calculate-0-pi-2-dx-cos-2-x-3-sin-2-x-

Question Number 83252 by mathmax by abdo last updated on 29/Feb/20

c a l c u l a t e \int_{0}^{\frac{π}{2}} \frac{d x}{{c o s}^{2} x + \sqrt{3} {s i n}^{2} x}

Commented by mathmax by abdo last updated on 29/Feb/20

I = \int_{0}^{\frac{π}{2}} \frac{d x}{\frac{1 + c o s (2 x)}{2} + \sqrt{3} \frac{1 - c o s (2 x)}{2}} = 2 \int_{0}^{\frac{π}{2}} \frac{d x}{(1 - \sqrt{3}) c o s (2 x) + 1 + \sqrt{3}}

=_{2 x = t} 2 \int_{0}^{π} \frac{d t}{2 {(1 - \sqrt{3}) c o s t + 1 + \sqrt{3})} = \int_{0}^{π} \frac{d t}{(1 - \sqrt{3}) c o s t + 1 + \sqrt{3}}

=_{t a n (\frac{t}{2}) = u} \int_{0}^{\infty} \frac{2 d u}{(1 + u^{2}) {(1 - \sqrt{3}) \times \frac{1 - u^{2}}{1 + u^{2}} + 1 + \sqrt{3}}}

= 2 \int_{0}^{\infty} \frac{d u}{(1 - \sqrt{3}) (1 - u^{2}) + (1 + \sqrt{3}) (1 + u^{2})}

= 2 \int_{0}^{\infty} \frac{d u}{1 - \sqrt{3} - (1 - \sqrt{3}) u^{2} + (1 + \sqrt{3}) + (1 + \sqrt{3}) u^{2}}

= 2 \int_{0}^{\infty} \frac{d u}{2 + (1 + \sqrt{3} - 1 + \sqrt{3}) u^{2}} = 2 \int_{0}^{\infty} \frac{d u}{2 + 2 \sqrt{3} u^{2}} = \int_{0}^{\infty} \frac{d u}{1 + \sqrt{3} u^{2}}

=_{\sqrt{\sqrt{3}} u = z} \int_{0}^{\infty} \frac{d z}{(^{4} \sqrt{3}) (1 + z^{2})} = \frac{1}{(^{4} \sqrt{3})} \times \frac{π}{2}

Answered by MJS last updated on 29/Feb/20

\int \frac{d x}{\cos^{2} x + \sqrt{3} \sin^{2} x} =

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

= \int \frac{d t}{\sqrt{3} t^{2} + 1} = \frac{\sqrt[4]{27}}{3} \arctan \sqrt[4]{3} t =

= \frac{\sqrt[4]{27}}{3} \arctan (\sqrt[4]{3} \tan x) + C

\underset{0}{\int^{\frac{π}{2}}} \frac{d x}{\cos^{2} x + \sqrt{3} \sin^{2} x} = \frac{\sqrt[4]{27}}{6} π

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