Mathematics

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

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

##### 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
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86

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