Q.Integrate the function $\cos^3 x\,e^{\log\sin x}$

Showing the 12 most recent of 32 on this concept.

1. KEAM 2025Set eng-2025-04284 marksMCQ
   Q.$\int \frac{dx}{\cos^{2/3}x \, \sin^{4/3}x}$ is (A) $3\tan^3 x + C$ (B) $3\tan^{1/3} x + C$ (C) $-3\tan^{1/3} x + C$ (D) $-3\tan^{-1/3} x + C$ (E) $3\tan^{-1/3} x + C$

   ›Reveal solutionSolution

   Rewrite as $\sec^2x\,\tan^{-4/3}x\,dx$; sub $t=\tan x$ to get $\int t^{-4/3}dt=-3t^{-1/3}=-3\tan^{-1/3}x+C$.

   The integrand $\dfrac{1}{\cos^{2/3}x\,\sin^{4/3}x}$ has denominator powers summing to $2$, so factor out $\cos^2x$. Multiplying numerator and denominator by $\cos^{4/3}x$ (equivalently dividing top and bottom by $\cos^2x$): …

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2. 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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3. KEAM 2024Set eng-2024-06074 marksMCQ
   Q.$\displaystyle\int x^5 e^{x^3}\,dx =$ (A) $\dfrac{e^{x^3}}{3}(x^3-1)+C$ (B) $\dfrac{e^{x^3}}{5}(x^5-1)+C$ (C) $\dfrac{e^{x^3}}{4}(x^4-1)+C$ (D) $\dfrac{e^{x^3}}{3}(x^5-1)+C$ (E) $\dfrac{x^3 e^{x^3}}{3}+C$

   ›Reveal solutionSolution

   Substitute $u=x^3$, then integrate $ue^u$ by parts.

   Let $u=x^3$, so $du=3x^2dx$ and $x^5dx=x^3\cdot x^2dx=\frac{u}{3}du$. Thus

   $\int x^5e^{x^3}dx=\frac13\int u\,e^u\,du.$

   By parts, $\int ue^u du=ue^u-e^u=e^u(u-1)$, so …

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4. KEAM 2022Set eng-2022-P2-B14 marksMCQ
   Q.$\int \frac{1}{x^3}\sqrt{1 - \frac{1}{x^2}} \, dx =$ (A) $\frac{-1}{6}\left(1 - \frac{1}{x^2}\right)^{\frac{3}{2}} + C$ (B) $\frac{1}{3}\left(1 - \frac{1}{x^2}\right)^{\frac{3}{2}} + C$ (C) $\frac{-1}{3}\left(1 - \frac{1}{x^2}\right)^{\frac{3}{2}} + C$ (D) $\frac{4}{3}\left(1 - \frac{1}{x^2}\right)^{\frac{3}{2}} + C$ (E) $\frac{-4}{3}\left(1 - \frac{1}{x^2}\right)^{\frac{3}{2}} + C$

   ›Reveal solutionSolution

   The integral equals $\frac13\left(1-\frac1{x^2}\right)^{3/2}+C$.

   Concept and Intuition

   The derivative of $1-\frac1{x^2}$ is $\frac2{x^3}$, which matches the $\frac1{x^3}$ factor outside the root, so a substitution linearizes the integral.

   Step-by-Step Solution

   1. Let $u=1-\frac1{x^2}$.
   2. $du=\frac{2}{x^3}\,dx\Rightarrow \frac{dx}{x^3}=\frac{du}{2}$.
   3. $\int \frac{1}{x^3}\sqrt{1-\frac1{x^2}}\,dx=\int \sqrt u\,\frac{du}{2}=\frac12\cdot\frac{u^{3/2}}{3/2}$.
   4. $=\frac12\cdot\frac23 u^{3/2}=\frac13 u^{3/2}=\frac13\left(1-\frac1{x^2}\right)^{3/2}+C$. …

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5. KEAM 2026Set eng-2026-04184 marksMCQ
   Q.$\displaystyle\int \dfrac{10e^x}{(2e^x + 5)^3}\, dx$ is equal to (A) $\dfrac{5}{2(2e^x + 5)^2} + C$ (B) $\dfrac{-5}{(2e^x + 5)^2} + C$ (C) $\dfrac{-10}{(2e^x + 5)^2} + C$ (D) $\dfrac{-5}{2(2e^x + 5)^2} + C$ (E) $\dfrac{5}{(2e^x + 5)^2} + C$

   ›Reveal solutionSolution

   Substitute $u=2e^x+5$, $du=2e^x\,dx$, giving $5\int u^{-3}du=-\dfrac{5}{2(2e^x+5)^2}+C$.

   Let $u=2e^x+5$, so $du=2e^x\,dx$, i.e. $e^x\,dx=\tfrac{du}{2}$. Then …

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6. 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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7. KEAM 2021Set eng-2021-P2-B14 marksMCQ
   Q.$\int \dfrac{\cos(\tan x)}{\cos^2 x} \, dx =$ (A) $(\tan x)\sin(\tan x) + C$ (B) $\sin(\tan x) + C$ (C) $\sec(\tan x) + C$ (D) $(\cos x)\sin(\tan x) + C$ (E) $\cos^2(\tan x) + C$

   ›Reveal solutionSolution

   $\int\dfrac{\cos(\tan x)}{\cos^2 x}\,dx=\sin(\tan x)+C$.

   Concept and Intuition

   The factor $1/\cos^2 x=\sec^2 x$ is exactly the derivative of $\tan x$, so substituting $u=\tan x$ linearizes the integral.

   Step-by-Step Solution

   1. Rewrite as $\int\cos(\tan x)\sec^2 x\,dx$.
   2. Let $u=\tan x$, $du=\sec^2 x\,dx$. …

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8. 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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9. 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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10. KEAM 2026Set eng-2026-04184 marksMCQ
    Q.$\displaystyle\int \left(27x^3\left(1 - x^3\right)\right)^{\frac{2}{3}}\, dx =$ (A) $-\dfrac{3}{4}\left(1 - x^3\right)^{\frac{4}{3}} + C$ (B) $-\dfrac{3}{5}\left(1 - x^3\right)^{\frac{5}{3}} + C$ (C) $-\dfrac{9}{3}\left(1 - x^3\right)^{\frac{1}{3}} + C$ (D) $-\dfrac{9}{4}\left(1 - x^3\right)^{\frac{4}{3}} + C$ (E) $-\dfrac{9}{5}\left(1 - x^3\right)^{\frac{5}{3}} + C$

    ›Reveal solutionSolution

    Simplify to $9x^2(1-x^3)^{2/3}$, then substitute $u=1-x^3$ to obtain $-\dfrac{9}{5}(1-x^3)^{5/3}+C$.

    Since $\bigl(27x^3(1-x^3)\bigr)^{2/3}=27^{2/3}(x^3)^{2/3}(1-x^3)^{2/3}=9x^2(1-x^3)^{2/3}$, let $u=1-x^3$, $du=-3x^2\,dx$, so $x^2\,dx=-\tfrac{du}{3}$: …

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11. 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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12. 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. …

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