handled by ordinary partial fractions.

   Step-by-Step Solution

   1. Let $s=\sin x$, so $ds=\cos x\,dx$, and $\cos^3x\,dx=(1-s^2)\,ds$.
   2. The integral becomes $\displaystyle\int\frac{1-s^2}{s^2+s^4}\,ds=\int\frac{1-s^2}{s^2(1+s^2)}\,ds$.
   3. Partial fractions (in $u=s^2$): $\dfrac{1-u}{u(1+u)}=\dfrac1u-\dfrac2{1+u}$, so the integrand is $\dfrac1{s^2}-\dfrac2{1+s^2}$.
   4. Integrating: $\displaystyle\int\left(\frac1{s^2}-\frac2{1+s^2}\right)ds=-\frac1s-2\,\mathrm{Tan}^{-1}s+C=-\mathrm{cosec}\,x-2\,\mathrm{Tan}^{-1}(\sin x)+C$.
   5. Comparing with $c-\mathrm{cosec}\,x-f(x)$: $f(x)=2\,\mathrm{Tan}^{-1}(\sin x)$.
   6. $f(\pi/2)=2\,\mathrm{Tan}^{-1}(\sin(\pi/2))=2\,\mathrm{Tan}^{-1}(1)=2\cdot\dfrac\pi4=\dfrac\pi2$.

   Common Mistakes

   • Sign slip when matching to the given form $c-\mathrm{cosec}\,x-f(x)$ (easy to flip the sign of $f$).
   • Forgetting $\mathrm{Tan}^{-1}(1)=\pi/4$, not $\pi/2$.

   ✓Final answer

   The correct option is (C) — $\dfrac{\pi}{2}$.

   ANSWER: C

9. 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

10. 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

11. 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

12. AP EAPCET 2021Set eng-2021-08-23-AN1 markMCQ
    Q.$\displaystyle\int_{2/e}^{1/e} \dfrac{1}{x(\log x)^{1/3}}\,dx =$ (A) $\dfrac{3}{2}\{1 + (\log(2) - 1)^{2/3}\}$ (B) $1$ (C) $\dfrac{3}{2}\{1 + (\log(2) + 1)^{3/2}\}$ (D) $\dfrac{3}{2}\{1 - (\log(2) - 1)^{2/3}\}$

    ›Reveal solutionSolution

    Substitute $u=\log x$ to turn the integral into $\int u^{-1/3}du$, then evaluate between the transformed limits $u=\log2-1$ and $u=-1$.

    Concept and Intuition

    An integrand of the form $\dfrac{1}{x\cdot g(\log x)}$ always calls for the substitution $u=\log x$, since $du = dx/x$ removes the $x$ and $1/x$ entirely, leaving a pure power of $u$.

    Step-by-Step Solution

    1. Let $u=\log x$, so $du = dx/x$.
    2. Limits: at $x=2/e$, $u=\log(2/e)=\log2-1$. At $x=1/e$, $u=\log(1/e)=-1$.
    3. The integral becomes $\displaystyle\int_{\log2-1}^{-1} u^{-1/3}\,du = \left[\frac{u^{2/3}}{2/3}\right]_{\log2-1}^{-1} = \frac32\Big[u^{2/3}\Big]_{\log2-1}^{-1}$.
    4. Using the real cube root, $(-1)^{2/3} = \big((-1)^{1/3}\big)^2 = (-1)^2 = 1$.
    5. So the value is $\dfrac32\big[1-(\log2-1)^{2/3}\big]$.

    Common Mistakes

    • Swapping the order of the transformed limits (the lower $x$-limit $2/e$ gives the LOWER $u$-value, which must stay as the lower limit of the $u$-integral).
    • Treating $(-1)^{2/3}$ as undefined instead of using the real cube-root convention, which gives $1$.

    ✓Final answer

    The correct option is (D) — $\dfrac{3}{2}\{1-(\log(2)-1)^{2/3}\}$.

    ANSWER: D