How do you integrate #int (4x)/sqrt(x^2-14x+40)dx# using trigonometric substitution?

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Answer 1 · 2018-06-28T05:26:45

Use the substitution #x-7=3sectheta#.

Let

#I=int(4x)/sqrt(x^2-14x+40)dx#

Complete the square in the denominator:

#I=4intx/sqrt((x-7)^2-9)dx#

Apply the substitution #x-7=3sectheta#:

#I=4int(3sectheta+7)/(3tantheta)(3secthetatanthetad theta)#

Simplify:

#L=4int(3sec^2theta+7sectheta)d theta#

Integrate term by term:

#L=4(3tantheta+7ln|3sectheta+3tantheta|)+C#

Reverse the substitution:

#L=4sqrt((x-7)^2-9)+28ln|x-7+sqrt((x-7)^2-9)|+C#