Question

Find
integral tan^-x​

Answer 1

Discussion of

(integral) tan x = - ln|cos x| + C.

1. Proof

Strategy: Make in terms of sin's and cos's; Use Substitution.

(integral) tan x dx = (integral) sin x

COs x dx

set

u = COs x.

then we find

du = - sin x dx

substitute du=-sin x, u=COs x

(integral) sin x

COs x dx = - (integral)

(-1) sin x dx

COs x

= - (integral) du

u

Solve the integral

= - ln |u| + C

substitute back u=COs x

= - ln |COs x| + C

Q.E.D.

2. Alternate Form of Result

(integral) tan x dx = - ln |COs x| + C

= ln | (COs x)-1 | + C

= ln |sec x| + C

Therefore:

(integral) tan x dx = - ln |COs x| + C = ln |sec x| + C

Answer 2

$\int\ {tan^{-1}x} \, dx$

$= \int\ {tan^{-1}x}*1 \, dx$

Applying Integration by parts, we get

$\int\ {tan^{-1}x}*1 \, dx$

$= tan^{-1}x*\int{dx} - \int{\frac{d}{dx}(tan^{-1}x)*\int{dx}$

$= xtan^{-1}x - \int{\frac{1}{1+x^2}*xdx$

$= xtan^{-1}x-\frac{1}{2} \int{\frac{2x}{1+x^2}dx$

$= xtan^{-1}x-\frac{1}{2} log|1+x^2|+c$

Hence

$\int{tan^{-1}x} \, dx= xtan^{-1}x-\frac{1}{2} log|1+x^2|+c$

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