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 32 on this concept.

1. KEAM 2025Set eng-2025-04264 marksMCQ
   Q.$\displaystyle\int_0^{\pi/2} \dfrac{1}{1+\sin x}\, dx =$ (A) $2$ (B) $\dfrac{1}{2}$ (C) $\dfrac{1}{4}$ (D) $1$ (E) $0$

   ›Reveal solutionSolution

   Rationalise the denominator, integrate $\sec^2x-\sec x\tan x$.

   Multiply numerator and denominator by $1-\sin x$:

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

   Antiderivative is $\tan x-\sec x=\dfrac{\sin x-1}{\cos x}$. Evaluating from $0$ to $\tfrac{\pi}{2}$: at $x\to\tfrac{\pi}{2}$ the value $\to 0$; at $x=0$ it is $0-1=-1$. Thus the integral $=0-(-1)=1$.

   ✓Final answer

   The correct option is (D).

2. KEAM 2026Set eng-2026-04194 marksMCQ
   Q.$\int \frac{\sin(\cot^{-1}x)}{1 + x^2} dx$ is equal to (A) $-\cos(\cot^{-1}x) + C$ (B) $\cos(\cot^{-1}x) + C$ (C) $\frac{\cos(\cot^{-1}x)}{1 + x^2} + C$ (D) $\frac{\cos(\cot^{-1}x)}{2} + C$ (E) $\frac{-\cos(\cot^{-1}x)}{1 + x^2} + C$

   ›Reveal solutionSolution

   Substitute $u=\cot^{-1}x$ so $du=-\tfrac{dx}{1+x^2}$.

   Let $u=\cot^{-1}x\Rightarrow du=-\dfrac{dx}{1+x^2}$, i.e. $\dfrac{dx}{1+x^2}=-du$.

   $\int\frac{\sin(\cot^{-1}x)}{1+x^2}\,dx=\int\sin u\,(-du)=-\left(-\cos u\right)+C=\cos u+C.$

   Back-substituting, $=\cos(\cot^{-1}x)+C$.

   ✓Final answer

   The correct option is (B).

3. KEAM 2025Set eng-2025-04234 marksMCQ
   Q.$\int \dfrac{\sin^{-1} x}{\sqrt{1 - x^2}}\, dx =$ (A) $\dfrac{1}{2}(\sin^{-1} x)^2 + C$ (B) $-(\sin^{-1} x)\sqrt{1 - x^2} + C$ (C) $(\sin^{-1} x)\sqrt{1 - x^2} + x + C$ (D) $(\sin^{-1} x)\sqrt{1 - x^2} - x + C$ (E) $(\sin^{-1} x)^2 + C$

   ›Reveal solutionSolution

   Substituting u=sin^{-1}x turns the integral into u du = (1/2)(sin^{-1}x)^2 + C.

   Concept and Intuition

   The factor 1/sqrt(1-x^2) is exactly the derivative of sin^{-1}x, so the substitution u = sin^{-1}x collapses the integral.

   Step-by-Step Solution

   1. Let u = sin^{-1}x, then du = dx/sqrt(1-x^2).
   2. The integral becomes integral of u du.
   3. That equals u^2/2 + C = (1/2)(sin^{-1}x)^2 + C.

   Common Mistakes

   • Forgetting the factor 1/2 from integrating u.

   ✓Final answer

   The correct option is (A) — (1/2)(sin^{-1}x)^2 + C.

   ANSWER: A

4. KEAM 2025Set eng-2025-04254 marksMCQ
   Q.The value of $\int_0^{\pi/3}\dfrac{\tan t}{\cos t}\,dt$ is equal to (A) $\dfrac{1}{2}$ (B) $\dfrac{-1}{2}$ (C) $2$ (D) $-2$ (E) $1$

   ›Reveal solutionSolution

   Write $\dfrac{\tan t}{\cos t}=\dfrac{\sin t}{\cos^{2}t}$; its antiderivative is $\sec t$.

   $\dfrac{\tan t}{\cos t}=\dfrac{\sin t}{\cos^{2}t}.$

   With $u=\cos t$, $du=-\sin t\,dt$, the antiderivative is $\dfrac{1}{\cos t}=\sec t$. Thus

   $\int_0^{\pi/3}\dfrac{\sin t}{\cos^{2}t}\,dt=\bigl[\sec t\bigr]_0^{\pi/3}=\sec\dfrac{\pi}{3}-\sec 0=2-1=1.$

   ✓Final answer

   The correct option is (E).

