Aspire Faculty ID #13530 · Topic: JAMIA MILLIA ISLAMIA MCA 2024 · Just now
JAMIA MILLIA ISLAMIA MCA 2024

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

Solution

$\sin^2 x = \dfrac{1 - \cos 2x}{2} \Rightarrow \int_{0}^{\pi} \sin^2 x,dx = \dfrac{1}{2} \left[ x - \dfrac{\sin 2x}{2} \right]_{0}^{\pi} = \dfrac{1}{2}(\pi - 0) = \dfrac{\pi}{2}$
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