Q.Evaluate the definite integral $\int_{0}^{\pi/4}\frac{\sin x+\cos x}{9+16\sin 2x}\,dx$

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 $1-\sin x$:

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

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2. 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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3. 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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4. KEAM 2024Set eng-2024-06094 marksMCQ
   Q.$\int \dfrac{dx}{x^2 (x^4 + 1)^{3/4}} =$ (A) $-(x^4 + 1)^{1/4} + C$ (B) $(x^4 + 1)^{1/4} + C$ (C) $-\left(\dfrac{x^4 + 1}{x^4}\right)^{1/4} + C$ (D) $\left(\dfrac{x^4 + 1}{x^4}\right) + C$ (E) $\left(\dfrac{x^4 + 1}{x^4}\right)^{3/4} + C$

   ›Reveal solutionSolution

   Factor $x^4$ out of the radical; with $u=1+x^{-4}$, $du=-4x^{-5}dx$, the integral becomes $-u^{1/4}=-\left(\frac{x^4+1}{x^4}\right)^{1/4}+C$.

   Write $(x^4+1)^{3/4}=x^3(1+x^{-4})^{3/4}$, so

   $\int\frac{dx}{x^2(x^4+1)^{3/4}}=\int\frac{dx}{x^5(1+x^{-4})^{3/4}}.$ …

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5. 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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6. 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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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 …

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8. 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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9. 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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10. KEAM 2025Set eng-2025-04264 marksMCQ
    Q.$\displaystyle\int \dfrac{\sec x}{(\sec x + \tan x)^9} \, dx =$ (A) $\dfrac{1}{9}(\sec x + \tan x)^9 + C$ (B) $\dfrac{-1}{9}(\sec x + \tan x)^9 + C$ (C) $\dfrac{-1}{9}(\sec x + \tan x)^{-9} + C$ (D) $\dfrac{1}{9}(\sec x + \tan x)^{-9} + C$ (E) $(\sec x + \tan x)^{-9} + C$

    ›Reveal solutionSolution

    Substitute $u=\sec x+\tan x$; then $\sec x\,dx=du/u$.

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

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11. 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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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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