int_0^pi x sin^3 x cos^2 x , dx = | vedclass

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

Question

(A) ધારો કે $I = \int_0^\pi x \sin^3 x \cos^2 x \, dx$  ...$(i)$ગુણધર્મ $\int_0^a f(x) \, dx = \int_0^a f(a-x) \, dx$ નો ઉપયોગ કરતા:$I = \int_0^\pi (\

Vedclass · Accepted Answer

$\frac{2\pi}{15}$

Vedclass · Answer

(A) ધારો કે $I = \int_0^\pi x \sin^3 x \cos^2 x \, dx$  ...$(i)$ગુણધર્મ $\int_0^a f(x) \, dx = \int_0^a f(a-x) \, dx$ નો ઉપયોગ કરતા:$I = \int_0^\pi (\pi - x) \sin^3(\pi - x) \cos^2(\pi - x) \, dx$કારણ કે $\sin(\pi - x) = \sin x$ અને $\cos(\pi - x) = -\cos x$,તેથી:$I = \int_0^\pi (\pi - x) \sin^3 x \cos^2 x \, dx$  ...(ii)$(i)$ અને (ii) નો સરવાળો કરતા:$2I = \int_0^\pi \pi \sin^3 x \cos^2 x \, dx = \pi \int_0^\pi (1 - \cos^2 x) \cos^2 x \sin x \, dx$ધારો કે $t = \cos x$,તો $dt = -\sin x \, dx$. સીમાઓ $0 	o \pi$ થી બદલાઈને $1 	o -1$ થશે:$2I = \pi \int_1^{-1} (1 - t^2) t^2 (-dt) = \pi \int_{-1}^1 (t^2 - t^4) \, dt$$2I = \pi \left[ \frac{t^3}{3} - \frac{t^5}{5} ight]_{-1}^1 = \pi \left( (\frac{1}{3} - \frac{1}{5}) - (-\frac{1}{3} + \frac{1}{5}) ight) = \pi \left( \frac{2}{15} + \frac{2}{15} ight) = \frac{4\pi}{15}$$I = \frac{2\pi}{15}$

$\int_0^\pi x \sin^3 x \cos^2 x \, dx =$

Similar Questions

જો $I_{n}=\int_0^{\frac{\pi}{4}} \tan ^n x \, dx$ હોય,તો $I_{13}+I_{11}=$

સંકલન $\int_{0}^{\pi / 2} \frac{1}{1+(\tan x)^{-101}} d x$ નું મૂલ્ય કેટલું થાય?

$\int_{0}^{1} \log \left(\frac{1}{x}-1\right) d x$ ની કિંમત શોધો.

$\int_0^2 x^8\left(\frac{4}{x^2}-1\right)^{5 / 2} d x=$

ધારો કે $u = \int\limits_0^1 {\frac{{\ln (x + 1)}}{{{x^2} + 1}}} \,dx$ અને $v = \int\limits_0^{\frac{\pi }{2}} {\ln (\sin 2x)} \,dx$,તો:

Vedclass Test Series

Exam Paper Generator

Online Exam Module