find-0-2pi-dx-1-cosx-3sinx-2-

Question Number 137055 by mathmax by abdo last updated on 29/Mar/21

find \int_{0}^{2 π} \frac{dx}{{(1 + cosx + 3 sinx)}^{2}}

Commented by MJS_new last updated on 30/Mar/21

0 ⩽ {(1 + \cos x + 3 \sin x)}^{2} ⩽ 11 + 2 \sqrt{10}

\Rightarrow

\frac{11 - 2 \sqrt{10}}{81} ⩽ \frac{1}{{(1 + \cos x + 3 \sin x)}^{2}} ⩽ + \infty

\Rightarrow

integral diverges

Answered by Dwaipayan Shikari last updated on 30/Mar/21

\int_{0}^{2 π} \frac{d x}{(2 {c o s}^{2} \frac{x}{2} + 6 s i n \frac{x}{2} c o s \frac{x}{2})}

= \frac{1}{4} \int \frac{{s e c}^{2} \frac{x}{2}}{{(c o s \frac{x}{2} + 3 s i n \frac{x}{2})}^{2}} d x = \frac{1}{4} \int \frac{{s e c}^{4} \frac{x}{2}}{{(1 + 3 t a n \frac{x}{2})}^{2}} d x

= \frac{1}{2} \int \frac{{s e c}^{4} u}{{(1 + 3 t a n u)}^{2}} d u = \frac{1}{2} \int \frac{(1 + t^{2})}{{(1 + 3 t)}^{2}} d t

= \frac{1}{18} \int \frac{d t}{{(t + \frac{1}{3})}^{2}} + \frac{1}{18} \int \frac{t^{2} + \frac{2 t}{3} + \frac{1}{9}}{{(t + \frac{1}{3})}^{2}} - \frac{1}{18} \int \frac{\frac{2 t}{3} + \frac{1}{9}}{{(t + \frac{1}{3})}^{2}} d t

= - \frac{1}{18} . \frac{1}{t + \frac{1}{3}} + \frac{1}{18} - \frac{1}{27} l o g (t + \frac{1}{3}) + \frac{1}{162 (t + \frac{1}{3})} - \frac{1}{81 (t + \frac{1}{3})}

= {[\frac{1}{t a n (\frac{x}{2})} (\frac{1}{162} - \frac{1}{18} - \frac{1}{81}) + \frac{1}{18} - \frac{1}{27} l o g (t a n \frac{x}{2} + \frac{1}{3})]}_{0}^{2 π}

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