Q.Evaluate the integral using substitution $\int_{0}^{2}x\sqrt{x+2}\,dx$ (Put $x+2=t^2$)

Showing the 12 most recent of 32 on this concept.

1. 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 …

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2. 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 …

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3. 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 …

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4. KEAM 2022Set eng-2022-P2-B14 marksMCQ
   Q.$\int \frac{e^{\frac{1}{\sqrt{t}}}}{t\sqrt{t}} \, dt =$ (A) $\frac{1}{2}e^{\frac{1}{\sqrt{t}}} + C$ (B) $\frac{-1}{2}e^{\frac{1}{\sqrt{t}}} + C$ (C) $2e^{\frac{1}{\sqrt{t}}} + C$ (D) $-2e^{\frac{1}{\sqrt{t}}} + C$ (E) $e^{\frac{1}{\sqrt{t}}} + C$

   ›Reveal solutionSolution

   The integral equals $-2e^{1/\sqrt t}+C$.

   Concept and Intuition

   Recognize that the exponent's derivative appears (up to a constant) in the integrand, so a direct substitution works.

   Step-by-Step Solution

   1. Let $u=\frac1{\sqrt t}=t^{-1/2}$.
   2. $du=-\frac12 t^{-3/2}\,dt=-\frac12\cdot\frac{1}{t\sqrt t}\,dt$, so $\frac{dt}{t\sqrt t}=-2\,du$.
   3. $\int \frac{e^{1/\sqrt t}}{t\sqrt t}\,dt=\int e^{u}(-2)\,du=-2e^{u}+C$. …

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5. KEAM 2024Set eng-2024-06054 marksMCQ
   Q.$\int \frac{\sec x}{(\sec x + \tan x)^2}\,dx =$ (A) $\frac{2}{5(\sec x + \tan x)^4} + C$ (B) $\frac{-1}{2(\sec x + \tan x)^2} + C$ (C) $\frac{2}{3(\sec x + \tan x)^{3/2}} + C$ (D) $\frac{-2}{3(\sec x + \tan x)^3} + C$ (E) $(\sec x + \tan x)^2 + C$

   ›Reveal solutionSolution

   The substitution $u=\sec x+\tan x$ turns it into $\int u^{-3}\,du$.

   Let $u=\sec x+\tan x$. Then $du=(\sec x\tan x+\sec^2 x)\,dx=\sec x(\tan x+\sec x)\,dx=\sec x\,u\,dx$, so $\sec x\,dx=\dfrac{du}{u}$. Hence …

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6. 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. …

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7. KEAM 2025Set eng-2025-04254 marksMCQ
   Q.$\int\dfrac{\sec^2\left(\sqrt{2x+5}\right)}{\sqrt{2x+5}}\,dx=$ (A) $2\tan\left(\sqrt{2x+5}\right)+C$ (B) $\dfrac{1}{2}\tan\left(\sqrt{2x+5}\right)+C$ (C) $\tan\left(2x+5\right)+C$ (D) $\tan\left(\sqrt{2x+5}\right)+C$ (E) $2\tan\left(2x+5\right)+C$

   ›Reveal solutionSolution

   Substitute $u=\sqrt{2x+5}$; the differential cancels the $\dfrac{1}{\sqrt{2x+5}}$ factor.

   Let $u=\sqrt{2x+5}$. Then

   $du=\dfrac{1}{2\sqrt{2x+5}}\cdot 2\,dx=\dfrac{dx}{\sqrt{2x+5}}.$

   So the integral becomes …

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8. 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: $x=0\to u=0$, $x=\frac{\pi}{4}\to u=1$. …

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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 $a=\sqrt2$: …

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10. 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 $u=\sqrt t$, $t=u^2$, $dt=2u\,du$: …

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11. KEAM 2026Set eng-2026-04174 marksMCQ
    Q.$\displaystyle\int \dfrac{\sqrt{\sqrt{x} + 1}}{\sqrt{x}}\, dx =$ (A) $\dfrac{4}{3}(\sqrt{x} + 1)^{\frac{3}{2}} + C$ (B) $\dfrac{2}{3}(\sqrt{x} + 1)^{\frac{3}{2}} + C$ (C) $\dfrac{4}{3}(\sqrt{x} + 1)^{\frac{3}{4}} + C$ (D) $\dfrac{1}{3}(\sqrt{x} + 1)^{\frac{3}{2}} + C$ (E) $\dfrac{3}{4}(\sqrt{x} + 1)^{\frac{3}{2}} + C$

    ›Reveal solutionSolution

    Substitute $u=\sqrt{x}+1$.

    With $u=\sqrt{x}+1$, $du=\dfrac{1}{2\sqrt{x}}\,dx$, i.e. $\dfrac{dx}{\sqrt{x}}=2\,du$. Then …

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12. KEAM 2024Set eng-2024-06064 marksMCQ
    Q.$\int\frac{4x\cos\sqrt{4x^{2}+7}}{\sqrt{4x^{2}+7}}\,dx =$ (A) $\frac{1}{2}\sin\sqrt{4x^{2}+7}+C$ (B) $\frac{7}{2}\sin\sqrt{4x^{2}+7}+C$ (C) $\sin\sqrt{4x^{2}+7}+C$ (D) $\frac{1}{4}\sin\sqrt{4x^{2}+7}+C$ (E) $\frac{7}{4}\sin\sqrt{4x^{2}+7}+C$

    ›Reveal solutionSolution

    With $u=\sqrt{4x^2+7}$ the integrand is exactly $\cos u\,du$, giving $\sin\sqrt{4x^2+7}+C$.

    Let $u=\sqrt{4x^2+7}$. Then $\frac{du}{dx}=\frac{8x}{2\sqrt{4x^2+7}}=\frac{4x}{\sqrt{4x^2+7}}$, so $du=\frac{4x}{\sqrt{4x^2+7}}\,dx$. …

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