int_{0}^{frac{pi}{2}} frac{tan ^{7} x}{cot ^ | vedclass

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(B) Let $ I = \int_{0}^{\pi / 2} \frac{	an ^{7} x}{\cot ^{7} x+	an ^{7} x} d x $. Using the property $ \int_{0}^{a} f(x) d x = \int_{0}^{a} f(a-x) d

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$ \frac{\pi}{4} $

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(B) Let $ I = \int_{0}^{\pi / 2} \frac{	an ^{7} x}{\cot ^{7} x+	an ^{7} x} d x $. Using the property $ \int_{0}^{a} f(x) d x = \int_{0}^{a} f(a-x) d x $,we get: $ I = \int_{0}^{\pi / 2} \frac{	an ^{7}(\frac{\pi}{2}-x)}{\cot ^{7}(\frac{\pi}{2}-x)+	an ^{7}(\frac{\pi}{2}-x)} d x $. Since $ 	an(\frac{\pi}{2}-x) = \cot x $ and $ \cot(\frac{\pi}{2}-x) = 	an x $,we have: $ I = \int_{0}^{\pi / 2} \frac{\cot ^{7} x}{	an ^{7} x+\cot ^{7} x} d x $. Adding the two expressions for $ I $: $ 2I = \int_{0}^{\pi / 2} \frac{	an ^{7} x + \cot ^{7} x}{	an ^{7} x+\cot ^{7} x} d x = \int_{0}^{\pi / 2} 1 d x $. $ 2I = [x]_{0}^{\pi / 2} = \frac{\pi}{2} $. Therefore,$ I = \frac{\pi}{4} $.

$ \int_{0}^{\frac{\pi}{2}} \frac{\tan ^{7} x}{\cot ^{7} x+\tan ^{7} x} d x $ is equal to

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