Mathematics

$$\displaystyle \int \dfrac{1}{3\sin x +4\cos x} dx$$ equal, where $$\tan \alpha = \dfrac{4}{3}$$


ANSWER

$$\dfrac{1}{5} ln \left|\tan \left(\dfrac{x}{2} + \dfrac{\alpha}{2}\right)\right|+K$$


SOLUTION
$$\displaystyle \int \dfrac{1}{3\sin x +4\cos x} dx = \int 5\dfrac{1}{\left(\dfrac{3}{5}\sin x + \dfrac{4}{5} \cos x\right)}dx$$ 
$$=\displaystyle \dfrac{1}{5} \int \dfrac{1}{\sin x\cdot \cos \alpha + \cos x \cdot \sin \alpha}dx$$    $$\tan \alpha = \dfrac{4}{3}$$
$$=\displaystyle \int \frac{1}{\sin (x+\alpha)}dx$$                            $$\sin \alpha = \dfrac{4}{5}$$
$$=\displaystyle \frac{1}{5} \int cose (x+\alpha) dx$$                      $$\cos \alpha  = \dfrac{3}{5}$$
$$=\displaystyle \frac{1}{5} ln \left|\tan \left(\frac{\pi}{2} + \frac{d}{2}\right)\right| + k$$
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Single Correct Medium Published on 17th 09, 2020
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