Mathematics

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


ANSWER

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


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
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