Mathematics

Evaluate the following integrals
$$\int { \cfrac { 1 }{ \sin { x } +\cos { x }  }  } dx$$


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

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

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

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

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

So $$I=-\displaystyle\int \dfrac{2 d t}{ t^2-2 t-1}$$

         $$=-2\displaystyle\int \dfrac{d t}{t^2-2(t)(1)-1}$$

        $$=-2\displaystyle\int \dfrac{d t}{t^2-2(t)(1)+(1)^2-1-(1)^2}$$        

        $$=-2\displaystyle\int \dfrac{d t}{(t-1)^2-2}$$                    $$(\because (a-b)^2=a^2-2 a b+b^2\ )$$

         $$=-2\displaystyle\int \dfrac{d t}{(t-1)^2-(\sqrt{2})^2}$$   

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

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