If int_0^pi frac{d x}{1+2 sin ^2 x}=k,then th | vedclass

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

Question

(C) Let $I = \int_0^\pi \frac{d x}{1+2 \sin ^2 x}$.Since the integrand $f(x) = \frac{1}{1+2 \sin ^2 x}$ satisfies $f(\pi - x) = f(x)$,we can write:$I

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(C) Let $I = \int_0^\pi \frac{d x}{1+2 \sin ^2 x}$.Since the integrand $f(x) = \frac{1}{1+2 \sin ^2 x}$ satisfies $f(\pi - x) = f(x)$,we can write:$I = 2 \int_0^{\pi / 2} \frac{d x}{1+2 \sin ^2 x}$.Dividing the numerator and denominator by $\cos ^2 x$:$I = 2 \int_0^{\pi / 2} \frac{\sec ^2 x}{\sec ^2 x + 2 	an ^2 x} d x = 2 \int_0^{\pi / 2} \frac{\sec ^2 x}{1 + 	an ^2 x + 2 	an ^2 x} d x = 2 \int_0^{\pi / 2} \frac{\sec ^2 x}{1 + 3 	an ^2 x} d x$.Let $	an x = t$,then $\sec ^2 x d x = d t$. When $x = 0, t = 0$ and when $x 	o \pi / 2, t 	o \infty$.$I = 2 \int_0^{\infty} \frac{d t}{1 + 3 t^2} = 2 \int_0^{\infty} \frac{d t}{1 + (\sqrt{3} t)^2}$.Using the formula $\int \frac{dx}{1+a^2x^2} = \frac{1}{a} 	an^{-1}(ax)$:$I = 2 \left[ \frac{1}{\sqrt{3}} 	an^{-1}(\sqrt{3} t) ight]_0^{\infty} = \frac{2}{\sqrt{3}} \left( \frac{\pi}{2} - 0 ight) = \frac{\pi}{\sqrt{3}}$.Given $k = \frac{\pi}{\sqrt{3}} \approx \frac{3.14159}{1.732} \approx 1.81$.The greatest integer less than or equal to $k$ is $\lfloor 1.81 floor = 1$.

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

Similar Questions

The integral $\int_{\frac{-1}{2}}^{\frac{1}{2}} \left([x] + \log_{e}\left(\frac{1+x}{1-x}\right)\right) dx$,where $[x]$ represents the greatest integer function,equals:

$\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\cos^{\frac{3}{2}} x}{\cos^{\frac{3}{2}} x + \sin^{\frac{3}{2}} x} \, dx = $ . . . . . . .

$\int_{-1}^{1} \log \left( \frac{1+x}{1-x} \right) \, dx = $

$ \int_{0}^{\frac{\pi}{2}} \frac{\tan ^{7} x}{\cot ^{7} x+\tan ^{7} x} d x $ is equal to

If $\int_{-a}^a f(x) dx = \int_0^a f(x) dx + \int_0^a g(x) dx$, then $g(x) =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module