Q.$\int_{0}^{\pi} \dfrac{x}{1+\sin x}\,dx$

Showing the 12 most recent of 28 on this concept.

1. KEAM 2026Set eng-2026-04184 marksMCQ
   Q.The value of $\displaystyle\int_0^{2\pi} \dfrac{\sin 2x}{x(2\pi - x)}\, dx$ is equal to (A) $\pi$ (B) $\dfrac{\pi}{2}$ (C) $\dfrac{\pi}{4}$ (D) $\dfrac{\pi}{8}$ (E) $0$

   ›Reveal solutionSolution

   The substitution $x\to 2\pi-x$ flips the sign of $\sin 2x$ while leaving $x(2\pi-x)$ unchanged, so the integrand is antisymmetric about $x=\pi$ and the integral vanishes. …

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2. KEAM 2026Set eng-2026-04224 marksMCQ
   Q.The value of $\int_{0}^{\pi/2} \frac{\cos^{11} x}{\cos^{11} x + \sin^{11} x}\, dx$ is equal to (A) $\pi$ (B) $\frac{3\pi}{2}$ (C) $\frac{\pi}{2}$ (D) $\frac{\pi}{4}$ (E) $2\pi$

   ›Reveal solutionSolution

   Apply $\int_0^{a}f(x)dx=\int_0^{a}f(a-x)dx$ with $a=\pi/2$; adding gives twice the value $=\pi/2$.

   Let $I=\int_0^{\pi/2}\frac{\cos^{11}x}{\cos^{11}x+\sin^{11}x}dx$.

   Replacing $x\to\frac\pi2-x$ swaps sine and cosine: $I=\int_0^{\pi/2}\frac{\sin^{11}x}{\sin^{11}x+\cos^{11}x}dx$. …

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3. KEAM 2024Set eng-2024-06054 marksMCQ
   Q.The value of $\int_{0}^{\pi/2} \frac{\cos^{2024} x}{\sin^{2024} x + \cos^{2024} x}\,dx$ is equal to (A) $\frac{\pi}{4}$ (B) $\frac{\pi}{2}$ (C) $2\pi$ (D) $\pi$ (E) $\frac{\pi}{3}$

   ›Reveal solutionSolution

   Apply $x\to\frac{\pi}{2}-x$ and add to get $2I=\frac{\pi}{2}$.

   Let $I=\displaystyle\int_0^{\pi/2}\frac{\cos^{2024}x}{\sin^{2024}x+\cos^{2024}x}\,dx$. Using $x\to\frac{\pi}{2}-x$ swaps sine and cosine:

   $I=\int_0^{\pi/2}\frac{\sin^{2024}x}{\sin^{2024}x+\cos^{2024}x}\,dx.$ …

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4. KEAM 2026Set eng-2026-04204 marksMCQ
   Q.$\int_{\pi/6}^{\pi/3} \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} \, dx$ is equal to (A) $0$ (B) $\frac{\pi}{6}$ (C) $\frac{\pi}{3}$ (D) $\frac{\pi}{12}$ (E) $\frac{\pi}{2}$

   ›Reveal solutionSolution

   Apply $\int_a^b f(x)\,dx=\int_a^b f(a+b-x)\,dx$ with $a+b=\frac{\pi}{2}$, which swaps $\sin$ and $\cos$.

   Let $I=\displaystyle\int_{\pi/6}^{\pi/3}\dfrac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}\,dx$. …

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5. KEAM 2025Set eng-2025-04274 marksMCQ
   Q.$\int_{-\pi/2}^{\pi/2} (x^5 + x^3 + x)\cos x\,dx =$ (A) $\dfrac{\pi}{4}$ (B) $\pi$ (C) $\dfrac{2\pi}{3}$ (D) $\dfrac{\pi}{2}$ (E) $0$

   ›Reveal solutionSolution

   The integrand is an odd function on a symmetric interval, so the integral is $0$.

   Let $g(x)=(x^5+x^3+x)\cos x$. Here $x^5+x^3+x$ is odd and $\cos x$ is even, so

   $g(-x)=(-x^5-x^3-x)\cos x=-g(x),$

   i.e. $g$ is odd. …

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6. KEAM 2024Set eng-2024-06054 marksMCQ
   Q.$\int_{-\pi/2}^{\pi/2} \frac{\cos^2 x}{1 + 2^{-x}}\,dx$ is equal to (A) $\frac{\pi}{3}$ (B) $\frac{\pi}{4}$ (C) $1$ (D) $\frac{1}{2}$ (E) $\frac{\pi}{2}$

   ›Reveal solutionSolution

   The $\frac{1}{1+2^{-x}}$ factor with an even numerator halves the plain integral of $\cos^2 x$: $I=\tfrac12\int_{-\pi/2}^{\pi/2}\cos^2x\,dx=\tfrac{\pi}{4}$.

   Let $I=\displaystyle\int_{-\pi/2}^{\pi/2}\frac{\cos^2 x}{1+2^{-x}}\,dx$. Replace $x\to -x$ (limits symmetric, $\cos^2$ even):

   $I=\displaystyle\int_{-\pi/2}^{\pi/2}\frac{\cos^2 x}{1+2^{x}}\,dx$.

