Q.Prove that $\int_{0}^{\pi/2}\sin^3 x\,dx=\frac{2}{3}$

Showing the 12 most recent of 32 on this concept.

1. 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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2. 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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3. 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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4. 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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5. 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. $\dfrac{dx}{1+x^2}=-du$. …

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6. KEAM 2026Set eng-2026-04194 marksMCQ
   Q.$\int \left(\frac{\sin t + \cos t}{13 + 36\sin 2t}\right) dt$ is equal to (A) $\frac{1}{84}\log\left|\frac{7 + 6(\sin t - \cos t)}{7 - 6(\sin t - \cos t)}\right| + C$ (B) $\frac{1}{81}\log\left|\frac{7 + 6(\sin t - \cos t)}{7 - 6(\sin t - \cos t)}\right| + C$ (C) $\frac{1}{84}\log\left|\frac{7 - 6(\sin t - \cos t)}{7 + 6(\sin t - \cos t)}\right| + C$ (D) $\frac{1}{48}\log\left|\frac{7 + 6(\sin t - \cos t)}{7 - 6(\sin t - \cos t)}\right| + C$ (E) $\frac{1}{64}\log\left|\frac{7 + 6(\sin t - \cos t)}{7 - 6(\sin t - \cos t)}\right| + C$

   ›Reveal solutionSolution

   Let $u=\sin t-\cos t$; the numerator becomes $du$ and $13+36\sin 2t=49-36u^2$, giving a standard $\tfrac{du}{a^2-k^2u^2}$ integral.

   Set $u=\sin t-\cos t$. Then

   $du=(\cos t+\sin t)\,dt,$

   which is exactly the numerator, and

   $u^2=1-2\sin t\cos t=1-\sin 2t\;\Rightarrow\;\sin 2t=1-u^2.$

   So the denominator is

   $13+36\sin 2t=13+36(1-u^2)=49-36u^2.$

   Hence

   $\int\frac{du}{49-36u^2}=\int\frac{du}{7^2-(6u)^2}.$ …

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7. 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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8. KEAM 2025Set eng-2025-04234 marksMCQ
   Q.$\int x^7 (x^8 + 1)^{-3/4}\, dx =$ (A) $\dfrac{1}{2}\left(1 + \dfrac{1}{x^8}\right)^{1/4} + C$ (B) $4\left(1 + \dfrac{1}{x^8}\right)^{1/4} + C$ (C) $(x^8 + 1)^{1/4} + C$ (D) $4(x^8 + 1)^{1/4} + C$ (E) $\dfrac{1}{2}(x^8 + 1)^{1/4} + C$

   ›Reveal solutionSolution

   Substituting u=x^8+1 gives (1/2)(x^8+1)^{1/4} + C.

   Concept and Intuition

   The presence of x^7 alongside x^8 signals the substitution u = x^8 + 1, whose differential absorbs x^7 dx.

   Step-by-Step Solution

   1. Let u = x^8 + 1, then du = 8 x^7 dx, so x^7 dx = du/8.
   2. Integral = (1/8) integral of u^{-3/4} du.
   3. = (1/8) * u^{1/4}/(1/4) = (1/8)4u^{1/4} = (1/2)u^{1/4}. …

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9. 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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10. 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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11. 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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12. KEAM 2025Set eng-2025-04264 marksMCQ
    Q.If $\displaystyle\int \dfrac{1}{x^7 \left(\dfrac{1}{x^6}+1\right)^{2/3}} \, dx = -\dfrac{1}{2}\left(\dfrac{1}{\dfrac{1}{x^6}+1}\right)^{p} + c$, then $p =$ (A) $\dfrac{2}{3}$ (B) $\dfrac{-1}{3}$ (C) $\dfrac{1}{3}$ (D) $\dfrac{-2}{3}$ (E) $\dfrac{1}{6}$

    ›Reveal solutionSolution

    Substitute $u=\tfrac1{x^6}+1$; the result is $-\tfrac12u^{1/3}$, matched to the given form $-\tfrac12(1/u)^{p}$ gives $p=-1/3$.

    Let $u=\dfrac{1}{x^6}+1$, so $du=-\dfrac{6}{x^7}\,dx$, i.e. $\dfrac{dx}{x^7}=-\dfrac{du}{6}$.

    $\int \frac{u^{-2/3}}{x^7}\,dx=\int u^{-2/3}\left(-\frac{du}{6}\right)=-\frac16\cdot\frac{u^{1/3}}{1/3}=-\frac12u^{1/3}+c.$ …

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