Mathematics

Evaluate the following integrals:
$$\int { \cfrac { 1 }{ 4\cos { x } -1 }  } dx\quad $$


SOLUTION
Let $$\text{I}=\displaystyle\int \dfrac{1}{4\cos x-1} d x$$

        $$=\displaystyle\int \dfrac{1}{4\bigg(\dfrac{1-\tan^2 (x/2)}{1+\tan^2 (x/2)}\bigg)-1}d x$$                       $$\bigg(\because \cos 2 A=\dfrac{1-\tan^2A}{1+\tan^2 A}\bigg)$$

        $$=\displaystyle\int \dfrac{1+\tan^2 \dfrac{x}{2}}{-\bigg(1+\tan^2 \dfrac{x}{2}\bigg)+4\bigg(1-\tan^2 \dfrac{x}{2}\bigg)}d x$$     

        $$=\displaystyle\int \dfrac{\text{sec}^2 \dfrac{x}{2}}{3-5\tan^2 \dfrac{x}{2}}d x$$               $$(\because \text{sec}^2 A-\tan^2 A=1)$$

Let $$t=\tan \dfrac{x}{2}\implies d t=\dfrac{1}{2}\text{sec}^2 \dfrac{x}{2} d x$$

So $$I=\displaystyle\int \dfrac{2 d t}{3-5 t^2}$$

         $$=-\dfrac{2}{5}\displaystyle\int \dfrac{d t}{t^2-\bigg(\sqrt{\dfrac{3}{5}}\bigg)^2}$$

         $$=-\dfrac{2}{5}\times \dfrac{1}{2\sqrt{\dfrac{3}{5}}}\log \bigg| \dfrac{t-\sqrt{\dfrac{3}{5}}}{t+\sqrt{\dfrac{3}{5}}}\bigg|+c$$                                        $$\bigg(\because \displaystyle\int \dfrac{d x}{x^2-a^2}=\dfrac{1}{2 a}\log \bigg|\dfrac{x-a}{x+a}\bigg|+c\ \bigg)$$

         $$=-\dfrac{\sqrt{5}}{5\sqrt{3}}\log \bigg|  \dfrac{\sqrt{5} t-\sqrt{3}}{\sqrt{5} t+\sqrt{3}}\bigg|+c$$                      

         $$=-\dfrac{1}{\sqrt{15}}\log \bigg|\dfrac{\sqrt{5}\tan \dfrac{x}{2}-\sqrt{3}}{\sqrt{5}\tan \dfrac{x}{2}+\sqrt{3}}\bigg|+c$$                                       $$\bigg(\because t=\tan \dfrac{x}{2}\bigg)$$
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Subjective Medium Published on 17th 09, 2020
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