Q.Choose the correct answer: The value of $\int_{0}^{\pi/2}\log\left(\frac{4+3\sin x}{4+3\cos x}\right)dx$ is (A) 2 (B) $\frac{3}{4}$ (C) 0 (D) $-2$

Showing the 12 most recent of 48 on this concept.

1. 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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2. 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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3. 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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4. 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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5. AP EAPCET 2025Set eng-2025-05-23-FN1 markMCQ
   Q.$\displaystyle\int_0^{3\pi/2} \dfrac{\cos^3 x}{\cos^3 x+\sin^3 x}\,dx=$ (A) 0 (B) 1 (C) $\dfrac{\pi}{4}$ (D) $\dfrac{3\pi}{4}$

   ›Reveal solutionSolution

   Using the substitution $x\to\frac{3\pi}{2}-x$ swaps the roles of $\cos^3x$ and $\sin^3x$ in the integrand, showing the given integral equals its "sine" counterpart; since the two together give the full interval length, each equals $\frac{3\pi}{4}$.

   Concept and Intuition

   This is a King's-rule-style symmetry trick: $\int_0^{a}f(x)dx=\int_0^af(a-x)dx$. Applying it with $a=3\pi/2$ swaps $\cos x\leftrightarrow-\sin x$ and $\sin x\leftrightarrow -\cos x$, which exactly interchanges the roles of $\cos^3x$ and $\sin^3x$ in the fraction (the minus signs cancel through the cube and the overall ratio), letting us equate the two "complementary" integrals.

   Step-by-Step Solution

   1. Define $I=\displaystyle\int_0^{3\pi/2}\frac{\cos^3x}{\cos^3x+\sin^3x}dx$ and $J=\displaystyle\int_0^{3\pi/2}\frac{\sin^3x}{\cos^3x+\sin^3x}dx$.
   2. Adding: $I+J=\displaystyle\int_0^{3\pi/2}1\,dx=\frac{3\pi}{2}$.
   3. Substitute $x\to\frac{3\pi}{2}-x$ in $I$: $\cos\left(\frac{3\pi}{2}-x\right)=-\sin x$ and $\sin\left(\frac{3\pi}{2}-x\right)=-\cos x$. …

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6. 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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7. 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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8. 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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9. 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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10. 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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11. 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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12. AP EAPCET 2021Set eng-2021-08-24-AN1 markMCQ
    Q.$\displaystyle\int_{-\pi/4}^{\pi/4} x\tan(1+x^2)\,dx=$ (A) 0 (B) $\dfrac{\pi}{4}$ (C) $\dfrac{-\pi}{4}$ (D) 1

    ›Reveal solutionSolution

    This is a symmetric-limits definite integral where checking odd/even parity instantly gives the answer without doing any actual antiderivative work: it is $0$.

    Concept and Intuition

    For $\displaystyle\int_{-a}^{a} f(x)\,dx$: if $f$ is odd ($f(-x)=-f(x)$) the integral is $0$; if $f$ is even it equals $2\int_0^a f(x)dx$. Any function built as (odd function of $x$) $\times$ (function of $x^2$) is automatically odd, because a function of $x^2$ never changes sign under $x\to -x$.

    Step-by-Step Solution

    1. Let $f(x) = x\tan(1+x^2)$.
    2. Replace $x$ by $-x$: $f(-x) = (-x)\tan(1+(-x)^2) = -x\tan(1+x^2) = -f(x)$.
    3. So $f$ is odd on the symmetric interval $\left[-\dfrac{\pi}{4},\dfrac{\pi}{4}\right]$. …

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