Q.Evaluate: $\int_{0}^{\pi} x\sin x\cos^2 x\,dx$

Showing the 12 most recent of 48 on this concept.

1. 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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2. 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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3. 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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4. 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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5. AP EAPCET 2021Set eng-2021-08-24-FN1 markMCQ
   Q.$\displaystyle\int_0^{\pi} \dfrac{x\tan x}{\sec x + \tan x}\, dx =$ (A) $\dfrac{\pi(\pi-2)}{2}$ (B) $\dfrac{\pi+2}{2}$ (C) $\dfrac{\pi(\pi+2)}{2}$ (D) $\dfrac{\pi-2}{2}$

   ›Reveal solutionSolution

   The integrand times $x$ over $[0,\pi]$ is handled with the symmetry trick $\int_0^a xf(x)dx=\frac{a}{2}\int_0^af(x)dx$ when $f(a-x)=f(x)$.

   Concept and Intuition

   $\tan x/(\sec x+\tan x)$ simplifies to $\sin x/(1+\sin x)$, a function symmetric under $x\to\pi-x$. This symmetry lets us replace the troublesome factor of $x$ with a constant $\pi/2$, reducing the problem to a standard definite integral.

   Step-by-Step Solution

   1. Simplify: $\dfrac{\tan x}{\sec x+\tan x}=\dfrac{\sin x/\cos x}{(1+\sin x)/\cos x}=\dfrac{\sin x}{1+\sin x}=f(x)$.
   2. Check symmetry: $f(\pi-x)=\dfrac{\sin(\pi-x)}{1+\sin(\pi-x)}=\dfrac{\sin x}{1+\sin x}=f(x)$. ✓
   3. By the King's rule, $I=\displaystyle\int_0^\pi xf(x)dx=\dfrac{\pi}{2}\int_0^\pi f(x)dx$.
   4. Compute $J=\displaystyle\int_0^\pi\dfrac{\sin x}{1+\sin x}dx=\int_0^\pi\left[1-\dfrac{1}{1+\sin x}\right]dx=\pi-\int_0^\pi\dfrac{dx}{1+\sin x}$. …

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6. 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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7. 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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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 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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10. AP EAPCET 2026Set eng-2026-05-18-FN1 markMCQ
    Q.$\displaystyle\int_0^{\pi/2} \dfrac{x\sin 2x}{1 + 4\cos^2 2x}\, dx =$ (A) $\dfrac{\pi}{8}\left(\mathrm{Tan}^{-1} 2\right)$ (B) $\dfrac{\pi}{4}\left(\mathrm{Tan}^{-1} 2\right)$ (C) $\dfrac{\pi}{2}\left(\mathrm{Tan}^{-1} 2\right)$ (D) $\dfrac{\pi}{8}\left(\mathrm{Tan}^{-1} 4\right)$

    ›Reveal solutionSolution

    Use the King's-rule symmetry $\int_0^a f(x)dx=\int_0^a f(a-x)dx$ to eliminate the explicit $x$, then substitute; the result is $\dfrac\pi8\mathrm{Tan}^{-1}2$.

    Concept and Intuition

    Whenever an integrand is $x$ times a function that is itself symmetric (or anti-symmetric) under $x\to a-x$, replacing $x$ by $a-x$ and adding to the original kills the explicit $x$-dependence, leaving a much simpler integral to evaluate.

    Step-by-Step Solution

    1. Let $I=\displaystyle\int_0^{\pi/2}\frac{x\sin2x}{1+4\cos^22x}\,dx$.
    2. Replace $x\to\frac\pi2-x$: since $\sin(\pi-2x)=\sin2x$ and $\cos(\pi-2x)=-\cos2x$ (so $\cos^2$ is unchanged),

    $I=\int_0^{\pi/2}\frac{\left(\frac\pi2-x\right)\sin2x}{1+4\cos^22x}\,dx=\frac\pi2\int_0^{\pi/2}\frac{\sin2x}{1+4\cos^22x}\,dx - I.$

    3. So $2I=\dfrac\pi2 J$ where $J=\displaystyle\int_0^{\pi/2}\frac{\sin2x}{1+4\cos^22x}\,dx$.
    4. Let $w=\cos2x$, $dw=-2\sin2x\,dx$. Limits: $x=0\Rightarrow w=1$; $x=\pi/2\Rightarrow w=-1$.

    $J=\int_{1}^{-1}\frac{-dw/2}{1+4w^2}=\int_{-1}^{1}\frac{dw/2}{1+4w^2}=\int_0^1\frac{dw}{1+4w^2}$

    (using the evenness of the integrand over the symmetric interval). …

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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 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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