Q.By using the properties of definite integrals, evaluate the integral $\int_{0}^{\pi/2}(2\log\sin x-\log\sin 2x)\,dx$

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 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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4. AP EAPCET 2022Set eng-2022-07-05-AN1 markMCQ
   Q.$\displaystyle\int_{-\pi}^{\pi} \dfrac{2x(1 + \sin x)}{1 + \cos^2 x}\,dx =$ (A) $2\pi$ (B) $\pi^2$ (C) $\pi + 2$ (D) $\pi/2$

   ›Reveal solutionSolution

   Splitting the integrand by parity kills the odd piece immediately; the remaining even piece is handled with the classic "$\int_0^\pi x f(\sin x, \cos x)dx=\frac{\pi}{2}\int_0^\pi f(\sin x,\cos x)dx$"-style symmetry trick, collapsing to a simple arctangent integral.

   Concept and Intuition

   Whenever a definite integral is over a symmetric interval like $[-\pi,\pi]$, always check parity first: odd integrands vanish, and even integrands can be doubled and computed over $[0,\pi]$ only. For integrals of the form $\int_0^\pi x\,g(\sin x,\cos x)\,dx$, the substitution $x\to\pi-x$ often converts the "$x$" factor into "$\pi - x$", letting you solve for the integral in terms of a simpler one without the $x$ multiplier.

   Step-by-Step Solution

   1. Write $\dfrac{2x(1+\sin x)}{1+\cos^2x}=\dfrac{2x}{1+\cos^2x}+\dfrac{2x\sin x}{1+\cos^2x}$.
   2. The first term, $\dfrac{2x}{1+\cos^2x}$, is odd (odd/even = odd), so $\int_{-\pi}^\pi$ of it is $0$.
   3. The second term, $\dfrac{2x\sin x}{1+\cos^2x}$, is even (odd$\times$odd = even, divided by even), so

   $\int_{-\pi}^{\pi}\frac{2x\sin x}{1+\cos^2x}dx = 2\int_0^\pi\frac{2x\sin x}{1+\cos^2x}dx=4\int_0^\pi\frac{x\sin x}{1+\cos^2x}dx.$

   4. Let $J=\displaystyle\int_0^\pi\frac{x\sin x}{1+\cos^2x}dx$. Substituting $x\to\pi-x$ (using $\sin(\pi-x)=\sin x,\ \cos(\pi-x)=-\cos x$, and $\cos^2$ unchanged): $J=\int_0^\pi\frac{(\pi-x)\sin x}{1+\cos^2x}dx=\pi\int_0^\pi\frac{\sin x}{1+\cos^2x}dx - J.$ …

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5. AP EAPCET 2023Set eng-2023-05-15-AN1 markMCQ
   Q.$\displaystyle\int_0^{\pi/2} \dfrac{\sin(\pi/4+x)+\sin(3\pi/4+x)}{\cos x+\sin x}dx =$ (A) $\dfrac{\pi}{\sqrt{2}}$ (B) $\dfrac{\pi}{2\sqrt{2}}$ (C) $\dfrac{\pi}{3\sqrt{2}}$ (D) $\dfrac{\pi}{4\sqrt{2}}$

   ›Reveal solutionSolution

   Simplify the numerator using angle-addition identities, then use the classic $x\to\pi/2-x$ symmetry trick (King's rule) to evaluate the definite integral without direct antidifferentiation.

   Concept and Intuition

   Symmetric definite integrals over $[0,\pi/2]$ often simplify beautifully using the substitution $x\to\pi/2-x$, which swaps $\sin$ and $\cos$. Adding the original and transformed integrals frequently collapses to something trivial.

   Step-by-Step Solution

   1. Expand: $\sin(\pi/4+x)=\dfrac{\sqrt2}{2}(\cos x+\sin x)$ and $\sin(3\pi/4+x)=\dfrac{\sqrt2}{2}(\cos x-\sin x)$.
   2. Sum $= \dfrac{\sqrt2}{2}\big[(\cos x+\sin x)+(\cos x-\sin x)\big] = \sqrt2\cos x$.
   3. So $I=\displaystyle\int_0^{\pi/2}\dfrac{\sqrt2\cos x}{\cos x+\sin x}dx$.
   4. Substitute $x\to\pi/2-x$: $I=\displaystyle\int_0^{\pi/2}\dfrac{\sqrt2\sin x}{\sin x+\cos x}dx = J$ (same value, by the King's-rule symmetry). …

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6. 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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7. AP EAPCET 2026Set eng-2026-05-18-FN1 markMCQ
   Q.$\displaystyle\int_0^{\pi/2} \dfrac{x\sin 2x}{1 + 4\cos^2 2x}\, dx =$ (A) $\dfrac{\pi}{8}\left(\mathrm{Tan}^{-1} 2\right)$ (B) $\dfrac{\pi}{4}\left(\mathrm{Tan}^{-1} 2\right)$ (C) $\dfrac{\pi}{2}\left(\mathrm{Tan}^{-1} 2\right)$ (D) $\dfrac{\pi}{8}\left(\mathrm{Tan}^{-1} 4\right)$

   ›Reveal solutionSolution

   Use the King's-rule symmetry $\int_0^a f(x)dx=\int_0^a f(a-x)dx$ to eliminate the explicit $x$, then substitute; the result is $\dfrac\pi8\mathrm{Tan}^{-1}2$.

   Concept and Intuition

   Whenever an integrand is $x$ times a function that is itself symmetric (or anti-symmetric) under $x\to a-x$, replacing $x$ by $a-x$ and adding to the original kills the explicit $x$-dependence, leaving a much simpler integral to evaluate.

   Step-by-Step Solution

   1. Let $I=\displaystyle\int_0^{\pi/2}\frac{x\sin2x}{1+4\cos^22x}\,dx$.
   2. Replace $x\to\frac\pi2-x$: since $\sin(\pi-2x)=\sin2x$ and $\cos(\pi-2x)=-\cos2x$ (so $\cos^2$ is unchanged),

   $I=\int_0^{\pi/2}\frac{\left(\frac\pi2-x\right)\sin2x}{1+4\cos^22x}\,dx=\frac\pi2\int_0^{\pi/2}\frac{\sin2x}{1+4\cos^22x}\,dx - I.$

   3. So $2I=\dfrac\pi2 J$ where $J=\displaystyle\int_0^{\pi/2}\frac{\sin2x}{1+4\cos^22x}\,dx$.
   4. Let $w=\cos2x$, $dw=-2\sin2x\,dx$. Limits: $x=0\Rightarrow w=1$; $x=\pi/2\Rightarrow w=-1$.

   $J=\int_{1}^{-1}\frac{-dw/2}{1+4w^2}=\int_{-1}^{1}\frac{dw/2}{1+4w^2}=\int_0^1\frac{dw}{1+4w^2}$

   (using the evenness of the integrand over the symmetric interval). …

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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 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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10. 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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11. 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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12. 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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