If f(t) = int_0^t tan^{(2n-1)} x , dx,n in N, | vedclass

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(C) Given $f(t) = \int_0^t 	an^{(2n-1)} x \, dx$. We need to evaluate $f(t+\pi) = \int_0^{t+\pi} 	an^{(2n-1)} x \, dx$. Using the property of defini

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$f(t) + f(\pi)$

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(C) Given $f(t) = \int_0^t 	an^{(2n-1)} x \, dx$. We need to evaluate $f(t+\pi) = \int_0^{t+\pi} 	an^{(2n-1)} x \, dx$. Using the property of definite integrals,$\int_0^{t+\pi} = \int_0^t + \int_t^{t+\pi}$. So,$f(t+\pi) = \int_0^t 	an^{(2n-1)} x \, dx + \int_t^{t+\pi} 	an^{(2n-1)} x \, dx$. Since $	an x$ has a period of $\pi$,$\int_t^{t+\pi} 	an^{(2n-1)} x \, dx = \int_0^{\pi} 	an^{(2n-1)} x \, dx$. Thus,$f(t+\pi) = f(t) + \int_0^{\pi} 	an^{(2n-1)} x \, dx$. Since $f(\pi) = \int_0^{\pi} 	an^{(2n-1)} x \, dx$,we have $f(t+\pi) = f(t) + f(\pi)$.

If $f(t) = \int_0^t \tan^{(2n-1)} x \, dx$,$n \in N$,then $f(t+\pi) =$

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