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

If $\int \frac{1}{(1 + x)\sqrt{x}} \, dx = f(x) + A$,where $A$ is any arbitrary constant,then the function $f(x)$ is

Integrate the function $\frac{x^{3} \sin \left(\tan ^{-1} x^{4}\right)}{1+x^{8}}$.

$\int \frac{e^x \, dx}{\sqrt{1 - e^{2x}}} = $

$\int \frac{\cos x+x \sin x}{x(x+\cos x)} d x = \_\_\_\_$

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module