Q.Evaluate $\int_0^{\pi/2} \dfrac{\sin^4 x}{\sin^4 x + \cos^4 x}\, dx$

Showing the 12 most recent of 48 on this concept.

1. 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 $\sin\left(\frac{3\pi}{2}-x\right)=-\cos x$. …

   🔒 Start your 14-day free trial to see the full solution →
2. 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 $\displaystyle\int_0^{\pi/2}(1-\sin x\cos x)\,dx$. …

   🔒 Start your 14-day free trial to see the full solution →
3. 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 $J=\displaystyle\int_0^\pi\dfrac{\sin x}{1+\cos^2x}dx$. …

   🔒 Start your 14-day free trial to see the full solution →
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 $2I=\pi\displaystyle\int_0^\pi\frac{\sin x}{1+\cos^2 x}dx$. …

   🔒 Start your 14-day free trial to see the full solution →
5. 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 $J=\displaystyle\int_0^\pi\frac{\sin x}{1+\cos^2x}\,dx$. …

   🔒 Start your 14-day free trial to see the full solution →
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). …

   🔒 Start your 14-day free trial to see the full solution →
7. 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 $\left[-\dfrac{\pi}{4},\dfrac{\pi}{4}\right]$. …

   🔒 Start your 14-day free trial to see the full solution →
8. AP EAPCET 2023Set eng-2023-05-19-FN1 markMCQ
   Q.$\displaystyle\int_0^{\pi/2} \dfrac{\sum_{n=0}^{4} \sin\left(\dfrac{n\pi}{4} + x\right)}{\cos x + \sin x}\, dx =$ (A) $\dfrac{\pi}{2}$ (B) $\dfrac{\pi}{2\sqrt{2}}$ (C) $\dfrac{3\pi}{\sqrt{2}}$ (D) $(\sqrt{2} + 1)\dfrac{\pi}{4}$

   ›Reveal solutionSolution

   The numerator collapses to $(1+\sqrt2)\cos x$ via angle-sum identities, and the resulting integral is a standard "cos/(cos+sin)" symmetric integral equal to $\pi/4$.

   Concept and Intuition

   Sums like $\sum\sin(n\pi/4+x)$ over a few terms are best expanded using known values of $\sin,\cos$ at multiples of $\pi/4$, and the resulting $\cos x/(\cos x+\sin x)$ type integral is solved by the standard symmetric-pair trick: define $I,J$ with numerators swapped, use $I+J$ and $I-J$.

   Step-by-Step Solution

   1. Expand: $n=0$: $\sin x$; $n=1$: $\sin(\pi/4+x)$; $n=2$: $\sin(\pi/2+x)=\cos x$; $n=3$: $\sin(3\pi/4+x)$; $n=4$: $\sin(\pi+x)=-\sin x$.
   2. Sum $=\sin x-\sin x+\cos x+\sin(\pi/4+x)+\sin(3\pi/4+x)=\cos x+[\sin(\pi/4+x)+\sin(3\pi/4+x)]$.
   3. Sum-to-product on the bracket: $2\sin\left(\dfrac{(\pi/4+x)+(3\pi/4+x)}{2}\right)\cos\left(\dfrac{(3\pi/4+x)-(\pi/4+x)}{2}\right)=2\sin\left(\dfrac\pi2+x\right)\cos\left(\dfrac\pi4\right)=2\cos x\cdot\dfrac{\sqrt2}{2}=\sqrt2\cos x$.
   4. Total numerator $=\cos x+\sqrt2\cos x=(1+\sqrt2)\cos x$. …

   🔒 Start your 14-day free trial to see the full solution →
9. 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 $\cos(\pi/2-x)=\sin x$). …

   🔒 Start your 14-day free trial to see the full solution →
10. 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 $K=\displaystyle\int_0^\pi \frac{\sin x}{1+\cos^2x}dx$. …

    🔒 Start your 14-day free trial to see the full solution →
11. AP EAPCET 2025Set eng-2025-05-26-AN1 markMCQ
    Q.$\displaystyle\int_0^{\pi} \left(\sin^5 x \cos^3 x + \sin^4 x \cos^4 x + \sin^3 x \cos^4 x\right) dx =$ (A) $\dfrac{873}{2240}$ (B) $\dfrac{3\pi}{128} + \dfrac{12}{35}$ (C) $\dfrac{1641}{4480}$ (D) $\dfrac{3\pi}{128} + \dfrac{4}{35}$

    ›Reveal solutionSolution

    Use the $x\to\pi-x$ symmetry to kill the odd-cos-power term and double the even-cos-power terms' half-range integrals; a Wallis-formula and direct-substitution computation gives $\frac{3\pi}{128}+\frac4{35}$.

    Concept and Intuition

    For $f(x)=\sin^a x\cos^b x$, substituting $x\to\pi-x$ gives $\sin(\pi-x)=\sin x$ but $\cos(\pi-x)=-\cos x$, so $f(\pi-x)=(-1)^b f(x)$. Splitting $\int_0^\pi = \int_0^{\pi/2}+\int_{\pi/2}^\pi$ and substituting in the second piece shows $\int_0^\pi f\,dx = [1+(-1)^b]\int_0^{\pi/2}f\,dx$ — zero if $b$ is odd, doubled if $b$ is even.

    Step-by-Step Solution

    1. Term 1: $\sin^5x\cos^3x$ has $b=3$ (odd) $\Rightarrow \int_0^\pi=0$.
    2. Term 2: $\sin^4x\cos^4x$ has $b=4$ (even) $\Rightarrow \int_0^\pi=2\int_0^{\pi/2}\sin^4x\cos^4x\,dx$. Using $\sin x\cos x=\tfrac12\sin2x$: $\sin^4x\cos^4x=\tfrac1{16}\sin^4(2x)$. $\int_0^{\pi/2}\sin^4(2x)dx=\tfrac12\int_0^\pi\sin^4t\,dt=\tfrac12\cdot2\int_0^{\pi/2}\sin^4t\,dt=\int_0^{\pi/2}\sin^4t\,dt=\dfrac{3!!}{4!!}\cdot\dfrac\pi2=\dfrac38\cdot\dfrac\pi2=\dfrac{3\pi}{16}$. So $\int_0^{\pi/2}\sin^4x\cos^4x\,dx=\tfrac1{16}\cdot\dfrac{3\pi}{16}=\dfrac{3\pi}{256}$, doubled gives $\dfrac{3\pi}{128}$. …

    🔒 Start your 14-day free trial to see the full solution →
12. 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 $I = \displaystyle\int_0^{\pi/4}\log\!\left(\frac{2}{1+\tan x}\right)dx = \int_0^{\pi/4}\log2\,dx - \int_0^{\pi/4}\log(1+\tan x)\,dx = \frac{\pi}{4}\log2 - I$. …

    🔒 Start your 14-day free trial to see the full solution →