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Miscellaneous Exercise 4 · Q52

Q.Choose the correct option from the given alternatives: ∫0π/2sin⁡2x(1+cos⁡x)2 dx=\int_0^{\pi/2} \dfrac{\sin^2 x}{(1+\cos x)^2}\,dx = (A) 4−π2\frac{4-\pi}{2} (B) π−42\frac{\pi-4}{2} (C) 4−π24-\frac{\pi}{2} (D) 4+π24+\frac{\pi}{2}

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✓ Free question

sin⁡2x(1+cos⁡x)2=(1−cos⁡x)(1+cos⁡x)(1+cos⁡x)2=1−cos⁡x1+cos⁡x=tan⁡2x2\dfrac{\sin^2x}{(1+\cos x)^2}=\dfrac{(1-\cos x)(1+\cos x)}{(1+\cos x)^2}=\dfrac{1-\cos x}{1+\cos x}=\tan^2\dfrac x2 (half-angle identity).

∫0π/2tan⁡2x2 dx=∫0π/2(sec⁡2x2−1)dx=[2tan⁡x2−x]0π/2=(2tan⁡π4−π2)−0=2−π2=4−π2.\int_0^{\pi/2}\tan^2\frac x2\,dx=\int_0^{\pi/2}\Big(\sec^2\frac x2-1\Big)dx=\Big[2\tan\frac x2-x\Big]_0^{\pi/2}=\big(2\tan\tfrac\pi4-\tfrac\pi2\big)-0=2-\frac\pi2=\frac{4-\pi}2.

✓Final answer

∫0π/2sin⁡2x(1+cos⁡x)2 dx=4−π2\displaystyle\int_0^{\pi/2}\frac{\sin^2x}{(1+\cos x)^2}\,dx=\frac{4-\pi}2 — option (A) (note: the source extraction showed two visually identical option strings for "4 − π/2"; the printed grouping is reconstructed here as 4−π2\frac{4-\pi}{2} since that is the value that is actually reached by direct computation)

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