Evaluate the integral: int frac{cos ^4 x}{lef | vedclass

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

Question

(C) Let $I = \int \frac{\cos ^4 x}{\left(\sin ^2 x+\sin ^{-3} x \cos ^5 x\right)^3} d x$. Factor out $\sin ^2 x$ from the denominator: $I = \int \frac

Vedclass · Accepted Answer

$\frac{1}{10}\left(1+\cot ^5 x\right)^{-2}+C$

Vedclass · Answer

(C) Let $I = \int \frac{\cos ^4 x}{\left(\sin ^2 x+\sin ^{-3} x \cos ^5 x\right)^3} d x$. Factor out $\sin ^2 x$ from the denominator: $I = \int \frac{\cos ^4 x}{\left(\sin ^2 x(1+\sin ^{-5} x \cos ^5 x)\right)^3} d x = \int \frac{\cos ^4 x}{\sin ^6 x(1+\cot ^5 x)^3} d x$. This simplifies to: $I = \int \frac{\cot ^4 x \operatorname{cosec}^2 x}{(1+\cot ^5 x)^3} d x$. Let $1+\cot ^5 x = y$. Then $5 \cot ^4 x(-\operatorname{cosec}^2 x) d x = d y$,which implies $\cot ^4 x \operatorname{cosec}^2 x d x = -\frac{1}{5} d y$. Substituting these into the integral: $I = -\frac{1}{5} \int y^{-3} d y = -\frac{1}{5} \left(\frac{y^{-2}}{-2}\right) + C = \frac{1}{10} y^{-2} + C$. Substituting back $y = 1+\cot ^5 x$: $I = \frac{1}{10}(1+\cot ^5 x)^{-2} + C$.

Evaluate the integral: $\int \frac{\cos ^4 x}{\left(\sin ^2 x+\sin ^{-3} x \cos ^5 x\right)^3} d x$

Similar Questions

If $\int e^x(1+x) \cdot \sec ^2(x e^x) \, dx = f(x) + \text{constant}$,then $f(x)$ is equal to

The value of $\int \frac{x^{2}-1}{x^{4}+3 x^{2}+1} d x$ for $x>0$ is

$\int {{e^{ - x}}{{\csc }^2}(2{e^{ - x}} + 5)} \,dx = $

$\int \frac{x^{3} \sin \left(\tan ^{-1}\left(x^{4}\right)\right)}{1+x^{8}} d x$ is equal to

$\int \frac{x}{1+x^4} \, dx =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module