int frac{cos^3 x}{sin^2 x + sin x} dx = | vedclass

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

Question

(A) माना $I = \int \frac{\cos^3 x}{\sin^2 x + \sin x} dx$ है।$\cos^2 x = 1 - \sin^2 x$ प्रतिस्थापित करने पर:$I = \int \frac{(1 - \sin^2 x) \cos x}{\si

Vedclass · Accepted Answer

$\log|\sin x| - \sin x + c$

Vedclass · Answer

(A) माना $I = \int \frac{\cos^3 x}{\sin^2 x + \sin x} dx$ है।$\cos^2 x = 1 - \sin^2 x$ प्रतिस्थापित करने पर:$I = \int \frac{(1 - \sin^2 x) \cos x}{\sin^2 x + \sin x} dx$।माना $u = \sin x$, तब $du = \cos x dx$ है।$I = \int \frac{1 - u^2}{u^2 + u} du = \int \frac{(1 - u)(1 + u)}{u(u + 1)} du$।$I = \int \frac{1 - u}{u} du = \int (\frac{1}{u} - 1) du$।$I = \log|u| - u + c$।$u = \sin x$ वापस रखने पर, हमें $I = \log|\sin x| - \sin x + c$ प्राप्त होता है।

$\int \frac{\cos^3 x}{\sin^2 x + \sin x} dx =$

Similar Questions

$\int \frac{d x}{\sin \left(x-\frac{\pi}{3}\right) \cos x} = $

$\int \frac{e^x(1+x)}{\sin ^2(x \cdot e^x)} dx = $ . . . . . . $+ C$.

यदि $0 < x < 1$ और $\int \frac{dx}{\sqrt{x^2-x^5}} = \frac{1}{3} \log |f(x)| + C$ है,तो $f\left(\frac{1}{2}\right) = $

$\int {\frac{{{e^{m{{\tan }^{ - 1}}x}}}}{{1 + {x^2}}}dx} $ का मान क्या है?

$\int \frac{\csc^2 x}{1 + \cot x} dx = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module