Q.By using the properties of definite integrals, evaluate the integral $\int_{0}^{\pi/2}\frac{\cos^5 x}{\sin^5 x+\cos^5 x}\,dx$

Showing the 12 most recent of 48 on this concept.

1. AP EAPCET 2025Set eng-2025-05-23-AN1 markMCQ
   Q.$\displaystyle\int_{-\pi/2}^{\pi/2} \sin^2 x \cos^2 x (\sin x + \cos x)\,dx =$ (A) 0 (B) $\dfrac{2}{15}$ (C) $\dfrac{4}{15}$ (D) $\dfrac{2}{5}$

   ›Reveal solutionSolution

   This tests using odd/even symmetry over a symmetric interval to kill half the integral, then a simple $u=\sin x$ substitution for the rest. The answer is $\dfrac{4}{15}$, option (C).

   Concept and Intuition

   Over a symmetric interval $[-a,a]$, an odd integrand integrates to zero and an even integrand integrates to twice the integral over $[0,a]$. Expanding $(\sin x+\cos x)$ splits the problem cleanly into one odd term and one even term, so only the even term survives — turning a seemingly complicated integral into a single elementary substitution.

   Step-by-Step Solution

   1. Expand: $\sin^2x\cos^2x(\sin x+\cos x)=\sin^3x\cos^2x+\sin^2x\cos^3x$.
   2. $g(x)=\sin^3x\cos^2x$: since $\sin^3(-x)=-\sin^3x$ and $\cos^2(-x)=\cos^2x$, $g(-x)=-g(x)$ — odd. So $\int_{-\pi/2}^{\pi/2}g(x)\,dx=0$.
   3. $h(x)=\sin^2x\cos^3x$: both factors are even, so $h$ is even, and $\int_{-\pi/2}^{\pi/2}h\,dx=2\int_0^{\pi/2}h\,dx$.
   4. Write $h(x)=\sin^2x\cos^2x\cos x=\sin^2x(1-\sin^2x)\cos x$. Let $u=\sin x$, $du=\cos x\,dx$; limits $0\to1$. …

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2. 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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3. 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 $\dfrac{3\pi}{128}$. …

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4. 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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5. AP EAPCET 2023Set eng-2023-05-18-FN1 markMCQ
   Q.$\displaystyle\int_0^{\pi} x\sin^3x\cos^2x\,dx=$ (A) $\dfrac{2\pi}{15}$ (B) $\dfrac{4\pi}{15}$ (C) $\dfrac{\pi}{30}$ (D) $\dfrac{2\pi}{5}$

   ›Reveal solutionSolution

   Using the symmetry property $\int_0^\pi x\,g(x)\,dx=\frac{\pi}{2}\int_0^\pi g(x)\,dx$ (valid since $g(\pi-x)=g(x)$ here) reduces the problem to a simple substitution integral, giving $\dfrac{2\pi}{15}$.

   Concept and Intuition

   Whenever the integrand has the form $x\cdot g(x)$ over $[0,\pi]$ and $g(\pi-x)=g(x)$ (i.e., $g$ is symmetric about $x=\pi/2$), we can use the standard trick: let $I=\int_0^\pi x\,g(x)\,dx$, substitute $x\to\pi-x$ to get $I=\int_0^\pi(\pi-x)g(x)\,dx=\pi\int_0^\pi g(x)dx - I$, so $2I=\pi\int_0^\pi g(x)dx$, i.e. $I=\dfrac{\pi}{2}\int_0^\pi g(x)\,dx$. This avoids ever integrating $x\sin^3x\cos^2x$ directly by parts.

   Step-by-Step Solution

   1. Let $g(x)=\sin^3x\cos^2x$. Check symmetry: $g(\pi-x)=\sin^3(\pi-x)\cos^2(\pi-x)=(\sin x)^3(-\cos x)^2=\sin^3x\cos^2x=g(x)$. ✓
   2. So $\displaystyle\int_0^\pi x\sin^3x\cos^2x\,dx=\dfrac{\pi}{2}\int_0^\pi\sin^3x\cos^2x\,dx$.
   3. Compute $\int_0^\pi \sin^3x\cos^2x\,dx=\int_0^\pi \sin x(1-\cos^2x)\cos^2x\,dx$. Substitute $u=\cos x$, $du=-\sin x\,dx$; limits $x:0\to\pi$ give $u:1\to-1$.
   4. This becomes $\displaystyle\int_{1}^{-1}(1-u^2)u^2(-du)=\int_{-1}^{1}(u^2-u^4)\,du$. …

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6. 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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7. 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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8. AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQ
   Q.$\int_{-\pi}^{\pi} \frac{\cos^2 x}{1+a^x}dx$, ($a > 0$) $=$ (A) $a\pi$ (B) $\frac{\pi}{a}$ (C) $\frac{\pi}{2}$ (D) $2\pi$

   ›Reveal solutionSolution

   This is the classic "King's rule" property $\int_{-a}^{a}\frac{f(x)}{1+b^x}dx=\frac12\int_{-a}^{a}f(x)dx$ for even $f$; here it directly gives $I=\pi/2$, independent of $a$.

