Question

integration of sin inverse (2x) dx.
​

Answer 1

Step-by-step explanation:

I think this answer is helpful to you

Answer 2

The integration of sin inverse (2x) is $xsin^{-1}2x-\frac{1}{2}\sqrt{1+4x^{2} }$

Given: f(x) = sin inverse(2x)

To Find: Integration of f(x)

Solution:

The function is $sin^{-1} 2x$

Now integrate it

$\int {sin^{-1}2x } \, dx$

put 2x = t

differentiate it with respect to x

2dx = dt

dx = dt/2

$\frac{1}{2} \int\limit{sin^{-1} } \, dt$

$\frac{1}{2}(tsin^{-1}t-[\int\ {\frac{dsin^{-1}t }{dt} } \, tdt]\\\frac{1}{2}( tsin^{-1}t-[\int\frac{t}{\sqrt{1+t^{2} } } dt]\\\frac{1}{2}( tsin^{-1}t-I$

$I = \int\limits {\frac{t}{\sqrt{1+t^{2} } } } \, dt$

Put 1+t²=z

2tdt = dz

tdt = dz/2

$\frac{1}{2} \int\limits{\frac{1}{\sqrt{z} } } \, dx \\\frac{1}{2}2\sqrt{z}= \sqrt{z}$

Put the value of z

I = $\sqrt{1+t^{2} }$

The integration of the equation is

$\frac{1}{2}(tsin^{-1}t-\sqrt{1+t^{2} } )$

Put the value of t

The final integration is $xsin^{-1}2x-\frac{1}{2}\sqrt{1+4x^{2} }$

Hence, the final integration is$xsin^{-1}2x-\frac{1}{2}\sqrt{1+4x^{2} }$

#SPJ3