If {I_n} = int_0^{pi /4} {{tan ^n}theta ,dthe | vedclass

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

Question

(A) Given ${I_n} = \int_0^{\pi /4} {{	an ^n}	heta \,d	heta }$.Consider ${I_{n + 1}} + {I_{n - 1}} = \int_0^{\pi /4} {{	an ^{n + 1}}	heta \,d	het

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(A) Given ${I_n} = \int_0^{\pi /4} {{	an ^n}	heta \,d	heta }$.Consider ${I_{n + 1}} + {I_{n - 1}} = \int_0^{\pi /4} {{	an ^{n + 1}}	heta \,d	heta } + \int_0^{\pi /4} {{	an ^{n - 1}}	heta \,d	heta }$.$= \int_0^{\pi /4} {{	an ^{n - 1}}	heta (	an ^2	heta + 1)d	heta }$.$= \int_0^{\pi /4} {{	an ^{n - 1}}	heta \sec ^2	heta \,d	heta }$.Let $u = 	an 	heta$,then $du = \sec ^2	heta \,d	heta$. When $	heta = 0, u = 0$ and when $	heta = \pi /4, u = 1$.$= \int_0^1 {{u^{n - 1}}du} = \left[ \frac{u^n}{n} ight]_0^1 = \frac{1}{n}$.Therefore,$n({I_{n - 1}} + {I_{n + 1}}) = 1$.

If ${I_n} = \int_0^{\pi /4} {{\tan ^n}\theta \,d\theta ,} $ then for any positive integer $n,$ the value of $n({I_{n - 1}} + {I_{n + 1}})$ is

Similar Questions

$\int_{5 \pi}^{25 \pi}|\sin 2 x+\cos 2 x| d x=$ (in $\sqrt{2}$)

Assertion $(A)$: $\int_{\frac{\pi}{2}}^{\frac{3 \pi}{2}} [2 \sin x] dx = 0$,where $[.]$ denotes the greatest integer function.
Reason $(R)$: $2 \sin x$ is a decreasing function in $\left[\frac{\pi}{2}, \frac{3 \pi}{2}\right]$.

The value of the integral $\int_{-2}^{2} \frac{|x^{3}+x|}{e^{x|x|}+1} dx$ is equal to

$\int_{\,\pi /6}^{\,\pi /3} {\,\frac{{dx}}{{1 + \sqrt {\cot x} }}} $ is equal to:

The value of $\int_{-1 / 2}^{1 / 2} \cos ^{-1} x \, dx$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module