How do you integrate #int cosxsqrt(9+25sin^2x)# using trig substitutions?
1 Answer
Explanation:
#intcosxsqrt(9+25sin^2x)dx#
First apply the non-trigonometric substitution
#=intsqrt(9+25u^2)du#
Now, we will apply the trigonometric substitution
#=intsqrt(9+9tan^2theta)(3/5sec^2thetad theta)#
#=9/5intsqrt(1+tan^2theta)(sec^2theta d theta)#
Recall that
#=9/5intsec^3thetad theta#
The process of finding
#=9/10(secthetatantheta+lnabs(sectheta+tantheta))+C#
Note that
The secant of this triangle is equal to the hypotenuse over the adjacent side, or
#=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
#=(5usqrt(9+25u^2)+9lnabs(sqrt(9+25u^2)+5u))/10+C#
Since
#=(5sinxsqrt(9+25sin^2x)+9lnabs(sqrt(9+25sin^2x)+5sinx))/10+C#