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Exercise 7.10 · Q11

Q.By using the properties of definite integrals, evaluate the integral ∫−π/2π/2sin⁡2x dx\int_{-\pi/2}^{\pi/2}\sin^2 x\,dx

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Appeared in past exams:COMEDK 2023· Set 2023-M· 1mexact
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sin⁡2x\sin^2 x is an even function, so the integral over [−π/2,π/2][-\pi/2,\pi/2] is twice the integral over [0,π/2][0,\pi/2]; using sin⁡2x=1−cos⁡2x2\sin^2 x=\dfrac{1-\cos 2x}{2} gives the value π2\dfrac{\pi}{2}.

We evaluate ∫−π/2π/2sin⁡2x dx\displaystyle\int_{-\pi/2}^{\pi/2}\sin^2 x\,dx.

1. Use evenness. Since sin⁡2(−x)=sin⁡2x\sin^2(-x)=\sin^2 x, the integrand is even, so

∫−π/2π/2sin⁡2x dx=2∫0π/2sin⁡2x dx.\int_{-\pi/2}^{\pi/2}\sin^2 x\,dx=2\int_{0}^{\pi/2}\sin^2 x\,dx.

2. Apply the identity sin⁡2x=1−cos⁡2x2\sin^2 x=\dfrac{1-\cos 2x}{2}: …

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