Answer to Question #265728 in Calculus for Roots

Question #265728

Integrals are not always easy to evaluate; sometimes we need to be clever! In this problem we study the definite integral


I=∫(π>0) xsin(x)/1+cos^2(x) dx.


At first, it appears difficult to evaluate this integral with substitution (you can try with simple substitution ideas...). But it can be done! Let's see how it goes.



(a) Use the substitution u=π−x to show that


I= π/2∫(π>0)sin(x)/1+cos^2(x) dx.


(Recall that sin(π−x)=sin(x) and cos(π−x)=−cos(x).)



(b) Evaluate the remaining definite integral using substitution to get:


I=π^2/4.

1
Expert's answer
2021-11-18T17:45:55-0500

Let "I = \\int_{0}^{\u03c0}\\frac{xsinx}{1+cos\u00b2x}dx"

Let us substitute u = π-x

So du = -dx and x=0 => u = π and x = π => u = 0

"I = \\int_{0}^{\u03c0}\\frac{xsinx}{1+cos\u00b2x}dx=-\\int_{\u03c0}^{0}\\frac{(\u03c0-u)sin(\u03c0-u)}{1+cos\u00b2(\u03c0-u)} du" = "- \\int_{\u03c0}^{0}\\frac{(\u03c0-u)sinu}{1+cos\u00b2u}du" since sin(π-u)=sinu and cos²(π-u)={-cos(u)}²=cos²u

So "I=\\int_{0}^{\u03c0}\\frac{(\u03c0-u)sinu}{1+cos\u00b2u}du" [by property of definite integral that "\\int_{a}^{b}f(x)dx = - \\int_{b}^{a}f(x)dx" ]

Therefore "I=\\int_{0}^{\u03c0}\\frac{\u03c0sinu}{1+cos\u00b2u}du-\\int_{0}^{\u03c0}\\frac{usinu}{1+cos\u00b2u}du"

=> "I=\u03c0\\int_{0}^{\u03c0}\\frac{sinu}{1+cos\u00b2u}du-I" , since "\\int_{0}^{\u03c0}\\frac{usinu}{1+cos\u00b2u}du = \\int_{0}^{\u03c0}\\frac{xsinx}{1+cos\u00b2x}dx"

=> "2I= \u03c0\\int_{0}^{\u03c0}\\frac{sinu}{1+cos\u00b2u}du"

=> "I= \\frac{\u03c0}{2}\\int_{0}^{\u03c0}\\frac{sinu}{1+cos\u00b2u}du"

Let -cos u = z

So sin u du = dz and also when u=0 , z= -1 and when u = π , z = 1

Therefore "I= \\frac{\u03c0}{2}\\int_{-1}^{1}\\frac{dz}{1+(-z)\u00b2} = \\frac{\u03c0}{2}\\int_{-1}^{1}\\frac{dz}{1+z\u00b2}"

=> "I = \\frac{\u03c0}{2}[tan^{-1}z]_{-1}^{1}"

=> "I = \\frac{\u03c0}{2}[tan^{-1}1 - tan^{-1}(-1)]"

=> "I = \\frac{\u03c0}{2}(\\frac{\u03c0}{4}- \\frac{-\u03c0}{4})"

=> "I = \\frac{\u03c0}{2}(\\frac{\u03c0}{4}+ \\frac{\u03c0}{4} )= \\frac{\u03c0}{2}.\\frac{\u03c0}{2} = \\frac{\u03c0\u00b2}{4}"


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