Q.The value of $\int_{-\pi}^{\pi} \sin 3x\,\cos 2x\,dx$ is _______.

Showing the 12 most recent of 48 on this concept.

1. AP EAPCET 2022Set eng-2022-07-05-AN1 markMCQ
   Q.$\displaystyle\int_{-\pi}^{\pi} \dfrac{2x(1 + \sin x)}{1 + \cos^2 x}\,dx =$ (A) $2\pi$ (B) $\pi^2$ (C) $\pi + 2$ (D) $\pi/2$

   ›Reveal solutionSolution

   Splitting the integrand by parity kills the odd piece immediately; the remaining even piece is handled with the classic "$\int_0^\pi x f(\sin x, \cos x)dx=\frac{\pi}{2}\int_0^\pi f(\sin x,\cos x)dx$"-style symmetry trick, collapsing to a simple arctangent integral.

   Concept and Intuition

   Whenever a definite integral is over a symmetric interval like $[-\pi,\pi]$, always check parity first: odd integrands vanish, and even integrands can be doubled and computed over $[0,\pi]$ only. For integrals of the form $\int_0^\pi x\,g(\sin x,\cos x)\,dx$, the substitution $x\to\pi-x$ often converts the "$x$" factor into "$\pi - x$", letting you solve for the integral in terms of a simpler one without the $x$ multiplier.

   Step-by-Step Solution

   1. Write $\dfrac{2x(1+\sin x)}{1+\cos^2x}=\dfrac{2x}{1+\cos^2x}+\dfrac{2x\sin x}{1+\cos^2x}$.
   2. The first term, $\dfrac{2x}{1+\cos^2x}$, is odd (odd/even = odd), so $\int_{-\pi}^\pi$ of it is $0$.
   3. The second term, $\dfrac{2x\sin x}{1+\cos^2x}$, is even (odd$\times$odd = even, divided by even), so

   $\int_{-\pi}^{\pi}\frac{2x\sin x}{1+\cos^2x}dx = 2\int_0^\pi\frac{2x\sin x}{1+\cos^2x}dx=4\int_0^\pi\frac{x\sin x}{1+\cos^2x}dx.$

   4. Let $J=\displaystyle\int_0^\pi\frac{x\sin x}{1+\cos^2x}dx$. Substituting $x\to\pi-x$ (using $\sin(\pi-x)=\sin x,\ \cos(\pi-x)=-\cos x$, and $\cos^2$ unchanged): $J=\int_0^\pi\frac{(\pi-x)\sin x}{1+\cos^2x}dx=\pi\int_0^\pi\frac{\sin x}{1+\cos^2x}dx - J.$ …

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2. 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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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 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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5. 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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6. AP EAPCET 2026Set eng-2026-05-13-AN1 markMCQ
   Q.$\displaystyle\int_{-3\pi/2}^{-\pi/2} \left((x+\pi)^3 + \cos^2(x+3\pi)\right) dx =$ (A) $\dfrac{\pi}{8}$ (B) $\dfrac{\pi}{2}$ (C) $\dfrac{\pi}{4}-1$ (D) $\dfrac{\pi^4}{32}$

   ›Reveal solutionSolution

   Shifting the variable to $u=x+\pi$ turns the limits into a symmetric interval about zero; the odd cubic term vanishes and only the even $\cos^2u$ term survives, giving $\pi/2$.

   Concept and Intuition

   Whenever an integral's limits and integrand both have a shift-symmetry, substituting to center the interval at 0 lets us exploit odd/even function properties: odd functions integrate to zero over $[-a,a]$, and even functions can be doubled over $[0,a]$.

   Step-by-Step Solution

   1. Let $u = x+\pi$, so $du=dx$. When $x=-3\pi/2$, $u=-\pi/2$; when $x=-\pi/2$, $u=\pi/2$.
   2. $\cos^2(x+3\pi) = \cos^2((x+\pi)+2\pi) = \cos^2(u+2\pi) = \cos^2u$ (cosine has period $2\pi$).
   3. The integral becomes $\displaystyle\int_{-\pi/2}^{\pi/2}\left(u^3+\cos^2u\right)du$.
   4. $u^3$ is an odd function, so $\displaystyle\int_{-\pi/2}^{\pi/2}u^3\,du = 0$.
   5. $\cos^2u$ is even, so $\displaystyle\int_{-\pi/2}^{\pi/2}\cos^2u\,du = 2\int_0^{\pi/2}\cos^2u\,du$. …

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7. 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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8. 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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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 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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11. 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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12. 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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