int tan ^{-1}left(sqrt{frac{1-x}{1+x}}right) | vedclass

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

Question

(A) Let $I = \int 	an ^{-1} \sqrt{\frac{1-x}{1+x}} d x$. Substitute $x = \cos 2	heta$,so $dx = -2\sin 2	heta d	heta$. Then $\sqrt{\frac{1-x}{1+x}}

Vedclass · Accepted Answer

$\frac{1}{2}\left(x \cos ^{-1} x-\sqrt{1-x^2}\right)+c$

Vedclass · Answer

(A) Let $I = \int 	an ^{-1} \sqrt{\frac{1-x}{1+x}} d x$. Substitute $x = \cos 2	heta$,so $dx = -2\sin 2	heta d	heta$. Then $\sqrt{\frac{1-x}{1+x}} = \sqrt{\frac{1-\cos 2	heta}{1+\cos 2	heta}} = \sqrt{\frac{2\sin^2 	heta}{2\cos^2 	heta}} = 	an 	heta$. Thus,$I = \int 	heta (-2\sin 2	heta) d	heta = -2 \int 	heta \sin 2	heta d	heta$. Using integration by parts: $I = -2 \left[ 	heta \left( -\frac{\cos 2	heta}{2} ight) - \int 1 \cdot \left( -\frac{\cos 2	heta}{2} ight) d	heta ight] = 	heta \cos 2	heta - \int \cos 2	heta d	heta = 	heta \cos 2	heta - \frac{\sin 2	heta}{2} + c$. Since $x = \cos 2	heta$,then $	heta = \frac{1}{2} \cos^{-1} x$ and $\sin 2	heta = \sqrt{1-x^2}$. Substituting these back: $I = \frac{1}{2} x \cos^{-1} x - \frac{1}{2} \sqrt{1-x^2} + c = \frac{1}{2} \left( x \cos^{-1} x - \sqrt{1-x^2} ight) + c$.

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

Similar Questions

Which of the following is paramagnetic?

Let $f(x) = \begin{cases} \frac{x^4 - 5x^2 + 4}{|(x - 1)(x - 2)|}, & x \ne 1, 2 \\ 6, & x = 1 \\ 12, & x = 2 \end{cases}$. Then $f(x)$ is continuous on the set

The shortest distance between the line $y - x = 1$ and the curve $x = y^2$ is:

Copper sulphate is prepared by blowing a current of air through copper scrap and dilute $H_{2}SO_{4}$. Dilute $HNO_{3}$ is also added:

Leaf blade is spinous in case of

Vedclass Test Series

Exam Paper Generator

Online Exam Module