int sqrt{frac{cos x - cos^3 x}{1 - cos^3 x}} | vedclass

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

Question

(A) Let $I = \int \sqrt{\frac{\cos x(1 - \cos^2 x)}{1 - \cos^3 x}} \, dx = \int \sqrt{\frac{\cos x \sin^2 x}{1 - \cos^3 x}} \, dx$. Since $\sin x = \s

Vedclass · Accepted Answer

$\frac{2}{3} \cos^{-1}(\cos^{3/2} x)$

Vedclass · Answer

(A) Let $I = \int \sqrt{\frac{\cos x(1 - \cos^2 x)}{1 - \cos^3 x}} \, dx = \int \sqrt{\frac{\cos x \sin^2 x}{1 - \cos^3 x}} \, dx$. Since $\sin x = \sqrt{1 - \cos^2 x}$,we have $I = \int \sin x \sqrt{\frac{\cos x}{1 - \cos^3 x}} \, dx$. Let $u = \cos^{3/2} x$. Then $du = \frac{3}{2} \cos^{1/2} x (-\sin x) \, dx$,which implies $\sin x \sqrt{\cos x} \, dx = -\frac{2}{3} du$. Also,$u^2 = \cos^3 x$. Substituting these into the integral: $I = \int \sqrt{\frac{1}{1 - u^2}} \left(-\frac{2}{3} du\right) = -\frac{2}{3} \int \frac{1}{\sqrt{1 - u^2}} \, du$. This evaluates to $-\frac{2}{3} \sin^{-1}(u) + C$. Since $\sin^{-1}(u) + \cos^{-1}(u) = \frac{\pi}{2}$,we have $-\frac{2}{3} (\frac{\pi}{2} - \cos^{-1}(u)) = \frac{2}{3} \cos^{-1}(u) - \frac{\pi}{3}$. Absorbing the constant $-\frac{\pi}{3}$ into $C$,we get $\frac{2}{3} \cos^{-1}(\cos^{3/2} x) + C$.

$\int \sqrt{\frac{\cos x - \cos^3 x}{1 - \cos^3 x}} \, dx = $ . . . . . . $+ C$.

Similar Questions

The integral $\int \sec^{2/3} x \csc^{4/3} x \, dx$ is equal to: (Here $C$ is a constant of integration)

$\int \cos ^3 x e^{\log (\sin x)^2} d x=$

For $x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$,if $y(x) = \int \frac{\operatorname{cosec} x + \sin x}{\operatorname{cosec} x \sec x + \tan x \sin^2 x} \, dx$ and $\lim_{x \rightarrow (\frac{\pi}{2})^-} y(x) = 0$,then $y\left(\frac{\pi}{4}\right)$ is equal to:

$\int \frac{x^3}{\sqrt{x^2 + 2}} \, dx = $

$\int \frac{x^{e-1}+e^{x-1}}{x^e+e^x} d x=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module