If {I_1} = int {{{sin }^{ - 1}}x,dx} and {I_ | vedclass

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

Question

(C) Given ${I_1} = \int {{{\sin }^{ - 1}}x\,dx} $ and ${I_2} = \int {{{\sin }^{ - 1}}\sqrt {1 - {x^2}} } dx$. Using the identity ${\sin ^{ - 1}}\sqrt

Vedclass · Accepted Answer

${I_1} + {I_2} = \frac{\pi }{2}x$

Vedclass · Answer

(C) Given ${I_1} = \int {{{\sin }^{ - 1}}x\,dx} $ and ${I_2} = \int {{{\sin }^{ - 1}}\sqrt {1 - {x^2}} } dx$. Using the identity ${\sin ^{ - 1}}\sqrt {1 - {x^2}} = {\cos ^{ - 1}}x$,we have ${I_2} = \int {{{\cos }^{ - 1}}x\,dx} $. Now,consider the sum ${I_1} + {I_2} = \int {{{\sin }^{ - 1}}x\,dx} + \int {{{\cos }^{ - 1}}x\,dx} $. Since ${\sin ^{ - 1}}x + {\cos ^{ - 1}}x = \frac{\pi }{2}$,we get ${I_1} + {I_2} = \int {(\sin ^{ - 1}x + \cos ^{ - 1}x)\,dx} $. ${I_1} + {I_2} = \int {\frac{\pi }{2}\,dx} = \frac{\pi }{2}x + C$. Assuming the constant of integration $C = 0$ for the relation,we get ${I_1} + {I_2} = \frac{\pi }{2}x$.

If ${I_1} = \int {{{\sin }^{ - 1}}x\,dx} $ and ${I_2} = \int {{{\sin }^{ - 1}}\sqrt {1 - {x^2}} } dx$,then:

Similar Questions

If $\int \frac{\cos x+x}{1+\sin x} d x=f(x)+\int \frac{3 \cos \frac{x}{2}-\sin \frac{x}{2}}{\cos \frac{x}{2}+\sin \frac{x}{2}} d x+c_r$ then $f(x)=$

$\int \frac{\cos 2 x \cdot \sin 4 x}{\cos ^4 x\left(1+\cos ^2 2 x\right)} d x=$

If $\int \frac{\cos (13 x)-\cos (14 x)}{1+2 \cos (9 x)} d x=\frac{\sin (4 x)}{a}-\frac{\sin (5 x)}{b}+c$,then $a^b=$

If $f(x) = \frac{\sin^2 \pi x}{1+\pi^x}$,then $\int (f(x) + f(-x)) \, dx$ is equal to

$\int \frac{\sqrt{1-x^2} \sin ^{-1} x+x}{\sqrt{1-x^2}} d x=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module