Question

How do you integrate `int cosxsqrt(9+25sin^2x)` using trig substitutions?

Answer 1

`(5sinxsqrt(9+25sin^2x)+9lnabs(sqrt(9+25sin^2x)+5sinx))/10+C`

Explanation:

`intcosxsqrt(9+25sin^2x)dx`

First apply the non-trigonometric substitution `sinx=u` such that `cosxdx=du`.

`=intsqrt(9+25u^2)du`

Now, we will apply the trigonometric substitution `u=3/5tantheta`. This implies that `du=3/5sec^2thetad theta`. These yield:

`=intsqrt(9+9tan^2theta)(3/5sec^2thetad theta)`

`=9/5intsqrt(1+tan^2theta)(sec^2theta d theta)`

Recall that `1+tan^2theta=sec^2theta`:

`=9/5intsec^3thetad theta`

The process of finding `intsec^3thetad theta` is fairly complex, so click here to see how it's done.

`=9/10(secthetatantheta+lnabs(sectheta+tantheta))+C`

Note that `tantheta=(5u)/3`. This means we have a right triangle with an opposite side of `5u` and an adjacent side of `3`, resulting in a hypotenuse of `sqrt(9+25u^2)`.

The secant of this triangle is equal to the hypotenuse over the adjacent side, or `sqrt(9+25u^2)/3`. Replace these values in the solved integral to simplify:

`=9/10(sqrt(9+25u^2)/3((5u)/3)+lnabs(sqrt(9+25u^2)/3+(5u)/3))+C`

`=(5usqrt(9+25u^2))/10+9/10lnabs(1/3(sqrt(9+25u^2)+5u))+C`

Note that the `1/3` in the logarithm can be taken out as a constant and be absorbed with `C`, the constant of integration.

`=(5usqrt(9+25u^2)+9lnabs(sqrt(9+25u^2)+5u))/10+C`

Since `u=sinx`:

`=(5sinxsqrt(9+25sin^2x)+9lnabs(sqrt(9+25sin^2x)+5sinx))/10+C`