Mathematics

Evaluate the following integrals
$$\int { \cfrac { 1 }{ 2+\sin { x } +\cos { x }  }  } dx$$


SOLUTION
Let $$\text{I}=\displaystyle\int \dfrac{1}{2+\sin x+\cos x} d x$$

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

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

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

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

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

         $$=2\displaystyle\int \dfrac{d t}{t^2+2(t)(1)+3}$$

        $$=2\displaystyle\int \dfrac{d t}{t^2+2(t)(1)+(1)^2+3-(1)^2}$$        

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

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

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

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