   Adding the two forms: …

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7. KEAM 2021Set eng-2021-P2-B14 marksMCQ
   Q.The value of $\displaystyle\int_{\pi/8}^{3\pi/8} \dfrac{\sin^4 x}{\sin^4 x + \cos^4 x} \, dx$ is equal to (A) $\dfrac{\pi}{4}$ (B) $\dfrac{\pi}{8}$ (C) $\dfrac{\pi}{16}$ (D) $\dfrac{\pi}{2}$ (E) $1$

   ›Reveal solutionSolution

   $\displaystyle\int_{\pi/8}^{3\pi/8}\dfrac{\sin^4 x}{\sin^4 x+\cos^4 x}\,dx=\dfrac{\pi}{8}$.

   Concept and Intuition

   The king property $\int_a^b f(x)\,dx=\int_a^b f(a+b-x)\,dx$ with $a+b=\pi/2$ pairs the integrand with its cosine-counterpart, and the two add to $1$.

   Step-by-Step Solution

   1. Let $I=\int_{\pi/8}^{3\pi/8}\dfrac{\sin^4 x}{\sin^4 x+\cos^4 x}dx$.
   2. Replace $x\to\tfrac\pi2-x$: $I=\int_{\pi/8}^{3\pi/8}\dfrac{\cos^4 x}{\cos^4 x+\sin^4 x}dx$. …

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8. KEAM 2024Set eng-2024-06084 marksMCQ
   Q.$\displaystyle\int_{-1}^{1}x^2\sin x\,dx$ (A) $2\sin 1$ (B) $2$ (C) $4$ (D) $-2\sin 1$ (E) $0$

   ›Reveal solutionSolution

   $x^2\sin x$ is an odd function, so its integral over the symmetric interval $[-1,1]$ is $0$.

   Let $h(x)=x^2\sin x$. Then

   $h(-x)=(-x)^2\sin(-x)=x^2(-\sin x)=-h(x),$ …

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9. KEAM 2024Set eng-2024-06094 marksMCQ
   Q.$\int_{0}^{1} \log\left(\dfrac{1}{x} - 1\right) dx =$ (A) $\dfrac{1}{4}$ (B) $\dfrac{1}{2}$ (C) $-1$ (D) $1$ (E) $0$

   ›Reveal solutionSolution

   Split the log; both $\int_0^1\log(1-x)dx$ and $\int_0^1\log x\,dx$ equal $-1$, so their difference is $0$.

   Rewrite $\log\!\left(\frac1x-1\right)=\log\frac{1-x}{x}=\log(1-x)-\log x$. …

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10. KEAM 2024Set eng-2024-06064 marksMCQ
    Q.$\int_{\pi/5}^{3\pi/10}\frac{\sqrt{\tan x}}{1+\sqrt{\tan x}}\,dx =$ (A) $\frac{\pi}{4}$ (B) $\frac{\pi}{5}$ (C) $\frac{\pi}{10}$ (D) $\frac{\pi}{20}$ (E) $\frac{\pi}{2}$

    ›Reveal solutionSolution

    The limits sum to $\frac{\pi}{2}$, so the King property gives $2I=b-a=\frac{\pi}{10}$ and $I=\frac{\pi}{20}$.

    Let $I=\int_{\pi/5}^{3\pi/10}\frac{\sqrt{\tan x}}{1+\sqrt{\tan x}}\,dx$. Since $a+b=\frac{\pi}{5}+\frac{3\pi}{10}=\frac{\pi}{2}$, apply $x\to\frac{\pi}{2}-x$ so $\tan x\to\cot x$:

    $f\!\left(\tfrac{\pi}{2}-x\right)=\frac{\sqrt{\cot x}}{1+\sqrt{\cot x}}=\frac{1}{1+\sqrt{\tan x}}$. …

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11. KEAM 2024Set eng-2024-06094 marksMCQ
    Q.$\int_{-\pi/2}^{\pi/2} \sin^9 x \cos^2 x\, dx =$ (A) $\dfrac{2}{3}$ (B) $1$ (C) $\dfrac{1}{11}$ (D) $\dfrac{7\pi}{6}$ (E) $0$

    ›Reveal solutionSolution

    $\sin^9x$ is odd and $\cos^2x$ is even, so the product is odd; integrating an odd function over $[-\frac{\pi}{2},\frac{\pi}{2}]$ gives $0$.

    Let $g(x)=\sin^9x\cos^2x$. Then $g(-x)=\sin^9(-x)\cos^2(-x)=-\sin^9x\cos^2x=-g(x)$, so $g$ is odd. …

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12. KEAM 2025Set eng-2025-04234 marksMCQ
    Q.The value of $\displaystyle\int_{\pi/10}^{2\pi/5} \dfrac{\cot^3 x}{1 + \cot^3 x}\, dx$ is equal to (A) $\dfrac{\pi}{20}$ (B) $\dfrac{\pi}{10}$ (C) $\dfrac{3\pi}{20}$ (D) $\dfrac{\pi}{5}$ (E) $\dfrac{\pi}{4}$

    ›Reveal solutionSolution

    Using the King property with a+b=pi/2, the integral equals half of (b-a) = 3pi/20.

    Concept and Intuition

    The limits satisfy a+b = pi/10 + 2pi/5 = pi/2, and cot(pi/2 - x) = tan x, so the reflection x -> a+b-x pairs cot^3 with tan^3, and the two integrands add to 1.

    Step-by-Step Solution

    1. Let I = integral of cot^3 x/(1+cot^3 x) dx over [pi/10, 2pi/5].
    2. Replace x by pi/2 - x: cot -> tan, giving I = integral of tan^3 x/(1+tan^3 x) dx over the same limits. …

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