The value of I = int_{0}^{frac{pi}{4}} tan^{n | vedclass

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(A) Given $I = \int_{0}^{\pi/4} 	an^{n+1} x \, dx + \frac{1}{2} \int_{0}^{\pi/2} 	an^{n-1} \left( \frac{x}{2} ight) \, dx$. In the second integral

Vedclass · Accepted Answer

$\frac{1}{n}$

Vedclass · Answer

(A) Given $I = \int_{0}^{\pi/4} 	an^{n+1} x \, dx + \frac{1}{2} \int_{0}^{\pi/2} 	an^{n-1} \left( \frac{x}{2} ight) \, dx$. In the second integral, let $t = \frac{x}{2}$, then $dx = 2 \, dt$. When $x = 0$, $t = 0$ and when $x = \pi/2$, $t = \pi/4$. Substituting these into the second integral: $I = \int_{0}^{\pi/4} 	an^{n+1} x \, dx + \frac{1}{2} \int_{0}^{\pi/4} 	an^{n-1} t \cdot (2 \, dt) = \int_{0}^{\pi/4} 	an^{n+1} x \, dx + \int_{0}^{\pi/4} 	an^{n-1} x \, dx$. $I = \int_{0}^{\pi/4} 	an^{n-1} x (	an^2 x + 1) \, dx$. Since $	an^2 x + 1 = \sec^2 x$, we have $I = \int_{0}^{\pi/4} 	an^{n-1} x \sec^2 x \, dx$. Let $u = 	an x$, then $du = \sec^2 x \, dx$. When $x = 0$, $u = 0$ and when $x = \pi/4$, $u = 1$. $I = \int_{0}^{1} u^{n-1} \, du = \left[ \frac{u^n}{n} ight]_{0}^{1} = \frac{1}{n}$.

The value of $I = \int_{0}^{\frac{\pi}{4}} \tan^{n+1} x \, dx + \frac{1}{2} \int_{0}^{\frac{\pi}{2}} \tan^{n-1} \left( \frac{x}{2} \right) \, dx$ is

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