Mathematics

$$\displaystyle \int \dfrac { \sin { x }  }{ 1+4\cos { x }  } dx=\_ \_ \_ \_ \_ \quad +c$$


ANSWER

$$-\frac { 1 }{ 4 } \log { |1+4\cos { x } | } $$


SOLUTION
Given,

$$\displaystyle \int \dfrac{\sin \left(x\right)}{1+4\cos \left(x\right)}dx$$

Let $$u=1+4\cos \left(x\right)$$    $$du=-4sin(x)\ dx$$

$$\displaystyle=\int \:-\dfrac{1}{4u}du$$

$$\displaystyle=-\dfrac{1}{4}\log \left|u\right|$$

$$=-\dfrac{1}{4}\log \left|1+4\cos \left(x\right)\right|+C$$
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Single Correct Medium Published on 17th 09, 2020
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