Q.Prove that $\int_{-1}^{1}x^{17}\cos^4 x\,dx=0$

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 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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3. 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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4. 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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5. 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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6. AP EAPCET 2024Set eng-2024-05-21-AN1 markMCQ
   Q.$\displaystyle\int_{-1}^{1} \left(\sqrt{1+x+x^2} - \sqrt{1-x+x^2}\right) dx =$ (A) $2$ (B) $4$ (C) $0$ (D) $8$

   ›Reveal solutionSolution

   The integrand is an odd function of $x$, so its integral over the symmetric interval $[-1,1]$ vanishes.

   Concept and Intuition

   $\int_{-a}^a f(x)dx=0$ whenever $f$ is odd, i.e. $f(-x)=-f(x)$. Recognising this symmetry avoids a painful direct integration of the square roots.

   Step-by-Step Solution

   1. Let $g(x)=\sqrt{1+x+x^2}$. Then $g(-x)=\sqrt{1-x+x^2}$.
   2. The integrand is $f(x)=g(x)-g(-x)$.
   3. Check parity: $f(-x) = g(-x)-g(x) = -\big(g(x)-g(-x)\big) = -f(x)$ — so $f$ is odd.
   4. Hence $\int_{-1}^1 f(x)\,dx = 0$. …

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7. 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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8. 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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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 2024Set eng-2024-05-19-AN1 markMCQ
    Q.$\displaystyle\int_{0}^{\pi} x\sin^4 x\cos^6 x\,dx=$ (A) $\dfrac{3\pi^2}{512}$ (B) $\dfrac{3\pi^2}{256}$ (C) $\dfrac{\pi^2}{256}$ (D) $\dfrac{\pi^2}{512}$

    ›Reveal solutionSolution

    Using the King's-rule symmetry $\int_0^\pi xf(x)dx=\frac\pi2\int_0^\pi f(x)dx$ (valid because $\cos^6$ is unaffected by the sign flip under $x\to\pi-x$) reduces the problem to a standard Wallis-formula integral, giving $\dfrac{3\pi^2}{512}$.

    Concept and Intuition

    Whenever an integral has the form $\int_0^\pi x\,f(x)\,dx$ and $f(\pi-x)=f(x)$ (i.e. $f$ is symmetric about the midpoint $x=\pi/2$), substituting $x\to\pi-x$ shows the integral equals $\dfrac\pi2\int_0^\pi f(x)dx$ — the "$x$" essentially averages out to $\pi/2$. Here $f(x)=\sin^4x\cos^6x$ is such a function because raising $\cos x$ to an even power erases the sign flip from $\cos(\pi-x)=-\cos x$.

    Step-by-Step Solution

    1. Let $I=\displaystyle\int_0^\pi x\sin^4x\cos^6x\,dx$ and $f(x)=\sin^4x\cos^6x$.
    2. Substituting $x\to\pi-x$: $\sin(\pi-x)=\sin x$ and $\cos(\pi-x)=-\cos x$, so $f(\pi-x)=\sin^4x\,(-\cos x)^6=\sin^4x\cos^6x=f(x)$ (even power kills the sign).
    3. So $I = \displaystyle\int_0^\pi(\pi-x)f(x)dx = \pi\int_0^\pi f(x)dx - I \Rightarrow 2I=\pi\int_0^\pi f(x)dx \Rightarrow I=\dfrac\pi2\int_0^\pi f(x)dx$.
    4. By symmetry about $x=\pi/2$ (again since $f(\pi-x)=f(x)$), $\displaystyle\int_0^\pi f(x)dx = 2\int_0^{\pi/2}\sin^4x\cos^6x\,dx$. …

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12. AP EAPCET 2022Set eng-2022-07-05-AN1 markMCQ
    Q.$\displaystyle\int_{-1}^{1} \dfrac{\sin x - x^2}{3 - |x|}\,dx =$ (A) $7 + 18\log(3/2)$ (B) $18\log(9/4)$ (C) $7 + 9\log(9/4)$ (D) $7 - 18\log(3/2)$

    ›Reveal solutionSolution

    The $\sin x$ part is an odd function over a symmetric interval and vanishes; only the even $x^2$ part survives, reducing to a straightforward rational-function integral solved via polynomial division.

    Concept and Intuition

    $\sin x$ is odd, $x^2$ is even, and $3-|x|$ is even, so $\dfrac{\sin x}{3-|x|}$ is odd (integrates to $0$ over $[-1,1]$) while $\dfrac{x^2}{3-|x|}$ is even (double the integral over $[0,1]$, where $|x|=x$).

    Step-by-Step Solution

    1. Split: $\displaystyle\int_{-1}^1\frac{\sin x - x^2}{3-|x|}dx = \int_{-1}^1\frac{\sin x}{3-|x|}dx - \int_{-1}^1\frac{x^2}{3-|x|}dx$.
    2. First integral $=0$ (odd integrand over symmetric limits).
    3. Second integral: even integrand, so $=2\displaystyle\int_0^1\frac{x^2}{3-x}dx$ (using $|x|=x$ for $x\in[0,1]$).
    4. Polynomial division: $\dfrac{x^2}{3-x}=-x-3+\dfrac{9}{3-x}$ (check: $(-x-3)(3-x)=x^2-9$, so $x^2=(-x-3)(3-x)+9$).
    5. Integrate: $\displaystyle\int_0^1\left(-x-3+\frac{9}{3-x}\right)dx = \left[-\frac{x^2}{2}-3x-9\log(3-x)\right]_0^1$.
    6. At $x=1$: $-\dfrac12-3-9\log2 = -\dfrac72-9\log2$. At $x=0$: $-9\log3$. …

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