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

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 the integral of tanx

We know that $\tan x=\frac{\sin x}{\cos x}$, and we use substitution for integration. Let $u = \cos x$, then $du=-\sin xdx$. So $\int\tan xdx=-\int\frac{du}{u}$.

Step2: Integrate -du/u

The integral of $\frac{1}{u}$ is $\ln|u|+C$. So $-\int\frac{du}{u}=-\ln|u| + C$.

Step3: Substitute back u = cosx

Since $u = \cos x$, we have $-\ln|\cos x|+C$. And $-\ln|\cos x|=\ln|\frac{1}{\cos x}|=\ln|\sec x|$. So $\int\tan xdx=\ln|\sec x|+C$.

Answer:

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