   Concept and Intuition

   Whenever an even function $f(x)$ is divided by $1+b^x$ and integrated over a symmetric interval $[-a,a]$, replacing $x\to-x$ swaps $1/(1+b^x)$ with $b^x/(1+b^x)$ — and since these two fractions add to exactly $1$, averaging the original and transformed integral removes the exponential entirely, leaving half of $\int_{-a}^a f(x)dx$. This makes the $a^x$ term a red herring: the answer depends only on $\cos^2x$.

   Step-by-Step Solution

   1. Let $I=\displaystyle\int_{-\pi}^{\pi}\frac{\cos^2x}{1+a^x}dx$.
   2. Substitute $x\to -x$ (valid since limits are symmetric): $I=\displaystyle\int_{-\pi}^{\pi}\frac{\cos^2(-x)}{1+a^{-x}}dx=\int_{-\pi}^{\pi}\frac{\cos^2x}{1+a^{-x}}dx$ (using $\cos^2(-x)=\cos^2x$).
   3. Simplify $\dfrac{1}{1+a^{-x}} = \dfrac{a^x}{a^x+1}$, so $I=\displaystyle\int_{-\pi}^\pi\frac{\cos^2x\cdot a^x}{1+a^x}dx$.
   4. Add the two expressions for $I$: $2I=\displaystyle\int_{-\pi}^\pi\cos^2x\left[\frac{1}{1+a^x}+\frac{a^x}{1+a^x}\right]dx=\int_{-\pi}^\pi\cos^2x\,dx$. …

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9. AP EAPCET 2022Set eng-2022-07-07-AN1 markMCQ
   Q.$\displaystyle\int_{-\pi/4}^{\pi/4} \cos^{-8}x\,dx =$ (A) $\dfrac{14}{15}$ (B) $\dfrac{174}{35}$ (C) $\dfrac{192}{35}$ (D) $\dfrac{198}{35}$

   ›Reveal solutionSolution

   An even-power secant integral, reduced via $\sec^2x=1+\tan^2x$ and the substitution $u=\tan x$. Answer: $\dfrac{192}{35}$.

   Concept and Intuition

   $\cos^{-8}x=\sec^8x$ is an even function, so integrating it over the symmetric interval $[-\pi/4,\pi/4]$ is twice the integral over $[0,\pi/4]$. For even powers of secant, peel off one factor of $\sec^2x$ to pair with the $dx$ (since $d(\tan x)=\sec^2x\,dx$), and rewrite the remaining even power of $\sec x$ using $\sec^2x=1+\tan^2x$ — this turns the whole integral into a simple polynomial in $u=\tan x$.

   Step-by-Step Solution

   1. $\displaystyle\int_{-\pi/4}^{\pi/4}\sec^8x\,dx=2\int_0^{\pi/4}\sec^8x\,dx$ (even integrand, symmetric limits).
   2. Write $\sec^8x=\sec^6x\cdot\sec^2x=(1+\tan^2x)^3\sec^2x$.
   3. Substitute $u=\tan x$, $du=\sec^2x\,dx$; limits $x=0\to u=0$, $x=\pi/4\to u=1$: $\int_0^{\pi/4}\sec^8x\,dx=\int_0^1(1+u^2)^3\,du$.
   4. Expand: $(1+u^2)^3=1+3u^2+3u^4+u^6$.
   5. Integrate: $\int_0^1(1+3u^2+3u^4+u^6)du=\left[u+u^3+\dfrac35u^5+\dfrac17u^7\right]_0^1=1+1+\dfrac35+\dfrac17$. …

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10. AP EAPCET 2025Set eng-2025-05-24-FN1 markMCQ
    Q.$\displaystyle\int_{-2\pi}^{2\pi} \sin^4(2x)\cos^6(2x)\, dx =$ (A) $\dfrac{3\pi}{64}$ (B) $\dfrac{9\pi}{64}$ (C) $\dfrac{9\pi}{35}$ (D) $\dfrac{9\pi}{280}$

    ›Reveal solutionSolution

    Substitute $u=2x$, use periodicity (period $\pi$) to reduce to $8$ copies of a Wallis-formula integral over $[0,\pi/2]$, giving $\frac{3\pi}{64}$.

    Concept and Intuition

    $\sin^4(2x)\cos^6(2x)$ is a periodic function. Rather than grinding through a power-reduction expansion over the full range $[-2\pi,2\pi]$, it's far more efficient to (a) substitute to a clean variable, (b) exploit periodicity to shrink the domain to one period, and (c) use the standard Wallis reduction formula for $\int_0^{\pi/2}\sin^m\cos^n$.

