calculate-0-pi-dx-3-cosx-2-sinx-

Question Number 63711 by mathmax by abdo last updated on 07/Jul/19

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

Commented by mathmax by abdo last updated on 08/Jul/19

l e t A = \int_{0}^{π} \frac{d x}{\sqrt{3} c o s x + \sqrt{2} s i n x} c h a n g e m e n t t a n (\frac{x}{2}) = t g i v e

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

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

Δ^{^{'}} = {(- \sqrt{2})}^{2} + 3 = 5 \Rightarrow t_{1} = \frac{\sqrt{2} + \sqrt{5}}{\sqrt{3}} a n d t_{2} = \frac{\sqrt{2} - \sqrt{5}}{\sqrt{3}} \Rightarrow

A = \frac{- 2}{\sqrt{3}} \int_{0}^{\infty} \frac{d t}{(t - t_{1}) (t - t_{2})} = \frac{- 2}{\sqrt{3} (t_{1} - t_{2})} \int_{0}^{\infty} {\frac{1}{t - t_{1}} - \frac{1}{t - t_{2}}} d t

= \frac{- 2}{\sqrt{3} \frac{2 \sqrt{5}}{\sqrt{3}}} \int_{0}^{\infty} {\frac{1}{t - t_{1}} - \frac{1}{t - t_{2}}} d t = - \frac{1}{\sqrt{5}} {[l n ∣ \frac{t - t_{1}}{t - t_{2}} ∣]}_{0}^{+ \infty}

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

Answered by MJS last updated on 07/Jul/19

= \frac{\sqrt{5}}{5} \int \frac{d x}{\sin (x + \arctan \frac{\sqrt{6}}{2})}

now {it}^{'} s easy again \dots

Commented by mathmax by abdo last updated on 07/Jul/19

t h a n k y o u s i r .