Q.Evaluate $\int_0^{\pi/2} \log \sin x\, dx$

Showing the 12 most recent of 48 on this concept.

1. AP EAPCET 2025Set eng-2025-05-21-AN1 markMCQ
   Q.$\displaystyle\int_{0}^{\pi/2} \log|\tan x + \cot x|\,dx =$ (A) $\pi \log 2$ (B) $-\pi \log 2$ (C) $\dfrac{\pi}{2}\log 2$ (D) $2\pi \log 2$

   ›Reveal solutionSolution

   Simplify $\tan x+\cot x=2/\sin2x$, split the log, and use the classical result $\int_0^\pi \log\sin u\,du=-\pi\log2$. Answer: $\pi\log2$.

   Concept and Intuition

   $\tan x + \cot x$ always simplifies to $\dfrac{2}{\sin 2x}$ via the Pythagorean identity — a very common simplification in definite-integral problems. This converts the problem into the well-known integral of $\log \sin$, whose value over a full "half-period" ($0$ to $\pi$) is a standard memorized result, $-\pi\log 2$.

   Step-by-Step Solution

   1. $\tan x + \cot x = \dfrac{\sin x}{\cos x}+\dfrac{\cos x}{\sin x} = \dfrac{\sin^2x+\cos^2x}{\sin x\cos x} = \dfrac{1}{\sin x\cos x} = \dfrac{2}{\sin 2x}$.
   2. So $\log|\tan x+\cot x| = \log 2 - \log(\sin 2x)$ (since $\sin2x>0$ on $(0,\pi/2)$).
   3. $\displaystyle\int_0^{\pi/2}\log|\tan x+\cot x|\,dx = \frac{\pi}{2}\log 2 - \int_0^{\pi/2}\log(\sin2x)\,dx$.
   4. Let $I=\displaystyle\int_0^{\pi/2}\log(\sin2x)\,dx$. Substitute $u=2x,\,du=2dx$: $I=\dfrac12\displaystyle\int_0^{\pi}\log(\sin u)\,du$. …

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2. AP EAPCET 2024Set eng-2024-05-22-FN1 markMCQ
   Q.$\displaystyle\int_0^{\pi/4} \log(1 + \tan x)\,dx =$ (A) $\pi \log 2 + 1$ (B) $\dfrac{\pi}{2}\log 2 + 1$ (C) $\dfrac{\pi}{4}\log 2$ (D) $\dfrac{\pi}{8}\log 2$

   ›Reveal solutionSolution

   A textbook symmetry trick ($x \to \pi/4-x$) doubles the integral into something trivial to evaluate. Answer: $\dfrac{\pi}{8}\log2$.

   Concept and Intuition

   Whenever the limits are $0$ to $a$ and the integrand involves $\tan x$, trying the substitution $x \to a-x$ is a classic move — here it converts $\log(1+\tan x)$ into $\log\!\left(\dfrac{2}{1+\tan x}\right)$, and adding the original and transformed integrals collapses most of the complexity.

   Step-by-Step Solution

   1. Let $I=\displaystyle\int_0^{\pi/4}\log(1+\tan x)\,dx$.
   2. Substitute $x\to \dfrac{\pi}{4}-x$: $\tan\!\left(\dfrac{\pi}{4}-x\right)=\dfrac{1-\tan x}{1+\tan x}$, so $1+\tan\!\left(\dfrac\pi4-x\right) = \dfrac{2}{1+\tan x}$.
   3. So $I = \displaystyle\int_0^{\pi/4}\log\!\left(\frac{2}{1+\tan x}\right)dx = \int_0^{\pi/4}\log2\,dx - \int_0^{\pi/4}\log(1+\tan x)\,dx = \frac{\pi}{4}\log2 - I$. …

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3. AP EAPCET 2024Set eng-2024-05-20-AN1 markMCQ
   Q.$\displaystyle\int_{-1/24}^{1/24} \sec x \, \log\left(\dfrac{1-x}{1+x}\right) dx =$ (A) $\dfrac{\pi}{2}$ (B) $\pi$ (C) $1$ (D) $0$

   ›Reveal solutionSolution

   The integrand is odd (even $\times$ odd), and it's integrated over a symmetric interval $[-1/24,1/24]$, so the integral is $0$ — (D).

