find-dx-1-cosx-cos-2x-

Question Number 144213 by Mathspace last updated on 23/Jun/21

f i n d \int \frac{d x}{1 + c o s x + c o s (2 x)}

Answered by bemath last updated on 23/Jun/21

\int \frac{dx}{1 + \cos x + 2 \cos^{2} x - 1}

= \int \frac{dx}{2 \cos^{2} x + \cos x}

= \int \frac{dx}{\cos x (2 \cos x + 1)}

let \tan (\frac{x}{2}) = u \Rightarrow du = \frac{1}{2} \sec^{2} (\frac{x}{2}) dx

dx = 2 \cos^{2} (\frac{x}{2}) du = \frac{2}{1 + u^{2}} du

= \int \frac{\frac{2}{1 + u^{2}}}{(\frac{1 - u^{2}}{1 + u^{2}}) (\frac{3 - u^{2}}{1 + u^{2}})} du

= \int \frac{2 (1 + u^{2})}{(1 - u^{2}) (3 - u^{2})} du

partial fraction

\frac{2 + {2 u}^{2}}{(1 + u) (1 - u) (\sqrt{3} + u) (\sqrt{3} - u)} =

\frac{a}{1 + u} + \frac{b}{1 - u} + \frac{c}{\sqrt{3} + u} + \frac{d}{\sqrt{3} - u}

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