Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Find the value of  if <img alt="\\\int \sin^{-1}x\cos^{-1}xdx" src="https://learn.careers360.com/latex-imag

Find the value of f(x) if \\\int \sin^{-1}x\cos^{-1}xdx=f^{-1}(x)\left [ \frac{\pi}{2}x-xf^{-1}(x)-2\sqrt{1-x^2} \right ]+\frac{\pi}{2}\sqrt{1-x^2}+2x+C

Option: 1

\sin x


Option: 2

\sin^{-1}x


Option: 3

\cos^{-1}x


Option: 4

\cos x


Answers (1)

best_answer

 

 

Integration by Parts -

To evaluate  Integration function which is product of two function, use the method of integration by parts.

If two functions of x, u and v are given, then

 \mathrm{\int\mathit{uv\;dx=u\times\int v\;dx-\int\left \{ \frac{du}{dx}\int v\;dx \right \}dx}}

i.e.

The integral of product of two functions = ( first function) X ( integral of second function ) - integral of
(differential of first function X integral of second function).

-

I=\int \sin^{-1}x\cos^{-1}xdx

I=\sin^{-1}x\int \cos^{-1}xdx-\int \left [ \left ( \frac{d}{dx}\sin^{-1}x \right )\int \cos^{-1}xdx \right ]dx

I=\sin^{-1}x\left [ x\cos^{-1}x-\sqrt{1-x^2} \right ]-\int \left [ \frac{1}{\sqrt{1-x^2}}\left ( x\cos^{-1}x-\sqrt{1-x^2} \right) \right ]dx

As Integration of \int cos^{-1}x=x\cos^{-1}x-\sqrt{1-x^2} 

I=\sin^{-1}x\left [ x\cos^{-1}x-\sqrt{1-x^2} \right ]-\int \ \frac{x\cos^{-1}x}{\sqrt{1-x^2}}dx+\int1 dx

I=\sin^{-1}x\left [ x\cos^{-1}x-\sqrt{1-x^2} \right ]+\int1 dx-\int \frac{x}{\sqrt{1-x^2}}\cdot \cos^{-1}xdx

Using By Parts \int \frac{x}{\sqrt{1-x^2}}\cos^{-1}xdx= \cos^{-1}x\int \frac{x}{\sqrt{1-x^2}}dx-\int \left [ \left ( \frac{d}{dx}cos^{-1}x \right ) \int \frac{x}{\sqrt{1-x^2}}dx\right ]dx

\int \frac{x}{\sqrt{1-x^2}}\cos^{-1}xdx= -\cos^{-1}x\sqrt{1-x^2}dx\;-\int \left [ \frac{-1}{\sqrt{1-x^2}}\cdot \left(-\sqrt{1-x^2}\right)dx\right ]dx

\int \frac{x}{\sqrt{1-x^2}}\cos^{-1}xdx= -\cos^{-1}x\sqrt{1-x^2}dx-\int 1dx

we get

\\I=\sin^{-1}x\left [ x\cos^{-1}x-\sqrt{1-x^2} \right ]+\int1 dx\;+\left [\sqrt{1-x^2}\cos^{-1}x \right ]+\int 1dx\\I=\sin^{-1}x\left [ x\cos^{-1}x-\sqrt{1-x^2} \right ]\;+\left [\sqrt{1-x^2}\cos^{-1}x \right ]+2x+CNow use \sin^{-1}x+\cos^{-1}x=\frac\pi2

\\I=\sin^{-1}x\left [ x\left (\frac\pi2-\sin^{-1}x \right )-\sqrt{1-x^2} \right ]\;+\left [\sqrt{1-x^2}\left (\frac\pi2-\sin^{-1}x \right ) \right ]+2x+C

\\I=\sin^{-1}x\left [ \frac\pi2x-x\sin^{-1}x -\sqrt{1-x^2} \right ]+2x+\left [\frac\pi2\sqrt{1-x^2}-\sqrt{1-x^2}\sin^{-1}x \right ]

\\I=\sin^{-1}x\left [ \frac\pi2x-x\sin^{-1}x -2\sqrt{1-x^2} \right ]+\frac\pi2\sqrt{1-x^2}+2x+C

hence f^{-1}(x)=\sin^{-1}x or f(x)=\sin x

Posted by

Pankaj

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions