How do I evaluate the indefinite integral $\int {sec}^{2} (x) \cdot tan (x) d x$ $intsec^2(x)*tan(x)dx$ ?

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Answer 1 · 2018-06-08T12:20:21

The answer is $= \frac{1}{2} {sec}^{2} x + C$ $=1/2sec^2x+C$

Perform this integral by substitution

Let $u = {sec}^{2} x$ $u=sec^2x$ , $\Rightarrow$ $=>$ , $d u = 2 {sec}^{2} x tan x d x$ $du=2sec^2xtanxdx$

Therefore, the integral is

$I = \int {sec}^{2} x tan x d x = \frac{1}{2} \int d u$ $I=intsec^2xtanxdx=1/2intdu$

$= \frac{1}{2} u$ $=1/2u$

$= \frac{1}{2} {sec}^{2} x + C$ $=1/2sec^2x+C$

How do I evaluate the indefinite integral intsec^2(x)*tan(x)dx∫sec2(x)⋅tan(x)dxintsec^2(x)*tan(x)dx ?