If {u_n} = int_0^{pi /4} {{tan ^n}x,dx,} the | vedclass

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

Question

(A) Given ${u_n} = \int_0^{\pi /4} {{	an ^n}x\,dx} $. We can write ${u_n} + {u_{n - 2}} = \int_0^{\pi /4} {{	an ^n}x\,dx} + \int_0^{\pi /4} {{	an ^

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Given ${u_n} = \int_0^{\pi /4} {{	an ^n}x\,dx} $. We can write ${u_n} + {u_{n - 2}} = \int_0^{\pi /4} {{	an ^n}x\,dx} + \int_0^{\pi /4} {{	an ^{n - 2}}x\,dx} $. $= \int_0^{\pi /4} {{	an ^{n - 2}}x(	an ^2x + 1)\,dx} $. Since $	an ^2x + 1 = \sec ^2x$,we have: $= \int_0^{\pi /4} {{	an ^{n - 2}}x\sec ^2x\,dx} $. Let $t = 	an x$,then $dt = \sec ^2x\,dx$. When $x = 0, t = 0$ and when $x = \pi /4, t = 1$. $= \int_0^1 {{t^{n - 2}}\,dt} $. $= \left[ \frac{t^{n - 1}}{n - 1} ight]_0^1 = \frac{1}{n - 1} - 0 = \frac{1}{n - 1}$. Thus,${u_n} + {u_{n - 2}} = \frac{1}{n - 1}$.

If ${u_n} = \int_0^{\pi /4} {{\tan ^n}x\,dx,} $ then ${u_n} + {u_{n - 2}} = $

Similar Questions

$\int_{\pi / 6}^{\pi / 3} \frac{\sin ^{3} x}{\sin ^{3} x+\cos ^{3} x} d x$ is equal to

For $0 \le x \le \frac{\pi}{2}$,the value of $\int_{0}^{\sin^{2}x} \sin^{-1}(\sqrt{t}) \, dt + \int_{0}^{\cos^{2}x} \cos^{-1}(\sqrt{t}) \, dt$ is equal to

If $\int_0^1 {{e^{{x^2}}}(x - \alpha )\,dx = 0,} $ then

By using the properties of definite integrals,evaluate the integral $\int_{0}^{\frac{\pi}{2}} \frac{\sin^{\frac{3}{2}} x}{\sin^{\frac{3}{2}} x + \cos^{\frac{3}{2}} x} dx$.

Let $f$ be a continuous function defined on $[0,1]$ such that $\int_0^1 f^2(x) dx = (\int_0^1 f(x) dx)^2$. Then,the range of $f$

Vedclass Test Series

Exam Paper Generator

Online Exam Module