5. KEAM 2024Set eng-2024-06064 marksMCQ
   Q.$\int_{0}^{\pi/4}\left(\tan^{3} x+\tan^{5} x\right)dx =$ (A) $\frac{5}{12}$ (B) $\frac{1}{3}$ (C) $\frac{1}{4}$ (D) $\frac{1}{6}$ (E) $\frac{1}{12}$

   ›Reveal solutionSolution

   Factor out $\tan^3x\sec^2x$ and substitute $u=\tan x$ to get $\frac{1}{4}$.

   $\tan^3x+\tan^5x=\tan^3x(1+\tan^2x)=\tan^3x\sec^2x$.

   Let $u=\tan x$, $du=\sec^2x\,dx$. Limits: $x=0\to u=0$, $x=\frac{\pi}{4}\to u=1$.

   $\int_0^{\pi/4}\tan^3x\sec^2x\,dx=\int_0^1u^3\,du=\left[\frac{u^4}{4}\right]_0^1=\frac{1}{4}$.

   ✓Final answer

   The correct option is (C).

6. KEAM 2024Set eng-2024-06054 marksMCQ
   Q.$\int \frac{\tan^5 x \; \sec^2 x}{\tan^{12} x + 1}\,dx$ is equal to (A) $\frac{1}{6}\tan^{-1}\left[\tan^6 x\right] + C$ (B) $\frac{1}{2}\tan^{-1}\left[\tan^6 x\right] + C$ (C) $\frac{1}{4}\tan^{-1}\left[\tan^4 x\right] + C$ (D) $\frac{1}{3}\tan^{-1}\left[\tan^3 x\right] + C$ (E) $\frac{1}{7}\tan^{-1}\left[\tan^7 x\right] + C$

   ›Reveal solutionSolution

   Substitute $u=\tan^6 x$; the integral reduces to $\tfrac16\int \frac{du}{u^2+1} = \tfrac16\tan^{-1}(\tan^6 x)+C$.

   Let $u = \tan^6 x$. Then $du = 6\tan^5 x \sec^2 x\,dx$, so $\tan^5 x\sec^2 x\,dx = \tfrac{du}{6}$.

   Also $\tan^{12}x = (\tan^6 x)^2 = u^2$.

   $\int \dfrac{\tan^5 x\sec^2 x}{\tan^{12}x+1}\,dx = \dfrac{1}{6}\int \dfrac{du}{u^2+1} = \dfrac{1}{6}\tan^{-1}u + C = \dfrac{1}{6}\tan^{-1}(\tan^6 x)+C$.

   ✓Final answer

   The correct option is (A).

7. KEAM 2023Set eng-2023-P2-B24 marksMCQ
   Q.$\int \dfrac{2\tan x + 3}{\sin^{2} x + 2\cos^{2} x}\, dx =$ (A) $\dfrac{3}{\sqrt{2}} \sin^{-1}\left(\dfrac{\sin x}{\sqrt{2}}\right) + \ln\left|\sin^{2} x + 2\right| + C$ (B) $\dfrac{3}{\sqrt{2}} \tan^{-1}\left(\dfrac{\tan x}{\sqrt{2}}\right) + \ln\left|\tan^{2} x + 2\right| + C$ (C) $\dfrac{1}{\sqrt{2}} \tan^{-1}\left(\dfrac{\tan x}{\sqrt{2}}\right) - \ln\left|\tan^{2} x + 2\right| + C$ (D) $\dfrac{3}{\sqrt{2}} \cos^{-1}\left(\dfrac{\cos x}{\sqrt{2}}\right) + \ln\left|\sin^{2} x + 2\right| + C$ (E) $\dfrac{1}{\sqrt{2}} \cos^{-1}\left(\dfrac{\cos x}{\sqrt{2}}\right) - \ln\left|\cos^{2} x + 2\right| + C$

   ›Reveal solutionSolution

   The integral is $\dfrac{3}{\sqrt2}\tan^{-1}\!\left(\dfrac{\tan x}{\sqrt2}\right)+\log|\tan^2x+2|+C$.

   Concept and Intuition

   Divide numerator and denominator by $\cos^2x$ to convert everything into $\tan x$, then substitute $t=\tan x$.

   Step-by-Step Solution

   1. Dividing by $\cos^2x$: integrand $=\dfrac{(2\tan x+3)\sec^2x}{\tan^2x+2}$.
   2. Let $t=\tan x,\;dt=\sec^2x\,dx$: $\int\dfrac{2t+3}{t^2+2}\,dt$.
   3. Split: $\int\dfrac{2t}{t^2+2}dt=\log(t^2+2)$ and $\int\dfrac{3}{t^2+2}dt=\dfrac{3}{\sqrt2}\tan^{-1}\dfrac{t}{\sqrt2}$.

