Mathematics

# If $\displaystyle \frac{dI}{dy}=3^{\cos y}.\sin y$ then $I$ is equal to

$\displaystyle - \dfrac{3^{\cos y}}{\log 3}+c$

##### SOLUTION
$\dfrac {dI}{dy}=3^{\cos y}\cdot \sin y$

$dI=3^{\cos y}\cdot \sin y dy$
Integrating both sides
$I=\int 3^{\cos y}.\sin y.dy$
Substitute $\cos y=t$

$-\sin y\cdot dy=dt$

$=-\int 3^t\cdot dt=\dfrac {-3^t}{\log 3}+c$

$=\dfrac {-3^{\cos y}}{\log 3}+c$

$I=\dfrac {-3^{\cos y}}{\log 3}+c$

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Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 109

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