   Concept and Intuition

   Rather than actually evaluating a messy integral, check the parity of the integrand first: if $f(-x)=-f(x)$ (odd) and the limits are symmetric about $0$, the positive and negative halves cancel exactly, giving $0$ — no computation needed.

   Step-by-Step Solution

   1. Let $g(x)=\log\!\left(\dfrac{1-x}{1+x}\right)$. Then $g(-x)=\log\!\left(\dfrac{1-(-x)}{1+(-x)}\right)=\log\!\left(\dfrac{1+x}{1-x}\right)=\log\!\left[\left(\dfrac{1-x}{1+x}\right)^{-1}\right]=-\log\!\left(\dfrac{1-x}{1+x}\right)=-g(x)$. So $g$ is odd.
   2. $\sec x$ is an even function ($\sec(-x)=\sec x$).
   3. The product $\sec x\cdot g(x)$ is even $\times$ odd $=$ odd.
   4. The limits of integration, $-\dfrac1{24}$ to $\dfrac1{24}$, are symmetric about $0$. …

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4. AP EAPCET 2025Set eng-2025-05-21-FN1 markMCQ
   Q.$\displaystyle\int_0^{\pi} \dfrac{x\sin x}{1+\cos^2 x}\,dx =$ (A) $\dfrac{\pi^2}{4}$ (B) $\dfrac{\pi}{2}$ (C) $\dfrac{\pi^2}{2}$ (D) $\dfrac{\pi}{4}$

   ›Reveal solutionSolution

   This is the classic $\int_0^\pi x f(\sin x,\cos x)\,dx$ trick using the substitution $x\to \pi-x$; the answer is (A).

   Concept and Intuition

   For integrals of the form $\int_0^\pi x\,g(\cos x)\sin x\,dx$, substituting $x\to\pi-x$ (using $\sin(\pi-x)=\sin x$ and $\cos(\pi-x)=-\cos x$, but here $\cos^2$ is unaffected) converts the $x$ factor into $\pi - x$, letting you solve for the integral algebraically in terms of a simpler integral without the $x$ weight.

   Step-by-Step Solution

   1. Let $I=\displaystyle\int_0^\pi\frac{x\sin x}{1+\cos^2 x}dx$.
   2. Substitute $x\to\pi-x$: $\sin(\pi-x)=\sin x$, $\cos^2(\pi-x)=\cos^2 x$, so

   $I=\int_0^\pi\frac{(\pi-x)\sin x}{1+\cos^2x}dx=\pi\int_0^\pi\frac{\sin x}{1+\cos^2x}dx-I.$

   3. So $2I=\pi\displaystyle\int_0^\pi\frac{\sin x}{1+\cos^2 x}dx$. …

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5. AP EAPCET 2024Set eng-2024-05-20-AN1 markMCQ
   Q.$\displaystyle\int_0^{\pi} \dfrac{x\sin x}{1+\cos^2 x} \, dx =$ (A) $0$ (B) $\dfrac{\pi}{2}$ (C) $\dfrac{\pi^2}{2}$ (D) $\dfrac{\pi^2}{4}$

   ›Reveal solutionSolution

   This is the classic "$x\,f(\sin x)$" King's-rule setup; it reduces to $\pi/2$ times a standard arctan integral, giving $\dfrac{\pi^2}4$ — (D).