   Common Mistakes

   • Not recognising the $2t\,dt$ pattern for the log term.
   • Missing the $1/\sqrt2$ factor in the arctan term.

   ✓Final answer

   The correct option is (B) — $\dfrac{3}{\sqrt2}\tan^{-1}\!\left(\dfrac{\tan x}{\sqrt2}\right)+\log|\tan^2x+2|+C$.

   ANSWER: B

8. KEAM 2025Set eng-2025-04234 marksMCQ
   Q.$\int e^x (x^2 - 2)\cos\left(e^x (x^2 - 2x)\right)\, dx =$ (A) $\sin\left(e^x (x^2 - 2x)\right) + C$ (B) $\sin\left(e^x (x^2 - 2)\right) + C$ (C) $x^2 e^x \sin\left(e^x (x^2 - 2)\right) + C$ (D) $e^x \sin\left(e^x (x^2 - 2)\right) + C$ (E) $e^x \sin\left(x^2 e^x - 2x e^x\right) + C$

   ›Reveal solutionSolution

   Since d/dx[e^x(x^2-2x)] = e^x(x^2-2), the integral is sin(e^x(x^2-2x)) + C.

   Concept and Intuition

   Look for the outer function's argument, g(x) = e^x(x^2-2x), and check whether the rest of the integrand is exactly g'(x); if so the integral is a direct substitution.

   Step-by-Step Solution

   1. Let g(x) = e^x(x^2 - 2x).
   2. g'(x) = e^x(x^2-2x) + e^x(2x-2) = e^x(x^2 - 2), which is exactly the prefactor.
   3. So the integrand is cos(g)*g' dx, and integral cos(g) dg = sin(g) + C.
   4. Result: sin(e^x(x^2-2x)) + C.

   Common Mistakes

   • Confusing the argument (x^2-2x) with the prefactor (x^2-2).

   ✓Final answer

   The correct option is (A) — sin(e^x(x^2-2x)) + C.

   ANSWER: A

9. KEAM 2025Set eng-2025-04274 marksMCQ
   Q.$\int \dfrac{\cos\theta}{2 - \sin^2\theta}\,d\theta =$ (A) $\dfrac{1}{2}\log\left|\dfrac{\sqrt{2} - \sin\theta}{\sqrt{2} + \sin\theta}\right| + C$ (B) $\dfrac{1}{2}\log\left|\dfrac{\sqrt{2} + \sin\theta}{\sqrt{2} - \sin\theta}\right| + C$ (C) $\log\left|\dfrac{\sqrt{2} + \sin\theta}{\sqrt{2} - \sin\theta}\right| + C$ (D) $\dfrac{1}{\sqrt{2}}\log\left|\dfrac{\sqrt{2} + \sin\theta}{\sqrt{2} - \sin\theta}\right| + C$ (E) $\dfrac{1}{2\sqrt{2}}\log\left|\dfrac{\sqrt{2} + \sin\theta}{\sqrt{2} - \sin\theta}\right| + C$

   ›Reveal solutionSolution

   Substitution $u=\sin\theta$ gives $\int\dfrac{du}{2-u^2}=\dfrac{1}{2\sqrt2}\log\left|\dfrac{\sqrt2+\sin\theta}{\sqrt2-\sin\theta}\right|+C$.

   Let $u=\sin\theta$, so $du=\cos\theta\,d\theta$. The integral becomes

   $\int\frac{du}{2-u^2}=\int\frac{du}{(\sqrt2)^2-u^2}.$

   Using the standard result $\displaystyle\int\frac{du}{a^2-u^2}=\frac{1}{2a}\log\left|\frac{a+u}{a-u}\right|+C$ with $a=\sqrt2$:

   $=\frac{1}{2\sqrt2}\log\left|\frac{\sqrt2+u}{\sqrt2-u}\right|+C=\frac{1}{2\sqrt2}\log\left|\frac{\sqrt2+\sin\theta}{\sqrt2-\sin\theta}\right|+C.$

   ✓Final answer

   The correct option is (E).

10. KEAM 2025Set eng-2025-04294 marksMCQ
    Q.$\int \dfrac{dx}{\cos x \sqrt{2 \sin 2x}} =$ (A) $\dfrac{1}{2}\sqrt{\tan x} + C$ (B) $\sqrt{\tan x} + C$ (C) $2\sqrt{\tan x} + C$ (D) $4\sqrt{\tan x} + C$ (E) $3\sqrt{\tan x} + C$

    ›Reveal solutionSolution

    Writing $2\sin2x=4\sin x\cos x$ reduces the integrand to $\frac12\sec^2x\,(\tan x)^{-1/2}$; with $u=\tan x$ this integrates to $\sqrt{\tan x}+C$.

