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

2 Answers
Mar 20, 2018

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

Mar 20, 2018

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