Question

find the integral of the tangent of x with respect to x

\[\int\tan(x)\cdot\text dx\]

Answer 1

Start off by changing it to:\[\int\limits (sinx/cosx)dx\]

We will use substitution to evaluate this integral.

Let u=cos x and let du=-sinx

So the integral becomes:
\[-\int\limits (1/u)du\]
Which becomes
\[-\ln \left| u \right|+c\]
Insert u back into the result:
\[-\ln \left| \cos x \right|+c\]

I believe you can also do it by integration by parts, where the two functions are tan x and 1

Answer 2

\[\frac{\text dy}{\text dx}-y\tan x=-y^2\sec x\]\[\text {let }y=\frac 1v\]\[\frac{\text dy}{\text dx}=-\frac 1{v^2}\frac{\text dv}{\text dx}\]\[-\frac 1{v^2}\frac{\text dv}{\text dx}-\frac{\tan x}v=-\frac{\sec x}{v^2}\]\[\frac{\text dv}{\text dx}+v\tan x =\sec x\]\[v^\prime+v\tan x=\sec x\]\[R(x)=e^{\int \tan x\cdot\text dx}=e^{-\int\frac{-\sin x}{\cos x}\text dx}=e^{-\ln(\cos x)}=\sec x\]\[\left(v\sec x\right)^\prime=\sec^2 x\]\[v\sec x=\int \sec^2 x\text dx\]\[v\sec x=\tan x+c\]\[v=\sin x+c\cos x\]\[\frac 1y=\sin x+c\cos x\]\[y=\frac1{\sin x+c\cos x}\]