How do you integrate int sqrt(-x^2-10x)/xdx using trigonometric substitution?
2 Answers
Use the substitution
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
I=int(5costheta)/(5sintheta-5)(5costhetad theta)
Simplify:
I=5intcos^2theta/(sintheta-1)d theta
Apply the identity
I=5int(1-sin^2theta)/(sintheta-1)d theta
Apply the difference of squares
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
The answer is
Explanation:
Complete the square :
Therefore, the integral is
Let
Let
Therefore,