How do you integrate #int cosxsqrt(9+25sin^2x)# using trig substitutions?

1 Answer
Sep 11, 2016

#(5sinxsqrt(9+25sin^2x)+9lnabs(sqrt(9+25sin^2x)+5sinx))/10+C#

Explanation:

#intcosxsqrt(9+25sin^2x)dx#

First apply the non-trigonometric substitution #sinx=u# such that #cosxdx=du#.

#=intsqrt(9+25u^2)du#

Now, we will apply the trigonometric substitution #u=3/5tantheta#. This implies that #du=3/5sec^2thetad theta#. These yield:

#=intsqrt(9+9tan^2theta)(3/5sec^2thetad theta)#

#=9/5intsqrt(1+tan^2theta)(sec^2theta d theta)#

Recall that #1+tan^2theta=sec^2theta#:

#=9/5intsec^3thetad theta#

The process of finding #intsec^3thetad theta# is fairly complex, so click here to see how it's done.

#=9/10(secthetatantheta+lnabs(sectheta+tantheta))+C#

Note that #tantheta=(5u)/3#. This means we have a right triangle with an opposite side of #5u# and an adjacent side of #3#, resulting in a hypotenuse of #sqrt(9+25u^2)#.

The secant of this triangle is equal to the hypotenuse over the adjacent side, or #sqrt(9+25u^2)/3#. Replace these values in the solved integral to simplify:

#=9/10(sqrt(9+25u^2)/3((5u)/3)+lnabs(sqrt(9+25u^2)/3+(5u)/3))+C#

#=(5usqrt(9+25u^2))/10+9/10lnabs(1/3(sqrt(9+25u^2)+5u))+C#

Note that the #1/3# in the logarithm can be taken out as a constant and be absorbed with #C#, the constant of integration.

#=(5usqrt(9+25u^2)+9lnabs(sqrt(9+25u^2)+5u))/10+C#

Since #u=sinx#:

#=(5sinxsqrt(9+25sin^2x)+9lnabs(sqrt(9+25sin^2x)+5sinx))/10+C#