find-the-value-of-0-2pi-dx-3-2sinx-cosx-

Question Number 67526 by mathmax by abdo last updated on 28/Aug/19

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

Commented by mathmax by abdo last updated on 31/Aug/19

l e t I = \int_{0}^{2 π} \frac{d x}{3 + 2 s i n x + c o s x} \Rightarrow I = \int_{0}^{π} \frac{d x}{3 + 2 s i n x + c o s x} + \int_{π}^{2 π} \frac{d x}{3 + 2 s i n x + c o s x}

= H + K

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

= \int_{0}^{\infty} \frac{2 d t}{3 + 3 t^{2} + 4 t + 1 - t^{2}} = \int_{0}^{\infty} \frac{2 d t}{2 t^{2} + 4 t + 4} = \int_{0}^{\infty} \frac{d t}{t^{2} + 2 t + 2}

Δ^{^{'}} = 1 - 2 = - 1 < 0 \Rightarrow

H = \int_{0}^{\infty} \frac{d t}{t^{2} + 2 t + 1 + 1} = \int_{0}^{\infty} \frac{d t}{{(t + 1)}^{2} + 1} =_{t + 1 = α} \int_{1}^{+ \infty} \frac{d α}{1 + α^{2}}

= {[a r c t a n (α)]}_{1}^{+ \infty} = \frac{π}{2} - \frac{π}{4} = \frac{π}{4}

K =_{x = π + t} \int_{0}^{π} \frac{d t}{3 - 2 s i n t - c o s t} =_{t a n (\frac{t}{2}) = u} \int_{0}^{\infty} \frac{2 d u}{(1 + u^{2}) (3 - 2 \frac{2 u}{1 + u^{2}} - \frac{1 - u^{2}}{1 + u^{2}})}

= \int_{0}^{\infty} \frac{2 d u}{3 + 3 u^{2} - 4 u - 1 + u^{2}} = \int_{0}^{\infty} \frac{2 d u}{4 u^{2} - 4 u + 2}

= \int_{0}^{\infty} \frac{d u}{2 u^{2} - 2 u + 1} = \frac{1}{2} \int_{0}^{\infty} \frac{d u}{u^{2} - u + \frac{1}{2}} = \frac{1}{2} \int_{0}^{\infty} \frac{d u}{{(u - \frac{1}{2})}^{2} + \frac{3}{4} + \frac{1}{2}}

= \frac{1}{2} \int_{0}^{\infty} \frac{d u}{{(u - \frac{1}{2})}^{2} + \frac{5}{4}} =_{u - \frac{1}{2} = \frac{\sqrt{5}}{2} α} \frac{1}{2} \times \frac{4}{5} \int_{- \frac{1}{\sqrt{5}}}^{+ \infty} \frac{1}{1 + u^{2}} \frac{\sqrt{5}}{2} d α

= \frac{1}{\sqrt{5}} {[a r c t a n α]}_{- \frac{1}{\sqrt{5}}}^{+ \infty} = \frac{1}{\sqrt{5}} {\frac{π}{2} + a r c t a n (\frac{1}{\sqrt{5}})} \Rightarrow

I = \frac{π}{4} + \frac{π}{2 \sqrt{5}} + \frac{1}{\sqrt{5}} a r c t a n (\frac{1}{\sqrt{5}}) .