   Concept and Intuition

   Whenever the integrand has the shape $x\cdot g(\sin x,\cos^2x)$ over $[0,\pi]$ and $g$ is invariant under $x\to\pi-x$ (true here since $\sin(\pi-x)=\sin x$ and $\cos^2(\pi-x)=\cos^2x$), the King's-rule substitution $x\to\pi-x$ lets us replace the "$x$" prefactor with "$\pi-x$", and adding the two forms eliminates the awkward $x$ entirely.

   Step-by-Step Solution

   1. Let $I=\displaystyle\int_0^\pi\dfrac{x\sin x}{1+\cos^2x}dx$ and $f(x)=\dfrac{\sin x}{1+\cos^2x}$.
   2. Check $f(\pi-x)=\dfrac{\sin(\pi-x)}{1+\cos^2(\pi-x)}=\dfrac{\sin x}{1+\cos^2x}=f(x)$ — confirmed invariant.
   3. By King's rule: $I=\displaystyle\int_0^\pi(\pi-x)f(x)\,dx=\pi\int_0^\pi f(x)dx - I$.
   4. So $2I=\pi\displaystyle\int_0^\pi f(x)\,dx=\pi J$, where $J=\displaystyle\int_0^\pi\dfrac{\sin x}{1+\cos^2x}dx$. …

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6. AP EAPCET 2024Set eng-2024-05-20-FN1 markMCQ
   Q.$\int_{-\pi}^{\pi}\dfrac{x\sin^3x}{4-\cos^2x}dx=$ (A) $2\pi(1-\log3)$ (B) $2\pi\left(1-\dfrac{3}{4}\log3\right)$ (C) $\pi\left(1-\dfrac{3}{4}\log3\right)$ (D) $4\pi(1-\log3)$

   ›Reveal solutionSolution

   Use symmetry (odd integrand times $x$, plus $f(\pi-x)=f(x)$) to strip the $x$ out of the integral, then finish with a $u=\cos x$ substitution and partial fractions.

   Concept and Intuition

   Integrals of the form $\int_{-a}^a x\,g(x)\,dx$ where $g$ is odd become $2\int_0^a x\,g(x)\,dx$ (since $x\,g(x)$ is even). If additionally $g(\pi-x)=g(x)$ on $[0,\pi]$, the classic "King's Rule" trick $\int_0^\pi x\,g(x)\,dx=\dfrac\pi2\int_0^\pi g(x)\,dx$ removes the $x$ factor entirely.

   Step-by-Step Solution

   1. Let $g(x)=\dfrac{\sin^3x}{4-\cos^2x}$. Since $\sin^3(-x)=-\sin^3x$ and $\cos^2(-x)=\cos^2x$, $g$ is odd, so $x\,g(x)$ is even: $\int_{-\pi}^\pi xg(x)dx=2\int_0^\pi xg(x)dx$.
   2. Also $g(\pi-x)=\dfrac{\sin^3(\pi-x)}{4-\cos^2(\pi-x)}=\dfrac{\sin^3x}{4-\cos^2x}=g(x)$ (since $\sin(\pi-x)=\sin x$, $\cos(\pi-x)=-\cos x$), so King's Rule gives $\int_0^\pi xg(x)dx=\dfrac\pi2\int_0^\pi g(x)dx$.
   3. Combining: original integral $=2\cdot\dfrac\pi2\int_0^\pi g(x)dx=\pi\int_0^\pi g(x)dx=\pi J$.
   4. Compute $J=\int_0^\pi \dfrac{\sin^3x}{4-\cos^2x}dx$ via $u=\cos x$: $J=\int_{-1}^{1}\dfrac{1-u^2}{4-u^2}du=2\int_0^1\left[1-\dfrac{3}{4-u^2}\right]du$ (using $1-u^2=(4-u^2)-3$). …

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7. AP EAPCET 2024Set eng-2024-05-22-FN1 markMCQ
   Q.$\displaystyle\int_{-\pi}^{\pi} \dfrac{x \sin x}{1 + \cos^2 x}\,dx =$ (A) $\dfrac{3\pi^2}{4}$ (B) $\dfrac{\pi}{2} + 1$ (C) $\dfrac{\pi^2}{4}$ (D) $\dfrac{\pi^2}{2}$

   ›Reveal solutionSolution

   Combine the even-function property over $[-\pi,\pi]$ with the classic $x\to\pi-x$ symmetry trick for integrals of $x\,g(\sin x,\cos x)$. Answer: $\pi^2/2$.

