Mathematics

$$\displaystyle \int \frac { e ^ { x } ( 1 + \sin x ) } { 1 + \cos x } d x =$$


ANSWER

$$e ^ { x } \tan \frac { x } { 2 } + c$$


SOLUTION
Let $$I=\displaystyle\int\dfrac{e^x(1+\sin x)}{1+\cos x}dx$$

         $$=\displaystyle\int e^x\left(\dfrac{1+2\sin\dfrac{x}{2}\cos\dfrac{x}{2}}{2\cos^2\dfrac{x}{2}}\right)$$

         $$=\displaystyle\int e^x\left(\dfrac{\sin^2\dfrac{x}{2}+\cos^2\dfrac{x}{2}+2\sin\dfrac{x}{2}.\cos\dfrac{x}{2}}{2\cos^2\dfrac{x}{2}}\right)$$

         $$=\displaystyle\int e^x\left(\dfrac{\left(\sin\dfrac{x}{2}+\cos\dfrac{x}{2}\right)^2}{2\cos^2\dfrac{x}{2}}\right)$$

         $$=\displaystyle\int \dfrac{e^x}{2}\left(\dfrac{\sin\dfrac{x}{2}+\cos\dfrac{x}{2}}{\cos\dfrac{x}{2}}\right)^2$$

         $$=\displaystyle\int\dfrac{e^x}{2}\left(\dfrac{\sin\dfrac{x}{2}}{\cos\dfrac{x}{2}}+\dfrac{\cos\dfrac{x}{2}}{\cos\dfrac{x}{2}}\right)^2$$

         $$=\displaystyle\int\dfrac{e^x}{2}\left(\tan\dfrac{x}{2}+1\right)^2$$

         $$=\displaystyle\int\dfrac{e^x}{2}\left(\tan^2\dfrac{x}{2}+1+2\tan\dfrac{x}{2}\right)$$

         $$=\displaystyle\int\dfrac{e^x}{2}\left(\sec^2\dfrac{x}{2}+2\tan\dfrac{x}{2}\right)$$

         $$=\displaystyle\int e^x\left(\dfrac{1}{2}.\sec^2\dfrac{x}{2}+\dfrac{2}{2}\tan\dfrac{x}{2}\right)$$

         $$=\displaystyle\int e^x\left(\tan\dfrac{x}{2}+\dfrac{1}{2}\sec^2\dfrac{x}{2}\right)$$

Above is in the form of $$e^x[f(x)+f'(x)]dx=e^xf(x)+c$$
         $$=e^x\tan\dfrac{x}{2}+c$$

$$\therefore$$  $$\displaystyle\int\dfrac{e^x(1+\sin x)}{1+\cos x}dx$$  $$=e^x\tan\dfrac{x}{2}+c$$


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Single Correct Medium Published on 17th 09, 2020
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