Question

How do you integrate `int sqrt(-x^2-10x)/xdx` using trigonometric substitution?

Answer 1

Use the substitution `x+5=5sintheta`.

Explanation:

Let

`I=intsqrt(-x^2-10x)/xdx`

Complete the square in the square root:

`I=intsqrt(25-(x+5)^2)/xdx`

Apply the substitution `x+5=5sintheta`:

`I=int(5costheta)/(5sintheta-5)(5costhetad theta)`

Simplify:

`I=5intcos^2theta/(sintheta-1)d theta`

Apply the identity `sin^2theta+cos^2theta=1`:

`I=5int(1-sin^2theta)/(sintheta-1)d theta`

Apply the difference of squares `a^2-b^2=(a-b)(a+b)`:

`I=-5int(1+sintheta)d theta`

Integrate directly:

`I=-5(theta-costheta)+C`

Reverse the substitution:

`I=sqrt(25-(x+5)^2)-5sin^(-1)((x+5)/5)+C`

Answer 2

The answer is `=-5arcsin(1/5(x+5))+5sqrt(1-((x+5)/5)^2)+C`

Explanation:

Complete the square :

`-x^2-10x=25-(x+5)^2`

Therefore, the integral is

`I=int(sqrt(-x^2-10x)dx)/(x)=int(sqrt(25-(x+5)^2)dx)/x`

Let `u=x+5`, `=>`, `du=dx`

`I=int(sqrt(25-u^2)du)/(u-5)`

Let `u=5sinv`, `=>`, `du=5cosvdv`

Therefore,

`I=int((5cosv)*5cosvdv)/(5(sinv-1))`

`=5int(cos^2vdv)/(-(1-sinv))`

`=-5int(1+sinv)dv`

`=-5v+5cosv`

`=-5arcsin(u/5)+5sqrt(1-(u/5)^2)`

`=-5arcsin(1/5(x+5))+5sqrt(1-((x+5)/5)^2)+C`