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

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 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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3. 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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4. 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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5. 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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6. 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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7. 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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8. 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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9. 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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10. AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
    Q.$\displaystyle\int_{\frac{\pi}{2}}^{\frac{4051\pi}{2}} \dfrac{\cos^2 2x}{1+\sin 2x}\,dx =$ (A) $2026\pi$ (B) $2047\pi$ (C) $2027\pi$ (D) $2025\pi$

    ›Reveal solutionSolution

    The Pythagorean identity collapses $\cos^2 2x/(1+\sin2x)$ to $1-\sin2x$; integrating this over the given huge range gives $2025\pi$ after the cosine boundary terms cancel.

    Concept and Intuition

    A fraction like $\dfrac{\cos^2\theta}{1+\sin\theta}$ almost always simplifies using $\cos^2\theta = 1-\sin^2\theta = (1-\sin\theta)(1+\sin\theta)$, cancelling the $(1+\sin\theta)$ factor. This turns a seemingly hard rational-trig integral into a trivial polynomial-in-trig integral. The huge integration range is a distractor meant to make direct integration look unpleasant — the simplification removes that entirely.

    Step-by-Step Solution

    1. Simplify the integrand:

    $\frac{\cos^2 2x}{1+\sin 2x} = \frac{1-\sin^2 2x}{1+\sin2x} = \frac{(1-\sin2x)(1+\sin2x)}{1+\sin2x} = 1-\sin2x$

    (this holds wherever $1+\sin2x\neq 0$, which is almost everywhere, so it doesn't affect the definite integral).

    2. Integrate: $\displaystyle\int (1-\sin2x)\,dx = x + \frac{1}{2}\cos2x + C$.

    3. Evaluate at the upper limit $x = \dfrac{4051\pi}{2}$: here $2x = 4051\pi$. Since $4051$ is odd, $\cos(4051\pi) = -1$.

    So the antiderivative's value is $\dfrac{4051\pi}{2} - \dfrac12$.

    4. Evaluate at the lower limit $x=\dfrac{\pi}{2}$: here $2x=\pi$, so $\cos\pi=-1$. …

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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 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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