If int_{0}^{pi/2} tan^{n}(x) dx = k int_{0}^{ | vedclass

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(A) Let $I = \int_{0}^{\pi/2} 	an^{n}(x) dx$. Using the property of definite integration $\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx$,we replace $

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$k = 1$

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(A) Let $I = \int_{0}^{\pi/2} 	an^{n}(x) dx$. Using the property of definite integration $\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx$,we replace $x$ with $\frac{\pi}{2} - x$. $I = \int_{0}^{\pi/2} 	an^{n}(\frac{\pi}{2} - x) dx$. Since $	an(\frac{\pi}{2} - x) = \cot(x)$,we have $I = \int_{0}^{\pi/2} \cot^{n}(x) dx$. Comparing this with the given equation $\int_{0}^{\pi/2} 	an^{n}(x) dx = k \int_{0}^{\pi/2} \cot^{n}(x) dx$,we see that $I = k \cdot I$. Therefore,$k = 1$.

If $\int_{0}^{\pi/2} \tan^{n}(x) dx = k \int_{0}^{\pi/2} \cot^{n}(x) dx$,then

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