How do you integrate #int x^3 sqrt(-x^2 - 16x-39)dx# using trigonometric substitution?

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Answer 1 · 2017-01-25T02:31:23

#int x^3 sqrt(-x^2-16x-39) dx = 25 int (5sint+8)^3( 1-sin^2t)dt#

Write:

#-x^2-16x-39 = -x^2 -16x -64+25 = 25 -(x-8)^2#

so that we get a difference of squares under the root.

Now substitute:

#x-8 = 5sint#

#dx = 5costdt#

As the function is defined only for:

#abs(x-8) <= 5#

we have #t in (-pi/2,pi/2)#

Performing the substitution we get:

#int x^3 sqrt(-x^2-16x-39) dx = 5 int (5sint+8)^3 sqrt (25-25sin^2t) costdt = 25 int (5sint+8)^3 cos^2tdt = 25 int (5sint+8)^3( 1-sin^2t)dt#

where in #cost =+- sqrt(1-sin^2t) # we took the plus sign as for #t in (-pi/2,pi/2)# the cosine is positive.

We have now to integrate a polynomial in #sint#, which can be easily calculated using reduction formulas.