    Step-by-Step Solution

    1. Let $u=2x\Rightarrow du=2\,dx$. As $x$ runs from $-2\pi$ to $2\pi$, $u$ runs from $-4\pi$ to $4\pi$.

    $I=\int_{-2\pi}^{2\pi}\sin^4(2x)\cos^6(2x)\,dx=\frac12\int_{-4\pi}^{4\pi}\sin^4u\cos^6u\,du.$

    2. $\sin^4u\cos^6u$ is unchanged under $u\to u+\pi$ (since $\sin(u+\pi)=-\sin u$, $\cos(u+\pi)=-\cos u$, and both powers are even), so it has period $\pi$.
    3. The interval $[-4\pi,4\pi]$ has length $8\pi=8\times\pi$, i.e. exactly $8$ full periods, so

    $\int_{-4\pi}^{4\pi}\sin^4u\cos^6u\,du = 8\int_{0}^{\pi}\sin^4u\cos^6u\,du.$

    4. On $[0,\pi]$, the function is symmetric about $u=\pi/2$ (since $\sin(\pi-u)=\sin u$ and $\cos(\pi-u)=-\cos u$, and $\cos^6$ is even in sign), so $\int_0^\pi = 2\int_0^{\pi/2}$.
    5. By the Wallis formula (both exponents even): …

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11. AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
    Q.$\displaystyle\int_0^{\pi/2} \dfrac{200\sin x + 100\cos x}{\sin x + \cos x}\, dx =$ (A) $50\pi$ (B) $25\pi$ (C) $75\pi$ (D) $150\pi$

    ›Reveal solutionSolution

    Using the King's rule $x\to a-x$ on $[0,\pi/2]$ and adding the two equal-value forms of the integral gives $I=75\pi$.

    Concept and Intuition

    For $\int_0^{a} f(x)dx$, the substitution $x\to a-x$ leaves the integral's value unchanged but can transform the integrand into a different-looking (but equal-valued) expression. Adding the original and transformed integrands often makes the sines and cosines combine into a constant, collapsing the whole problem to $\int$ of a constant.

    Step-by-Step Solution

    1. Let $I = \displaystyle\int_0^{\pi/2}\dfrac{200\sin x+100\cos x}{\sin x+\cos x}dx$.
    2. Substitute $x\to \dfrac{\pi}{2}-x$: $\sin x\to\cos x$, $\cos x\to\sin x$, and the limits/interval are unchanged, so $I = \displaystyle\int_0^{\pi/2}\dfrac{200\cos x+100\sin x}{\cos x+\sin x}dx$ as well.
    3. Add the two expressions for $I$: $2I = \displaystyle\int_0^{\pi/2}\dfrac{(200\sin x+100\cos x)+(200\cos x+100\sin x)}{\sin x+\cos x}dx = \int_0^{\pi/2}\dfrac{300\sin x+300\cos x}{\sin x+\cos x}dx$. …

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12. AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
    Q.$\displaystyle\int_{-2\pi}^{2\pi} (1+\cos x)^3(1-\cos x)^4\,dx =$ (A) $0$ (B) $5\pi$ (C) $\dfrac{5\pi}{2}$ (D) $\dfrac{5\pi}{4}$

    ›Reveal solutionSolution

    Factoring $(1+\cos x)^3(1-\cos x)^4$ into $\sin^6x(1-\cos x)$ and splitting the integral shows the $\cos x\sin^6x$ part vanishes by symmetry, leaving $4\int_0^\pi\sin^6x\,dx=\frac{5\pi}{4}$.

    Concept and Intuition

    Products of $(1\pm\cos x)$ raised to powers usually simplify via $(1+\cos x)(1-\cos x)=1-\cos^2x=\sin^2x$. Pairing three factors from each gives $\sin^6x$, leaving one leftover $(1-\cos x)$ factor. Splitting that leftover separates the integral into an even, periodic $\sin^6x$ piece and an odd-derivative $\sin^6x\cos x$ piece that integrates to zero over a range where $\sin x$ returns to the same value at both ends.

    Step-by-Step Solution

    1. Rewrite the product:

    $(1+\cos x)^3(1-\cos x)^4 = \big[(1+\cos x)(1-\cos x)\big]^3\cdot(1-\cos x) = (\sin^2x)^3(1-\cos x) = \sin^6x - \sin^6x\cos x$

    2. Split the integral:

    $\int_{-2\pi}^{2\pi}\sin^6x\,dx \;-\; \int_{-2\pi}^{2\pi}\sin^6x\cos x\,dx$

    3. Second integral: $\sin^6x\cos x = \frac{d}{dx}\left(\frac{\sin^7x}{7}\right)$, so

    $\int_{-2\pi}^{2\pi}\sin^6x\cos x\,dx = \left[\frac{\sin^7x}{7}\right]_{-2\pi}^{2\pi} = \frac{0-0}{7} = 0$

    since $\sin(2\pi)=\sin(-2\pi)=0$.

    4. First integral: $\sin^6x$ has period $\pi$ (because $\sin(x+\pi)=-\sin x \Rightarrow \sin^6(x+\pi)=\sin^6x$). The interval $[-2\pi,2\pi]$ has length $4\pi$, i.e. exactly $4$ periods of length $\pi$:

    $\int_{-2\pi}^{2\pi}\sin^6x\,dx = 4\int_0^\pi \sin^6x\,dx$ …

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