0-pi-2-cos-2-x-cos-2-x-4sin-2-x-dx-

Question Number 160767 by cortano last updated on 06/Dec/21

\int_{0}^{\frac{π}{2}} \frac{\cos^{2} x}{\cos^{2} x + 4 \sin^{2} x} dx = ?

Answered by MJS_new last updated on 06/Dec/21

\underset{0}{\int^{π / 2}} \frac{\cos^{2} x}{\cos^{2} x + {4 \sin}^{2} x} d x =

[t = \tan x \to d x = \frac{d t}{t^{2} + 1}]

= \underset{0}{\int^{\infty}} \frac{d t}{(t^{2} + 1) (4 t^{2} + 1)} = \frac{1}{3} \underset{0}{\int^{\infty}} (\frac{1}{t^{2} + \frac{1}{4}} - \frac{1}{t^{2} + 1}) d t =

= \frac{1}{3} {[2 \arctan 2 t - \arctan t]}_{0}^{\infty} = \frac{π}{6}

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