Question

Integration of sin^-1x -cos^-1 x/sin^-1x+ cos^-1 x dx

Answer 1

We have the identity,

$\sin^{-1} x+\cos^{-1} x =\frac{\pi}{2}$.

Now,

$\int \sin^{-1} x dx=x\sin^{-1} x-\int \frac{x}{\sqrt{1-x^2}} dx\\ \int \sin^{-1} x dx=x\sin^{-1} x+\sqrt{1-x^2}+C\\ \int \cos^{-1} x dx=x\cos^{-1} x+\int \frac{x}{\sqrt{1-x^2}} dx\\ \int \cos^{-1} x dx=x\cos^{-1} x-\sqrt{1-x^2}+C$

Thus the given integral is

$\int \frac{\sin^{-1} x dx-\cos^{-1} x dx}{\sin^{-1} x dx+\cos^{-1} x dx}=\frac{2}{\pi}[x\sin^{-1} x+\sqrt{1-x^2}-x\cos^{-1} x+\sqrt{1-x^2}]+C \\ \int \frac{\sin^{-1} x dx-\cos^{-1} x dx}{\sin^{-1} x dx+\cos^{-1} x dx}=\frac{2}{\pi}[x\sin^{-1} x-x\cos^{-1} x+2\sqrt{1-x^2}]+C$

Answer 2

Solution :

11- x²

* A * (sin?(1 – 2x2) + cos-(241 – 32)]da

use formula sin-1x + cos-lx =

78* val] - cos -1(1 – 2x2) + – sin -1(2xv1 – x2)]dx

va [1 – 2 cos +1(x) – 2 sin = 2(x)]dx

to

x= [1 – 2(cos –+(x) + sin –+(x)]dx

4x4 [1 – 2 x []dx

= 0