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

$\int \frac{\log (\cot x)}{\sin 2 x} \,d x=$

$\int \frac{(5 \sin \theta-2) \cos \theta}{(5-\cos ^2 \theta-4 \sin \theta)} d \theta=$

$\int \frac{x}{\sqrt{x+4}} \, dx = $ . . . . . . $+ C, x > -4$.

Integrate the function: $\frac{1}{x^{\frac{1}{2}}+x^{\frac{1}{3}}}$

Let the equation of the curve passing through the point $(0,1)$ be given by $y=\int x^3 e^{x^4} d x$. If the equation of the curve is written in the form $x=f(y)$,then $f(y)=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module