Question

`inte^x[((sin^-1x)sqrt(1-x^2)+1)/sqrt(1-x^2)]dx` ?

Answer 1

`e^x(sin^-1x)+C`

Explanation:

We have:

`inte^x[((sin^-1x)sqrt(1-x^2)+1)/sqrt(1-x^2)]dx`

Splitting up the fraction, this becomes:

`=inte^x[sin^-1x+1/sqrt(1-x^2)]dx`

Split up the integrals:

`=inte^x(sin^-1x)dx+inte^x/sqrt(1-x^2)dx`

Attempt to integrate `inte^x(sin^-1x)dx` via integration by parts, which takes the form `intuv=uv-intvdu`.

Let `u=sin^-1x`, so `du=1/sqrt(1-x^2)dx`, and `dv=e^xdx`, so `v=e^x`.

Thus we see that the original integral equals:

`=[e^x(sin^-1x)-inte^x/sqrt(1-x^2)dx]+inte^x/sqrt(1-x^2)dx`

The integral `inte^x/sqrt(1-x^2)dx` will cancel with itself, leaving just

`=e^x(sin^-1x)`

Add the constant of integration:

`=e^x(sin^-1x)+C`

Answer 2

`e^x(sin^-1x)+C`

Explanation:

There is a pattern surrounding integrals involving `e^x`, in cases like this problem (and all the others surrounding it on the page).

Since:

`d/dxe^xf(x)=e^xf(x)+e^xf'(x)=e^x[f(x)+f'(x)]`

Then:

`inte^x[f(x)+f'(x)]dx=e^xf(x)+C`

This is the case for the given problem, which simplifies to be:

`inte^x[sin^-1x+1/sqrt(1-x^2)]dx`

If `f(x)=sin^-1x`, then `f'(x)=1/sqrt(1-x^2)`.

Thus the integral equals

`=e^xf(x)+C=e^x(sin^-1x)+C`