   Concept and Intuition

   First check parity: since $\sin(-x)=-\sin x$ and $\cos^2(-x)=\cos^2x$, the integrand $f(x)=\dfrac{x\sin x}{1+\cos^2x}$ satisfies $f(-x)=f(x)$ — it's even, so the integral over $[-\pi,\pi]$ is twice the integral over $[0,\pi]$. Then, for integrals of $x$ times a function of $\sin x,\cos x$ over $[0,\pi]$, the substitution $x\to \pi-x$ is the standard tool to eliminate the explicit $x$.

   Step-by-Step Solution

   1. Since $f(-x)=f(x)$: $\displaystyle\int_{-\pi}^{\pi} f\,dx = 2\int_0^\pi f\,dx = 2J$, where $J=\displaystyle\int_0^\pi \frac{x\sin x}{1+\cos^2x}dx$.
   2. Substitute $x\to\pi-x$ in $J$: $\sin(\pi-x)=\sin x$, $\cos(\pi-x)=-\cos x$ (so $\cos^2$ unchanged): $J=\int_0^\pi \frac{(\pi-x)\sin x}{1+\cos^2x}dx = \pi\int_0^\pi\frac{\sin x}{1+\cos^2x}dx - J.$
   3. So $2J = \pi K$ where $K=\displaystyle\int_0^\pi \frac{\sin x}{1+\cos^2x}dx$. …

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8. AP EAPCET 2025Set eng-2025-05-22-FN1 markMCQ
   Q.$\displaystyle\int_{0}^{\pi}\frac{x\sin x}{\sin^2 x + 2\cos^2 x}dx =$ (A) $\frac{\pi}{2}$ (B) $\frac{\pi^2}{2}$ (C) $\frac{\pi^2}{4}$ (D) $\frac{\pi}{4}$

   ›Reveal solutionSolution

   Rewriting $\sin^2x+2\cos^2x=1+\cos^2x$ and applying the classical $\int_0^\pi x\,f(\sin x,\cos^2x)\,dx=\frac{\pi}{2}\int_0^\pi f\,dx$ symmetry reduces the problem to a standard arctangent integral, giving $\pi^2/4$.

   Concept and Intuition

   Whenever an integrand over $[0,\pi]$ depends on $x$ only through $\sin x$ and $\cos^2x$ (both invariant, or simply transformed, under $x\mapsto\pi-x$), the King's-rule substitution $I=\int_0^\pi x f(x)dx = \frac\pi2\int_0^\pi f(x)dx$ eliminates the explicit $x$ factor.

   Step-by-Step Solution

   1. Simplify the denominator: $\sin^2x+2\cos^2x=(\sin^2x+\cos^2x)+\cos^2x=1+\cos^2x$.
   2. Let $I=\displaystyle\int_0^\pi\frac{x\sin x}{1+\cos^2x}\,dx$.
   3. Apply $x\to\pi-x$: $\sin(\pi-x)=\sin x$, $\cos(\pi-x)=-\cos x\Rightarrow\cos^2(\pi-x)=\cos^2x$. So $I=\displaystyle\int_0^\pi\frac{(\pi-x)\sin x}{1+\cos^2x}\,dx=\pi\int_0^\pi\frac{\sin x}{1+\cos^2x}dx - I$.
   4. So $2I=\pi\,J$ where $J=\displaystyle\int_0^\pi\frac{\sin x}{1+\cos^2x}\,dx$. …

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9. AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
   Q.$\displaystyle\int_0^{\pi/2} \dfrac{\sin^3 x}{\sin x + \cos x}\, dx =$ (A) $\dfrac{\pi + 1}{4}$ (B) $\dfrac{\pi - 2}{2}$ (C) $\dfrac{\pi - 2}{4}$ (D) $\dfrac{\pi - 1}{4}$

   ›Reveal solutionSolution

   By symmetry $I=J$ (the $\cos^3$ analogue) and $I+J=\pi/2-1/2$ via a sum-of-cubes factorisation, giving $I=(\pi-1)/4$.

