int_{0}^{frac{pi}{2}} sin^{2} x , dx = | vedclass

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(D) We use the trigonometric identity $\sin^{2} x = \frac{1 - \cos 2x}{2}$.$\int_{0}^{\frac{\pi}{2}} \sin^{2} x \, dx = \int_{0}^{\frac{\pi}{2}} \frac

Vedclass · Accepted Answer

$\frac{\pi}{4}$

Vedclass · Answer

(D) We use the trigonometric identity $\sin^{2} x = \frac{1 - \cos 2x}{2}$.$\int_{0}^{\frac{\pi}{2}} \sin^{2} x \, dx = \int_{0}^{\frac{\pi}{2}} \frac{1 - \cos 2x}{2} \, dx$$= \frac{1}{2} \int_{0}^{\frac{\pi}{2}} (1 - \cos 2x) \, dx$$= \frac{1}{2} \left[ x - \frac{\sin 2x}{2} \right]_{0}^{\frac{\pi}{2}}$$= \frac{1}{2} \left[ \left( \frac{\pi}{2} - \frac{\sin(\pi)}{2} \right) - \left( 0 - \frac{\sin(0)}{2} \right) \right]$$= \frac{1}{2} \left[ \frac{\pi}{2} - 0 - 0 + 0 \right]$$= \frac{\pi}{4}$

$\int_{0}^{\frac{\pi}{2}} \sin^{2} x \, dx =$

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