Q.Prove that \int_0^\pi \frac{x\sin x}{1+\cos^2 x},dx = \frac{\pi^2}{4}

Showing the 12 most recent of 14 on this concept.

1. CBSE 2026Set A1 markMCQ
   Q.$\int_0^{\pi/2}\frac{\sin x}{\sin x + \cos x}\,dx =$

   (a) $\pi$

   (b) $\frac{\pi}{2}$

   (c) $0$

   (d) $\frac{\pi}{4}$

   ›Reveal solutionSolution

   Add the integral to its $x\to\tfrac{\pi}{2}-x$ image to get $2I=\tfrac{\pi}{2}$, so $I=\tfrac{\pi}{4}$.

   Let $I=\int_0^{\pi/2}\dfrac{\sin x}{\sin x+\cos x}\,dx$. Replacing $x$ by $\tfrac{\pi}{2}-x$ swaps $\sin$ and $\cos$:

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

   Adding the two expressions for $I$:

   …

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2. CBSE 2026Set A1 markMCQ
   Q.$\int_0^a\frac{\sqrt{x}}{\sqrt{a-x} + \sqrt{x}}\,dx =$

   (a) $a$

   (b) $\frac{a}{2}$

   (c) $2a$

   (d) $3a$

   ›Reveal solutionSolution

   Use $\int_0^a f(x)\,dx=\int_0^a f(a-x)\,dx$; the substitution swaps $\sqrt{x}$ and $\sqrt{a-x}$, giving $I=\tfrac{a}{2}$.

   Let $I=\int_0^a\dfrac{\sqrt{x}}{\sqrt{a-x}+\sqrt{x}}\,dx$. Replacing $x$ by $a-x$:

   $I=\int_0^a\dfrac{\sqrt{a-x}}{\sqrt{x}+\sqrt{a-x}}\,dx$.

   Adding the two forms:

   …

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3. CBSE 2026Set ANNUAL1 markMCQ
   Q.∫ (from π/6 to π/3) √(cos x) / (√(sin x) + √(cos x)) dx is equal to:

   (a) π/4

   (b) π/6

   (c) π/12

   (d) π/2

   ›Reveal solutionSolution

   Using the property $\int_a^b f(x)\,dx = \int_a^b f(a+b-x)\,dx$, the integral equals its own "partner" integral, so twice the integral equals the length of the interval.

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

   Using the property $\int_a^b f(x)dx = \int_a^b f(a+b-x)dx$ with $a=\pi/6, b=\pi/3$, so $a+b = \pi/2$:

   $I = \int_{\pi/6}^{\pi/3} \frac{\sqrt{\cos(\pi/2 - x)}}{\sqrt{\sin(\pi/2-x)}+\sqrt{\cos(\pi/2-x)}}\,dx = \int_{\pi/6}^{\pi/3} \frac{\sqrt{\sin x}}{\sqrt{\cos x}+\sqrt{\sin x}}\,dx$

   Call this second integral $J$. By relabeling, $I = J$.

   Adding the original definitions of $I$ and $J$: …

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4. CBSE 2025Set IX1 markMCQ
   Q.The value of $\displaystyle\int_{0}^{\pi/2} \dfrac{dx}{1 + \sqrt{\tan x}}$ will be

   (a) $0$

   (b) $\dfrac{\pi}{2}$

   (c) $\dfrac{\pi}{4}$

   (d) $\dfrac{\pi}{8}$

   ›Reveal solutionSolution

   By the king-property $\int_0^a f(x)dx=\int_0^a f(a-x)dx$, $I=\dfrac{\pi}{4}$; option (c).

   Concept. The property $\displaystyle\int_0^a f(x)\,dx=\int_0^a f(a-x)\,dx$ turns a hard integral into a solvable pair.

   Let $I=\displaystyle\int_0^{\pi/2}\frac{dx}{1+\sqrt{\tan x}}$. Replace $x$ by $\tfrac{\pi}{2}-x$; since $\tan\!\left(\tfrac{\pi}{2}-x\right)=\cot x$, …

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5. CBSE 2025Set E1 markMCQ
   Q.$\int_{0}^{\pi/2} \frac{\sqrt{\cos x}}{\sqrt{\sin x} + \sqrt{\cos x}}\,dx =$

   (a) $\pi$

   (b) $\pi/2$

   (c) $\pi/4$

   (d) $2\pi$

   ›Reveal solutionSolution

   Using $\int_0^{\pi/2}f(x)dx=\int_0^{\pi/2}f(\tfrac{\pi}{2}-x)dx$, add the two forms: $2I=\frac{\pi}{2}$, so $I=\frac{\pi}{4}$.

   Let $I=\int_0^{\pi/2}\frac{\sqrt{\cos x}}{\sqrt{\sin x}+\sqrt{\cos x}}\,dx.$ Replacing $x\to \tfrac{\pi}{2}-x$ swaps $\sin$ and $\cos$:

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

   Adding the two expressions for $I$:

   …

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6. CBSE 2025Set E1 markMCQ
   Q.$\int_{0}^{\pi/2} \log\tan x\,dx =$

   (a) $\pi/4$

   (b) $\pi/2$

   (c) $0$

   (d) $\pi$

   ›Reveal solutionSolution

   Replacing $x\to\tfrac{\pi}{2}-x$ turns $\log\tan x$ into $\log\cot x=-\log\tan x$, so $I=-I\Rightarrow I=0$.

