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Question Number 46853 by maxmathsup by imad last updated on 01/Nov/18

$$fnd\int \frac{dx}{1+cos\left(tx\right)}$$

Commented by maxmathsup by imad last updated on 01/Nov/18

$$letA\left(t\right)=\int \frac{dx}{1+cos\left(tx\right)}\Rightarrow A\left(t\right){=}_{tx=u}\int \frac{1}{1+cosu}\frac{du}{t}$$$$=\frac{1}{t}\int \frac{du}{1+cosu}{=}_{tan\left(\frac{u}{2}\right)=\alpha }\frac{1}{t}\int \frac{1}{1+\frac{1-{\alpha }^2}{1+{\alpha }^2}}\frac{2d\alpha }{1+{\alpha }^2}$$$$=\frac{1}{t}\int \frac{2d\alpha }{1+{\alpha }^2+1-{\alpha }^2}=\frac{1}{t}\int d\alpha =\frac{\alpha }{t}+c=\frac{tan\left(\frac{u}{2}\right)}{t}+c$$$$=\frac{tan\left(\frac{tx}{2}\right)}{t}+c.$$

Answered by tanmay.chaudhury50@gmail.com last updated on 01/Nov/18

$$\int \frac{1-cos\left(tx\right)}{{sin}^2\left(tx\right)}dx$$$$\int {cosec}^2\left(tx\right)-cot\left(tx\right)cosec\left(tx\right)dx$$$$\frac{-cot\left(tx\right)}{t}+\frac{cosec\left(tx\right)}{t}+c$$

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