    Use $\sin 2x=2\sin x\cos x$, so $2\sin 2x=4\sin x\cos x$ and $\sqrt{2\sin 2x}=2\sqrt{\sin x\cos x}$. Then

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

    Let $u=\tan x$, $du=\sec^2 x\,dx$:

    $\frac12\int u^{-1/2}\,du=\frac12\cdot2u^{1/2}=\sqrt{u}=\sqrt{\tan x}+C.$

    ✓Final answer

    The correct option is (B).

11. KEAM 2024Set eng-2024-06064 marksMCQ
    Q.$\int\frac{x^{2}-1}{x^{4}+3x^{2}+1}\,dx =$ (A) $\frac{1}{\sqrt{3}}\tan^{-1}\left(\frac{x^{2}+1}{\sqrt{3}x}\right)+C$ (B) $\tan^{-1}(x^{2}-1)+C$ (C) $\tan^{-1}\left(x-\frac{1}{x}\right)+C$ (D) $\frac{1}{\sqrt{5}}\tan^{-1}\left(\frac{x^{2}+1}{\sqrt{5}x}\right)+C$ (E) $\tan^{-1}\left(x+\frac{1}{x}\right)+C$

    ›Reveal solutionSolution

    The substitution $u=x+\frac{1}{x}$ turns it into $\int\frac{du}{u^2+1}=\tan^{-1}\left(x+\frac{1}{x}\right)+C$.

    Divide numerator and denominator by $x^2$:

    $\frac{x^2-1}{x^4+3x^2+1}=\frac{1-\frac{1}{x^2}}{x^2+3+\frac{1}{x^2}}$.

    Let $u=x+\frac{1}{x}$, so $du=\left(1-\frac{1}{x^2}\right)dx$ and $x^2+\frac{1}{x^2}=u^2-2$. The denominator becomes $u^2-2+3=u^2+1$.

    Thus $\int\frac{du}{u^2+1}=\tan^{-1}u+C=\tan^{-1}\left(x+\frac{1}{x}\right)+C$.

    ✓Final answer

    The correct option is (E).

12. KEAM 2026Set eng-2026-04184 marksMCQ
    Q.$\displaystyle\int \dfrac{\sec x\sqrt{\sec x}}{\sqrt{\sin x} + \sqrt{\cos x}}\, dx =$ (A) $2\left(\sqrt{\tan x} - \log_e\left(\sqrt{\tan x} + 1\right)\right) + C$ (B) $2\left(\sqrt{\tan x} + \log_e\left(\sqrt{\tan x} + 1\right)\right) + C$ (C) $2\sqrt{\tan x} - \log_e\left(\sqrt{\tan x} + 1\right) + C$ (D) $2\sqrt{\tan x} + \log_e\left(\sqrt{\tan x} + 1\right) + C$ (E) $\sqrt{\tan x} - \log_e\left(\sqrt{\tan x} + 1\right) + C$

    ›Reveal solutionSolution

    Factor $\sqrt{\sin x}+\sqrt{\cos x}=\sqrt{\cos x}(\sqrt{\tan x}+1)$ to reduce the integrand to $\dfrac{\sec^2 x}{\sqrt{\tan x}+1}$; substituting $u=\sqrt{\tan x}$ gives option (A).

    Write the denominator as $\sqrt{\sin x}+\sqrt{\cos x}=\sqrt{\cos x}\bigl(\sqrt{\tan x}+1\bigr)$. The numerator is $\sec x\sqrt{\sec x}=\cos^{-3/2}x$, so

    $\dfrac{\cos^{-3/2}x}{\sqrt{\cos x}\,(\sqrt{\tan x}+1)}=\dfrac{\cos^{-2}x}{\sqrt{\tan x}+1}=\dfrac{\sec^2 x}{\sqrt{\tan x}+1}.$

    Let $t=\tan x$, $dt=\sec^2 x\,dx$, then $u=\sqrt t$, $t=u^2$, $dt=2u\,du$:

    $\int\dfrac{dt}{\sqrt t+1}=\int\dfrac{2u\,du}{u+1}=2\int\left(1-\dfrac{1}{u+1}\right)du=2\bigl(u-\log(u+1)\bigr)+C.$

    Thus the integral is $2\bigl(\sqrt{\tan x}-\log(\sqrt{\tan x}+1)\bigr)+C$.

    ✓Final answer

    The correct option is (A).