How do you evaluate the integral of #(x + (15/(sin^2(x))))# from pi/6 to pi/2?

1 Answer
mason m
Mar 26, 2016

#pi^2/9+15sqrt3approx27.077#

Explanation:

We have

#int_(pi/6)^(pi/2)(x+15/sin^2(x))dx#

Which can be written as

#=int_(pi/6)^(pi/2)(x+15csc^2(x))dx#

Both of these terms can be evaluated separately (for now, indefinitely, we will evaluate the integral directly following this):

#intxdx=x^2/2+C#

#int15csc^2(x)dx=15intcsc^2(x)dx=-15cot(x)+C#

So, we want to evaluate

#=[x^2/2-15cot(x)]_(pi/6)^(pi/2)#

#=((pi/2)^2/2-15cot(pi/2))-((pi/6)^2/2-15cot(pi/6))#

#=(pi^2/8-15(0))-(pi^2/72-15(sqrt3))#

#=pi^2/8-pi^2/72+15sqrt3#

#=pi^2/9+15sqrt3approx27.077#

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