Question: "How to integrate∫0π22asin2xa2sin2x+b2cos⁡xdx"

Corbin Walton

Answered question

2022-02-26

How to integrate

\int_{0}^{\frac{π}{2}} \frac{2 a \sin^{2} x}{a^{2} \sin^{2} x + b^{2} \cos x} d x

vazen2bl

Beginner2022-02-27Added 9 answers

Essentially, we want to integrate

I = \int_{0}^{\frac{π}{2}} \frac{\sin^{2} (t) d t}{1 + c \sin^{2} (t)}

where

c < 1

. We have

I = \sum_{k = 0}^{\infty} {(- c)}^{k} \int_{0}^{\frac{π}{2}} \sin^{2 k + 2} (t) d t = \frac{π}{8} \cdot \sum_{k = 0}^{\infty} {(\frac{- c}{4})}^{k} (\begin{matrix} 2 k + 2 \\ k + 1 \end{matrix})

= \frac{π}{2 c} (\frac{\sqrt{1 + c} - 1}{\sqrt{1 + c}})

\frac{1}{1 + r} = \sum_{k = 0}^{\infty} {(- r)}^{k} \int_{0}^{2 π} \sin^{2 k} d x = (\begin{matrix} 2 k \\ k \end{matrix}) \frac{2 π}{2^{2 k}}

\frac{1}{\sqrt{1 - 4 x}} = \sum_{k = 0}^{\infty} (\begin{matrix} 2 k \\ k \end{matrix}) x^{k} \lor x \in [- \frac{1}{4}, \frac{1}{4})

The integral you have is

J = \frac{2 a}{b^{2}} \int_{0}^{\frac{π}{2}} \frac{\sin^{2} (x)}{1 + (\frac{a^{2} - b^{2}}{b^{2}}) \sin^{2} (x)} d x = \frac{2 a}{b^{2}} \cdot \frac{π}{2 c} (\frac{| \frac{a}{b} | - 1}{| \frac{a}{b} |})

= \frac{a π}{a^{2} - b^{2}} \frac{| a | - | b |}{| a |}

This gives you

J = \frac{sign (a)}{| a | + | b |} π

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