Question

if $\frac{dy}{dx}=\tan x$, then $y=$
a $\frac{1}{2}\tan ^{2}x + c$
c $\ln|\sec x| + c$
e $\sec x\tan x + c$

Explanation:

Step1: Integrate $\tan x$

$$\begin{align*} y&=\int\tan xdx\\ &=\int\frac{\sin x}{\cos x}dx \end{align*}$$

Let $u = \cos x$, then $du=-\sin xdx$.

Step2: Substitute and integrate

$$\begin{align*} y&=-\int\frac{du}{u}\\ &=-\ln|u| + C\\ &=-\ln|\cos x|+C\\ &=\ln|\cos x|^{-1}+C\\ &=\ln|\sec x| + C \end{align*}$$

Answer:

C. $\ln|\sec x| + C$