int_0^pi frac{x sin x}{1+cos ^2 x} d x= | vedclass

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

Question

(A) ધારો કે $I = \int_0^\pi \frac{x \sin x}{1+\cos ^2 x} d x$. ગુણધર્મ $\int_0^a f(x) dx = \int_0^a f(a-x) dx$ નો ઉપયોગ કરતા,આપણને મળે છે: $I = \int_0

Vedclass · Accepted Answer

$\frac{\pi^2}{4}$

Vedclass · Answer

(A) ધારો કે $I = \int_0^\pi \frac{x \sin x}{1+\cos ^2 x} d x$. ગુણધર્મ $\int_0^a f(x) dx = \int_0^a f(a-x) dx$ નો ઉપયોગ કરતા,આપણને મળે છે: $I = \int_0^\pi \frac{(\pi-x) \sin(\pi-x)}{1+\cos^2(\pi-x)} dx = \int_0^\pi \frac{(\pi-x) \sin x}{1+\cos^2 x} dx$. $I = \pi \int_0^\pi \frac{\sin x}{1+\cos^2 x} dx - \int_0^\pi \frac{x \sin x}{1+\cos^2 x} dx$. $I = \pi \int_0^\pi \frac{\sin x}{1+\cos^2 x} dx - I$. $2I = \pi \int_0^\pi \frac{\sin x}{1+\cos^2 x} dx$. ધારો કે $u = \cos x$,તો $du = -\sin x dx$. જ્યારે $x=0, u=1$; જ્યારે $x=\pi, u=-1$. $2I = \pi \int_1^{-1} \frac{-du}{1+u^2} = \pi \int_{-1}^1 \frac{du}{1+u^2}$. $2I = \pi [	an^{-1} u]_{-1}^1 = \pi (	an^{-1}(1) - 	an^{-1}(-1)) = \pi (\frac{\pi}{4} - (-\frac{\pi}{4})) = \pi (\frac{\pi}{2}) = \frac{\pi^2}{2}$. $I = \frac{\pi^2}{4}$.

$\int_0^\pi \frac{x \sin x}{1+\cos ^2 x} d x=$

Similar Questions

જો $g(x) = \int_0^x \cos^4 t \,dt$ હોય, તો $g(x+\pi)$ ની કિંમત શું થાય?

$\int_{e^{-1}}^{e^2} \left| \frac{\log_e x}{x} \right| dx$ નું મૂલ્ય શોધો.

$\int_{-a}^{a} \sin x \, f(\cos x) \, dx = $

$\int_0^1 \frac{\log x}{\sqrt{1 - x^2}} \, dx = $

$\int_0^1 \tan^{-1}(1-x+x^2) dx$ નું મૂલ્ય શોધો.

Vedclass Test Series

Exam Paper Generator

Online Exam Module