Rajasthan Boardहिन्दी माध्यमकक्षा 12 साइन्सगणितसमाकलन2 Marks
Question
$\int_{\frac{-\pi}{4}}^{\frac{\pi}{4}} \sin^2 x$ dx का मान ज्ञात कीजिए।
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Answer
हम प्रेक्षित करते हैं कि $\sin^2 x$ एक सम फलन है।
इसलिए $\int_{\frac{-\pi}{4}}^{\frac{\pi}{4}} \sin ^{2} x d x =2 \int_{0}^{\frac{\pi}{4}} \sin ^{2} x d x $
$=2 \int_{0}^{\frac{\pi}{4}} \frac{(1-\cos 2 x)}{2} d x =\int_{0}^{\frac{\pi}{4}}(1 - \cos 2x) dx$
$=\left[x-\frac{1}{2} \sin 2 x\right]_{0}^{\frac{\pi}{4}} =\left(\frac{\pi}{4}-\frac{1}{2} \sin \frac{\pi}{2}\right) -0=\frac{\pi}{4}-\frac{1}{2}$
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