   Concept and Intuition

   Define $J$ as the same integral with $\sin^3x$ replaced by $\cos^3x$. The substitution $x\to\pi/2-x$ shows $I=J$ exactly (not just numerically), because it swaps sine and cosine while leaving $\sin x+\cos x$ and the interval unchanged. Then the sum-of-cubes identity $\sin^3x+\cos^3x = (\sin x+\cos x)(1-\sin x\cos x)$ lets $I+J$ collapse to an elementary integral, and since $I=J$, each equals half of that sum.

   Step-by-Step Solution

   1. Let $I=\displaystyle\int_0^{\pi/2}\dfrac{\sin^3x}{\sin x+\cos x}dx$ and $J=\displaystyle\int_0^{\pi/2}\dfrac{\cos^3x}{\sin x+\cos x}dx$.
   2. Substituting $u=\pi/2-x$ in $I$: $\sin x\to\cos u$, $\cos x\to\sin u$, limits unchanged, giving $I=\displaystyle\int_0^{\pi/2}\dfrac{\cos^3u}{\cos u+\sin u}du = J$. So $I=J$.
   3. $I+J = \displaystyle\int_0^{\pi/2}\dfrac{\sin^3x+\cos^3x}{\sin x+\cos x}dx$. Using $\sin^3x+\cos^3x=(\sin x+\cos x)(\sin^2x-\sin x\cos x+\cos^2x)=(\sin x+\cos x)(1-\sin x\cos x)$, this simplifies to $\displaystyle\int_0^{\pi/2}(1-\sin x\cos x)\,dx$. …

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10. AP EAPCET 2022Set eng-2022-07-04-FN1 markMCQ
    Q.$\int_0^\pi x(\sin^2(\sin x) + \cos^2(\cos x)) \, dx =$ (A) $\pi^2$ (B) $\pi^2/2$ (C) $2\pi$ (D) $\pi/4$

    ›Reveal solutionSolution

    Uses the $\int_0^a xf(x)dx=\frac{a}{2}\int_0^af(x)dx$ trick (valid when $f(a-x)=f(x)$) plus a $\pi/2$-symmetry identity to collapse the integrand to a constant; the answer is $\pi^2/2$.

    Concept and Intuition

    Whenever $\int_0^a x\,f(x)\,dx$ appears with $f(a-x)=f(x)$, King's Property gives $\int_0^a xf(x)dx = \frac{a}{2}\int_0^a f(x)dx$ — the $x$ weight averages out. Separately, $\sin^2(\sin x)+\cos^2(\cos x)$ has a beautiful complementary-angle identity that makes its integral over a quarter period trivial.

    Step-by-Step Solution

    1. Let $h(x)=\sin^2(\sin x)+\cos^2(\cos x)$. Check $h(\pi-x)$: $\sin(\pi-x)=\sin x$ and $\cos(\pi-x)=-\cos x$, and since $\cos(-\cos x)=\cos(\cos x)$ (cosine is even), we get $h(\pi-x)=h(x)$.
    2. By King's property, $I=\int_0^\pi x\,h(x)\,dx = \dfrac{\pi}{2}\int_0^\pi h(x)\,dx$.
    3. Since $h(\pi-x)=h(x)$, $h$ is symmetric about $x=\pi/2$, so $\int_0^\pi h\,dx = 2\int_0^{\pi/2}h\,dx$.
    4. On $[0,\pi/2]$, substitute $x\to \pi/2-x$ in just the cosine part: $\int_0^{\pi/2}\cos^2(\cos x)\,dx = \int_0^{\pi/2}\cos^2(\sin x)\,dx$ (since $\cos(\pi/2-x)=\sin x$). …

