int_{0}^{frac{pi}{2}} frac{sin ^{1000} x}{si | vedclass

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(D) Let $ I = \int_{0}^{\frac{\pi}{2}} \frac{\sin ^{1000} x}{\sin ^{1000} x+\cos ^{1000} x} \, dx $. Using the property $ \int_{0}^{a} f(x) \, dx = \i

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$ \frac{\pi}{4} $

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(D) Let $ I = \int_{0}^{\frac{\pi}{2}} \frac{\sin ^{1000} x}{\sin ^{1000} x+\cos ^{1000} x} \, dx $. Using the property $ \int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a-x) \, dx $,we get: $ I = \int_{0}^{\frac{\pi}{2}} \frac{\sin ^{1000} (\frac{\pi}{2}-x)}{\sin ^{1000} (\frac{\pi}{2}-x)+\cos ^{1000} (\frac{\pi}{2}-x)} \, dx $ Since $ \sin(\frac{\pi}{2}-x) = \cos x $ and $ \cos(\frac{\pi}{2}-x) = \sin x $,we have: $ I = \int_{0}^{\frac{\pi}{2}} \frac{\cos ^{1000} x}{\cos ^{1000} x+\sin ^{1000} x} \, dx $ Adding the two expressions for $ I $: $ 2I = \int_{0}^{\frac{\pi}{2}} \frac{\sin ^{1000} x + \cos ^{1000} x}{\sin ^{1000} x+\cos ^{1000} x} \, dx $ $ 2I = \int_{0}^{\frac{\pi}{2}} 1 \, dx = [x]_{0}^{\frac{\pi}{2}} = \frac{\pi}{2} $ Therefore,$ I = \frac{\pi}{4} $.

$ \int_{0}^{\frac{\pi}{2}} \frac{\sin ^{1000} x}{\sin ^{1000} x+\cos ^{1000} x} \, dx $ is equal to

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