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Question of 373

Q.∫sin⁡2x2 dx=\int\sin^2\frac{x}{2}\,dx =

(a) 12x−12sin⁡x+k\frac{1}{2}x - \frac{1}{2}\sin x + k
(b) 12x−12cos⁡x+k\frac{1}{2}x - \frac{1}{2}\cos x + k
(c) 12sin⁡x+k\frac{1}{2}\sin x + k
(d) −12sin⁡x+k-\frac{1}{2}\sin x + k
Bihar BsebBihar Board Intermediate 2026MCQ· 1mImportance★★★★★
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Apply the half-angle identity sin⁡2x2=1−cos⁡x2\sin^2\tfrac{x}{2}=\tfrac{1-\cos x}{2}, then integrate.

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