The value of int frac{cos ^3 x}{sin ^2 x+sin | vedclass

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

Question

(A) Let $I = \int \frac{\cos ^3 x}{\sin ^2 x+\sin x} \,d x$. Using the identity $\cos^2 x = 1 - \sin^2 x$, we have: $I = \int \frac{(1-\sin^2 x) \cos

Vedclass · Accepted Answer

$\log (\sin x)-\sin x+C$, where $C$ is a constant of integration.

Vedclass · Answer

(A) Let $I = \int \frac{\cos ^3 x}{\sin ^2 x+\sin x} \,d x$. Using the identity $\cos^2 x = 1 - \sin^2 x$, we have: $I = \int \frac{(1-\sin^2 x) \cos x}{\sin x(1+\sin x)} \,d x$. Since $1-\sin^2 x = (1-\sin x)(1+\sin x)$, the expression simplifies to: $I = \int \frac{(1-\sin x)(1+\sin x) \cos x}{\sin x(1+\sin x)} \,d x = \int \frac{1-\sin x}{\sin x} \cos x \,d x$. Substitute $t = \sin x$, then $dt = \cos x \,d x$. $I = \int \left(\frac{1}{t} - 1\right) dt$. Integrating with respect to $t$: $I = \log |t| - t + C$. Substituting back $t = \sin x$: $I = \log |\sin x| - \sin x + C$.

The value of $\int \frac{\cos ^3 x}{\sin ^2 x+\sin x} \,d x$ is

Similar Questions

Integrate the function $\frac{e^{2x}-1}{e^{2x}+1}$.

If $\int \log \left(6 \sin ^2 x+17 \sin x+12\right)^{\cos x} d x=f(x)+c$ then,$f\left(\frac{\pi}{2}\right)=$

$\int \frac{d x}{x^{2 / 3}\left(1+x^{2 / 3}\right)}=$

$\int \frac{x^2-1}{x^3 \sqrt{2 x^4-2 x^2+1}} d x=$

The value of the integral $\int \frac{d x}{\left(e^x+e^{-x}\right)^2}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module