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11. AP EAPCET 2023Set eng-2023-05-17-AN1 markMCQ
    Q.$\displaystyle\int_{\pi/6}^{\pi/3} \dfrac{1}{1+\sqrt{\cot x}}\,dx=$ (A) $\pi/4$ (B) $\pi/2$ (C) $\pi/6$ (D) $\pi/12$

    ›Reveal solutionSolution

    Using the reflection property $\int_a^b f(x)dx=\int_a^b f(a+b-x)dx$ with $a+b=\pi/2$ makes the two forms of the integrand add up to exactly $1$, giving $I=\pi/12$.

    Concept and Intuition

    Integrals of the shape $\int \dfrac{dx}{1+\sqrt{h(x)}}$ over symmetric limits around $x=\pi/4$-type reflections often pair up with their "co-function" version to sum to a constant — a classic trick that avoids ever actually antidifferentiating $\cot$ or $\tan$ raised to a half power.

    Step-by-Step Solution

    1. Let $I=\displaystyle\int_{\pi/6}^{\pi/3}\dfrac{dx}{1+\sqrt{\cot x}}$.
    2. Since $\pi/6+\pi/3=\pi/2$, substitute $x\to\pi/2-x$: $\cot(\pi/2-x)=\tan x$, so $I=\displaystyle\int_{\pi/6}^{\pi/3}\dfrac{dx}{1+\sqrt{\tan x}}$ as well.
    3. Add the two expressions for $I$: $2I=\displaystyle\int_{\pi/6}^{\pi/3}\left[\dfrac1{1+\sqrt{\cot x}}+\dfrac1{1+\sqrt{\tan x}}\right]dx$. …

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12. AP EAPCET 2021Set eng-2021-08-24-FN1 markMCQ
    Q.$\displaystyle\int_0^{\pi} \dfrac{x\tan x}{\sec x + \tan x}\, dx =$ (A) $\dfrac{\pi(\pi-2)}{2}$ (B) $\dfrac{\pi+2}{2}$ (C) $\dfrac{\pi(\pi+2)}{2}$ (D) $\dfrac{\pi-2}{2}$

    ›Reveal solutionSolution

    The integrand times $x$ over $[0,\pi]$ is handled with the symmetry trick $\int_0^a xf(x)dx=\frac{a}{2}\int_0^af(x)dx$ when $f(a-x)=f(x)$.

    Concept and Intuition

    $\tan x/(\sec x+\tan x)$ simplifies to $\sin x/(1+\sin x)$, a function symmetric under $x\to\pi-x$. This symmetry lets us replace the troublesome factor of $x$ with a constant $\pi/2$, reducing the problem to a standard definite integral.

    Step-by-Step Solution

    1. Simplify: $\dfrac{\tan x}{\sec x+\tan x}=\dfrac{\sin x/\cos x}{(1+\sin x)/\cos x}=\dfrac{\sin x}{1+\sin x}=f(x)$.
    2. Check symmetry: $f(\pi-x)=\dfrac{\sin(\pi-x)}{1+\sin(\pi-x)}=\dfrac{\sin x}{1+\sin x}=f(x)$. ✓
    3. By the King's rule, $I=\displaystyle\int_0^\pi xf(x)dx=\dfrac{\pi}{2}\int_0^\pi f(x)dx$.
    4. Compute $J=\displaystyle\int_0^\pi\dfrac{\sin x}{1+\sin x}dx=\int_0^\pi\left[1-\dfrac{1}{1+\sin x}\right]dx=\pi-\int_0^\pi\dfrac{dx}{1+\sin x}$. …

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