Question

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

• A. $3^{\cos y}+c$
• B. $-\frac{3^{\cos y}}{\log 3}+c$
• C. $\sin y+c$
• D. $3^{\sin y}+c$

Answer 1

The correct option is B $-\frac{3^{\cos y}}{\log 3}+c$

$\frac{dI}{dy}=3^{\cos y}\cdot \sin y$

$dI=3^{\cos y}\cdot \sin ydy$

Integrating both sides

$I=\int 3^{\cos y}.\sin y.dy$

Substitute $\cos y=t$

$-\sin y\cdot dy=dt$

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

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

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