if $\frac{dy}{dx}=\tan x$, then $y =$ \na $\frac{1}{2}\tan^{2}x + c$\nb $sec^{2}x + c$\nc $ln|sec x|+c$\nd…

if $\frac{dy}{dx}=\tan x$, then $y =$ \na $\frac{1}{2}\tan^{2}x + c$\nb $sec^{2}x + c$\nc $ln|sec x|+c$\nd $ln|cos x| + c$\ne $sec x\tan x + c$

if $\frac{dy}{dx}=\tan x$, then $y =$ \na $\frac{1}{2}\tan^{2}x + c$\nb $sec^{2}x + c$\nc $ln|sec x|+c$\nd $ln|cos x| + c$\ne $sec x\tan x + c$

Answer

Explanation:

Step1: Recall integral of tangent

We know that $\tan x=\frac{\sin x}{\cos x}$, and we need to find $\int\tan xdx$.

Step2: Use substitution

Let $u = \cos x$, then $du=-\sin xdx$. So $\int\tan xdx=\int\frac{\sin x}{\cos x}dx=-\int\frac{du}{u}$.

Step3: Integrate $\frac{1}{u}$

The integral of $\frac{1}{u}$ is $\ln|u|+C$. Substituting back $u = \cos x$, we get $-\ln|\cos x|+C$. Since $-\ln|\cos x|=\ln|\frac{1}{\cos x}|=\ln|\sec x|$, then $\int\tan xdx=\ln|\sec x|+C$.

Answer:

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