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\( \int_{0}^{\frac{\pi}{2}} \frac{\tan ^{7} x}{\cot ^{7} x+\tan ^{7} x} d x \) is equal to
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The correct answer is:
\( \frac{I I}{4} \)
Given that, \( \int_{0}^{\pi / 2} \frac{\tan ^{7} x}{\cot ^{7} x+\tan ^{7} x} d x \)
\( =\int_{0}^{\pi / 2} \frac{\tan ^{7} x}{\tan ^{7}\left(\frac{\Pi}{2}-x\right)+\tan ^{7} x} d x \)
Since \( \int_{0}^{4} \frac{f(x)}{f(x f(a-x))} d x=\frac{a}{2} . \) So,
\( \int_{0}^{\pi / 2} \frac{\tan ^{7} x}{\cot ^{7} x+\tan ^{7} x} d x=\frac{\Pi}{4} \)
\( =\int_{0}^{\pi / 2} \frac{\tan ^{7} x}{\tan ^{7}\left(\frac{\Pi}{2}-x\right)+\tan ^{7} x} d x \)
Since \( \int_{0}^{4} \frac{f(x)}{f(x f(a-x))} d x=\frac{a}{2} . \) So,
\( \int_{0}^{\pi / 2} \frac{\tan ^{7} x}{\cot ^{7} x+\tan ^{7} x} d x=\frac{\Pi}{4} \)
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