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,