find-I-0-pi-dx-cosx-2sinx-

Question Number 27186 by abdo imad last updated on 02/Jan/18

f i n d I = \int_{0}^{π} \frac{d x}{c o s x + 2 s i n x} .

Commented by abdo imad last updated on 04/Jan/18

w e d o t h e c h a n g e m e n t t a n (\frac{x}{2}) = t

I = \int_{0}^{\infty} \frac{\frac{2 d t}{1 + t^{2}}}{\frac{1 - t^{2}}{1 + t^{2}} + \frac{4 t}{1 + t^{2}}} = \int_{0}^{\infty} \frac{2 d t}{- t^{2} + 4 t + 1}

= - 2 \int_{0}^{\infty} \frac{d t}{t^{2} - 4 t - 1} = - 2 \int_{0}^{\infty} \frac{d t}{t^{2} - 4 t + 4 - 5}

= - 2 \int_{0}^{\infty} \frac{d t}{{(t - 2)}^{2} - 5} = - 2 \int_{0}^{\infty} \frac{d t}{(t - 2 + \sqrt{5}) (t - 2 - \sqrt{5)}}

= \frac{1}{\sqrt{5}} \int_{0}^{\infty} (\frac{1}{t - 2 + \sqrt{} 5} - \frac{1}{t - 2 - \sqrt{5}})

= \frac{1}{\sqrt{5}} {[l n / \frac{t - 2 + \sqrt{5}}{t - 2 - \sqrt{5}} /]}_{0}^{\propto}

= \frac{1}{\sqrt{5}} (- l n / \frac{- 2 + \sqrt{5}}{- 2 - \sqrt{5}} /) = \frac{1}{\sqrt{5}} l n (\frac{2 + \sqrt{5}}{- 2 + \sqrt{5}}) .

Answered by mrW1 last updated on 03/Jan/18

I = \int_{0}^{π} \frac{d x}{c o s x + 2 s i n x}

= \frac{1}{\sqrt{5}} \int_{0}^{π} \frac{d x}{\frac{1}{\sqrt{5}} c o s x + \frac{2}{\sqrt{5}} s i n x}

= \frac{1}{\sqrt{5}} \int_{0}^{π} \frac{d x}{\sin α c o s x + \cos α s i n x}

= \frac{1}{\sqrt{5}} \int_{0}^{π} \frac{d x}{s i n (x + α)}

\dots \dots

= \frac{1}{\sqrt{5}} \ln (9 + 4 \sqrt{5})

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