Question

How do you evaluate the integral `int arcsin(5x-2)`?

Answer 1

`intarcsin(5x-2)dx=1/5((5x-2)arcsin(5x-2)+sqrt(1-(5x-2)²))+C`, `C in RR`

Explanation:

`intarcsin(5x-2)dx`
Let `t=5x-2`
`dt=5dx`
`intarcsin(5x-2)dx=1/5int5arcsin(5x-2)dx`
`=1/5intarcsin(t)dt`
`=1/5int1*arcsin(t)dt`
Using Integration by parts :
`intg'(t)f(t)dt=[g(t)f(t)]-intg(t)f'(t)dt`
There:
`g'(t) =1` `f(t)=arcsin(t)`
`g(t)=t`, `f'(t)=1/sqrt(1-t²)`
So: `1/5intarcsin(t)dt=1/5([tarcsin(t)]-intt/sqrt(1-t²)dt)`
Now let `u=sqrt(1-t²)`
`du=-t/sqrt(1-t²)dt`
So:`1/5([tarcsin(t)]-intt/sqrt(1-t²)dt)=1/5([tarcsin(t)]+int1du)`
`=1/5([tarcsin(t)]+u)`
`=1/5(tarcsin(t)+sqrt(1-t²))`
`=1/5((5x-2)arcsin(5x-2)+sqrt(1-(5x-2)²))+C`, `C in RR`
\0/ here's our answer!