Q.$\int_{0}^{2\pi}\operatorname{cosec}^{7}x\,dx=$
(A) $0$
(B) $1$
(C) $4$
(D) $2\pi$

Showing the 12 most recent of 48 on this concept.

1. 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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2. 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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3. 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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4. 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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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 2023Set eng-2023-05-18-AN1 markMCQ
   Q.$\int_{-4\pi}^{4\pi}\tan^9 x\sin^6 x\cos^3 x\,dx=$ (A) $16\times\frac{\pi}{2}$ (B) $8\times\frac{2}{3}$ (C) $16\times\frac{14}{17}\times\frac{12}{15}\times\ldots\times\frac{2}{3}$ (D) $0$

   ›Reveal solutionSolution

   Odd × even × even = odd, and the integral of any odd function over a symmetric interval is zero — no actual antiderivative work is needed.

   Concept and Intuition

   Before grinding through a nasty trig integral, always check parity. $\tan(-x)=-\tan x$ (odd), and raising an odd function to an odd power ($9$) keeps it odd. $\sin(-x)=-\sin x$ raised to an even power ($6$) becomes even, and $\cos(-x)=\cos x$ raised to any power stays even. Odd times even times even is odd, and an odd function's graph is antisymmetric about the origin, so equal positive and negative area cancels exactly over any interval symmetric about $0$.

   Step-by-Step Solution

   1. Let $g(x)=\tan^9x\sin^6x\cos^3x$.
   2. $g(-x)=\tan^9(-x)\sin^6(-x)\cos^3(-x)=(-\tan x)^9(\sin x)^6(\cos x)^3=-\tan^9x\sin^6x\cos^3x=-g(x)$.
   3. So $g$ is an odd function. …

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8. 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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9. 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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10. 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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11. 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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12. AP EAPCET 2022Set eng-2022-07-08-FN1 markMCQ
    Q.$\int_0^{2\pi} \cos mx \cos nx\,dx + \int_{-\pi}^{\pi} \sin mx \cos nx\,dx =$ (A) $0$, if $m = n$ and $m, n \in \mathbf{Z}$ (B) $\pi$ if $m = n$, $m, n \in \mathbf{Z}$ (C) $\pi$ if $m \ne n$, $m, n \in \mathbf{Z}$ (D) $2\pi\ \forall\ m, n \in \mathbf{R}$

    ›Reveal solutionSolution

    This tests Fourier orthogonality of $\cos$ and the odd/even symmetry trick for integrals over $[-\pi,\pi]$. The answer is $\pi$ when $m=n$.

    Concept and Intuition

    When you integrate a product of trig functions over a full period, the result depends on whether the functions are 'in phase' (same frequency) or not. $\cos mx$ and $\cos nx$ are orthogonal on a period unless $m=n$, in which case you're integrating $\cos^2(mx)$, which has a nonzero average. Separately, integrating an ODD function over a symmetric interval like $[-\pi,\pi]$ always gives zero by symmetry — the negative half exactly cancels the positive half — no computation needed.

    Step-by-Step Solution

    1. Second integral first (symmetry shortcut): $f(x)=\sin mx\cos nx$. Since $\sin(-mx)=-\sin mx$ and $\cos(-nx)=\cos nx$, we get $f(-x)=-f(x)$: $f$ is odd. Hence $\int_{-\pi}^{\pi} f(x)\,dx = 0$ for any $m,n$.
    2. First integral: Use the product-to-sum identity $\cos mx\cos nx = \tfrac12[\cos((m-n)x)+\cos((m+n)x)]$.
       • If $m\ne n$ (integers), both $\cos((m-n)x)$ and $\cos((m+n)x)$ complete a whole number of periods over $[0,2\pi]$ (unless $m+n=0$, but that case still integrates a nonzero-frequency cosine unless $m=n=0$), so the integral is $0$. …

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