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

$\int_0^{\frac{\pi}{4}} \log (1+\tan x) \, dx =$

$\int_{-\pi / 2}^{\pi / 2} \frac{\cos x}{1+e^{x}} d x$ is equal to

$\int_{-1}^{1} \sin^{11} x \, dx$ is equal to

The value of $\int_{-\pi}^{\pi} \frac{\cos^2 x}{1+\alpha^x} \, dx$ for $\alpha > 0$ is

$\int_{ - \pi }^{\pi } {\frac{{2x(1 + \sin x)}}{{1 + {{\cos }^2}x}}dx} $ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module