   Let $I=\int_0^{\pi/2}\log\tan x\,dx.$ Using $\int_0^{a}f(x)dx=\int_0^a f(a-x)dx$ with $a=\tfrac{\pi}{2}$:

   …

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7. CBSE 2025Set ANNUAL1 markMCQ
   Q.∫ (from π/6 to π/3) √(cos x) / (√(sin x) + √(cos x)) dx is equal to:

   (a) π/4

   (b) π/6

   (c) π/12

   (d) π/2

   ›Reveal solutionSolution

   Use the property $\int_a^b f(x)\,dx = \int_a^b f(a+b-x)\,dx$: adding the original integral to its "flipped" version gives a constant, and by symmetry the two halves are equal.

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

   Here $a=\pi/6,\ b=\pi/3$, so $a+b = \pi/2$. Replacing $x$ by $a+b-x = \dfrac{\pi}{2}-x$, and using $\cos\!\left(\dfrac\pi2-x\right)=\sin x$, $\sin\!\left(\dfrac\pi2-x\right)=\cos x$:

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

   …

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8. CBSE 2024Set D1 markMCQ
   Q.$\int_0^a \frac{\sqrt{x}}{\sqrt{x} + \sqrt{a-x}}\,dx =$

   (a) $a$

   (b) $\frac{a}{2}$

   (c) $2a$

   (d) $3a$

   ›Reveal solutionSolution

   Adding the integral to its $x\to a-x$ image gives $2I=a$, so $I=\frac{a}{2}$.

   Let $I = \displaystyle\int_0^a \dfrac{\sqrt{x}}{\sqrt{x}+\sqrt{a-x}}\,dx$.

   Using $\int_0^a f(x)\,dx = \int_0^a f(a-x)\,dx$:

   $I = \displaystyle\int_0^a \dfrac{\sqrt{a-x}}{\sqrt{a-x}+\sqrt{x}}\,dx$.

   …

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9. CBSE 2023Set ANNUAL1 markMCQ
   Q.If $f(a+b-x) = f(x)$, then $\displaystyle\int_a^b x\,f(x)\,dx$ is equal to-

   (a) $0$

   (b) $\dfrac{a+b}{2}\displaystyle\int_a^b f(b+x)\,dx$

   (c) $\dfrac{b-a}{2}\displaystyle\int_a^b f(x)\,dx$

   (d) $\dfrac{a+b}{2}\displaystyle\int_a^b f(x)\,dx$

   ›Reveal solutionSolution

   This is a standard King's-rule property of definite integrals.

   Let $I=\displaystyle\int_a^b x f(x)\,dx$. Using the property $\int_a^b g(x)dx=\int_a^b g(a+b-x)dx$:

   $I=\displaystyle\int_a^b (a+b-x)f(a+b-x)\,dx=\int_a^b (a+b-x)f(x)\,dx$ (since $f(a+b-x)=f(x)$)

   $I=(a+b)\displaystyle\int_a^b f(x)dx-\int_a^b x f(x)dx=(a+b)\int_a^b f(x)dx-I$

   …

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10. CBSE 2022Set FF1 markMCQ
    Q.The value of $\displaystyle\int_0^{\pi/2} \dfrac{dx}{1+\sqrt{\tan x}}$ will be:

    (a) 0

    (b) $\dfrac{\pi}{2}$

    (c) $\dfrac{\pi}{4}$

    (d) $\dfrac{\pi}{8}$

    ›Reveal solutionSolution

    By the King's-rule property, $I=\dfrac{\pi}{4}$ — option (c).

    Concept. Use $\displaystyle\int_0^a f(x)\,dx=\int_0^a f(a-x)\,dx$ with $a=\dfrac{\pi}{2}$.

    Let $I=\displaystyle\int_0^{\pi/2}\frac{dx}{1+\sqrt{\tan x}}$. Replacing $x\to \frac{\pi}{2}-x$ turns $\tan x$ into $\cot x$:

    $I=\int_0^{\pi/2}\frac{dx}{1+\sqrt{\cot x}}=\int_0^{\pi/2}\frac{\sqrt{\tan x}\,dx}{\sqrt{\tan x}+1}.$ …

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11. CBSE 2021Set I1 markMCQ
    Q.$\int_0^{\pi/2} \log\cot\theta\,d\theta =$

    (a) $\dfrac{\pi}{2}\log 2$

    (b) $\dfrac{\pi}{4}\log 2$

    (c) $2\pi\log 2$

    (d) $0$

    ›Reveal solutionSolution

    $\displaystyle\int_0^{\pi/2} \log\cot\theta\,d\theta = 0$.

    Let $I = \displaystyle\int_0^{\pi/2}\log\cot\theta\,d\theta$. Apply the King property $\int_0^a f(x)\,dx = \int_0^a f(a-x)\,dx$ with $a=\dfrac{\pi}{2}$:

    $I = \int_0^{\pi/2}\log\cot\!\left(\dfrac{\pi}{2}-\theta\right)d\theta = \int_0^{\pi/2}\log\tan\theta\,d\theta.$

    Adding the two forms: …

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12. CBSE 2021Set I1 markMCQ
    Q.$\int_0^{\pi/2} \dfrac{\cos\theta}{\cos\theta + \sin\theta}\,d\theta =$

    (a) $\pi$

    (b) $\dfrac{\pi}{2}$

    (c) $\dfrac{\pi}{3}$

    (d) $\dfrac{\pi}{4}$

    ›Reveal solutionSolution

    $\displaystyle\int_0^{\pi/2} \dfrac{\cos\theta}{\cos\theta + \sin\theta}\,d\theta = \dfrac{\pi}{4}$.

    Let $I = \displaystyle\int_0^{\pi/2}\dfrac{\cos\theta}{\cos\theta+\sin\theta}\,d\theta$. Apply the King property $\theta\to\dfrac{\pi}{2}-\theta$, which swaps $\sin\leftrightarrow\cos$:

    $I = \int_0^{\pi/2}\dfrac{\sin\theta}{\sin\theta+\cos\theta}\,d\theta.$

    Add the two forms: …

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