Mathematics

Solve $$\displaystyle\int {\dfrac{{dx}}{{2\sin x + \cos x + 3}}} $$


ANSWER

$$\tan^{-1}\left(\tan \dfrac{x}{2}+1\right)+c$$


SOLUTION
$$\displaystyle I=\int \dfrac{dx}{2\sin x+\cos x+3}$$

$$\displaystyle =\int \dfrac{dx}{2\left(\dfrac{2\tan\dfrac{x}{2}}{1+\tan^2\dfrac{x}{2}}\right)+\left(\dfrac{1-\tan^2\dfrac{x}{2}}{1+\tan^2\dfrac{x}{2}}\right)+3}$$

$$=\displaystyle \int \dfrac{1+\tan^2\dfrac{x}{2}dx}{4\tan\dfrac{x}{2}+1-\tan^2\dfrac{x}{2}+3+3\tan^2\dfrac{x}{2}}$$

$$=\displaystyle \int \dfrac{\sec^2\dfrac{x}{2}dx}{2\tan^2\dfrac{x}{2}+4\tan\dfrac{x}{2}+4}$$

Let $$\tan\dfrac{x}{2}=t$$

$$=\sec^2\dfrac{x}{2}.\dfrac{1}{2}dx=dt$$

$$=\boxed{\sec^2\dfrac{x}{2}dx=2dt}$$

$$=\displaystyle \int \dfrac{2dt}{2t^2+4t+4}=\int \dfrac{dt}{t^2+2t+2}$$

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

$$=\tan^{-1}\left(\tan \dfrac{x}{2}+1\right)+c$$
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Single Correct Medium Published on 17th 09, 2020
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