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

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: Integrate $\tan x$

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

Step2: Apply the integral formula

Using the formula $\int\frac{1}{t}dt=\ln|t| + C$, we have: $$-\int\frac{du}{u}=-\ln|u|+C$$ Substituting back $u = \cos x$, we get $-\ln|\cos x|+C$. Since $-\ln|\cos x|=\ln|\cos x|^{-1}=\ln|\sec x|$, so $\int\tan xdx=\ln|\sec x|+C$.

Answer:

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