int_{0}^{frac{pi}{4}} (tan^n x + tan^{n-2} x) | vedclass

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(A) Let $I = \int_{0}^{\pi/4} (	an^n x + 	an^{n-2} x) d(x - [x])$.Since $x - [x] = \{x\}$,where $\{x\}$ is the fractional part of $x$,we have $I = \

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$\frac{1}{n-1}$

Vedclass · Answer

(A) Let $I = \int_{0}^{\pi/4} (	an^n x + 	an^{n-2} x) d(x - [x])$.Since $x - [x] = \{x\}$,where $\{x\}$ is the fractional part of $x$,we have $I = \int_{0}^{\pi/4} (	an^n x + 	an^{n-2} x) d(\{x\})$.For $x \in [0, \pi/4]$,since $0 \le x Thus,$d(\{x\}) = dx$.Therefore,$I = \int_{0}^{\pi/4} (	an^n x + 	an^{n-2} x) dx$.Factoring out $	an^{n-2} x$,we get $I = \int_{0}^{\pi/4} 	an^{n-2} x (	an^2 x + 1) dx$.Using the identity $1 + 	an^2 x = \sec^2 x$,we have $I = \int_{0}^{\pi/4} 	an^{n-2} x \sec^2 x dx$.Let $u = 	an x$,then $du = \sec^2 x dx$.When $x = 0, u = 0$. When $x = \pi/4, u = 1$.So,$I = \int_{0}^{1} u^{n-2} du$.$I = \left[ \frac{u^{n-1}}{n-1} ight]_{0}^{1} = \frac{1^{n-1}}{n-1} - 0 = \frac{1}{n-1}$.

$\int_{0}^{\frac{\pi}{4}} (\tan^n x + \tan^{n-2} x) d(x - [x])$ is : (where $[.]$ denotes the greatest integer function)

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