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Q.Evaluate ∫ sin^2 x dx.

Chhattisgarh CgbseCGBSE Intermediate Board 2025Subjective· 2mImportance★★★★★
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Rewrite sin⁡2x\sin^2 x using the power-reduction identity sin⁡2x=1−cos⁡2x2\sin^2x = \frac{1-\cos2x}{2}, then integrate term by term.

sin⁡2x=1−cos⁡2x2\sin^2 x = \dfrac{1-\cos 2x}{2}

∫sin⁡2x dx=∫1−cos⁡2x2 dx=12∫1 dx−12∫cos⁡2x dx\int \sin^2 x\,dx = \int \dfrac{1-\cos 2x}{2}\,dx = \dfrac{1}{2}\int 1\,dx - \dfrac{1}{2}\int \cos 2x\,dx

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