How do you find the antiderivative of $\int {sec}^{2} x tan x d x$ $int sec^2xtanx dx$ ?

Question

2426 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-10-27T16:49:24

$\int {sec}^{2} x tan x d x = \frac{1}{2} {sec}^{2} x + c$ $intsec^2xtanxdx=1/2sec^2x+c$

Method 1

now $\frac{d}{d x} (sec x) = sec x tan x$ $d/(dx)(secx)=secxtanx$

so $\int {sec}^{2} x tan x d x = \frac{1}{2} {sec}^{2} x + c_{1}$ $intsec^2xtanxdx=1/2sec^2x+c_1$

since $\frac{d}{d x} ({sec}^{2} x) = 2 (sec x) \times sec x tan x$ $d/(dx)(sec^2x)=2(secx)xxsecxtanx$

Method 2

$\frac{d}{d x} ({tan}^{2} x) = 2 (tan x) \times {sec}^{2} x$ $d/(dx)(tan^2x)=2(tanx)xxsec^2x$

by the chain rule

so $\int {sec}^{2} x tan x d x = \frac{1}{2} {tan}^{2} x + c_{2}$ $intsec^2xtanxdx=1/2tan^2x+c_2$

the 2 results can be shown to be equivalent by use of

$1 + {tan}^{2} x = {sec}^{2} x$ $1+tan^2x=sec^2x$

How do you find the antiderivative of int sec^2xtanx dx∫sec2xtanxdxint sec^2xtanx dx?