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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

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

Vedclass · Answer

(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 $f(t+\pi) =$

Similar Questions

If $\int_0^\pi \frac{d x}{1+2 \sin ^2 x}=k$,then the greatest integer less than or equal to $k$ is

The value of $\int_{-\pi/6}^{\pi/6} \left( \frac{\pi + 4x^{11}}{1 - \sin(|x| + \pi/6)} \right) dx$ is equal to: (in $\pi$)

$\int_0^{\pi /2} \log(\sin x) \, dx = $

Let $f(x)$ be positive for all real $x$. If $I_1 = \int_{1-h}^{h} x f(x(1-x)) dx$ and $I_2 = \int_{1-h}^{h} f(x(1-x)) dx$,where $(2h-1) > 0$,then $\frac{I_1}{I_2}$ is

The function $F(x) = \int_0^x \log \left( \frac{1 - t}{1 + t} \right) \,dt$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module