How do you evaluate the integral $\int \frac{{sec}^{2} x}{1 + tan x} d x$ $int sec^2x/(1+tanx)dx$ ?

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How do you evaluate the integral $\int \frac{{sec}^{2} x}{1 + tan x} d x$ $int sec^2x/(1+tanx)dx$ ?

1 Answer

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Answer 1 · 2017-01-02T15:56:30

$\int \frac{{sec}^{2} x}{1 + tan x} d x = ln | 1 + tan x | + C$ $intsec^2x/(1 + tanx)dx = ln|1 + tanx| + C$

Explanation:

This is a u-subsitution problem. Our goal is to cancel out the numerator. Let $u = 1 + tan x$ $u = 1 + tanx$ . Then $d u = {sec}^{2} x d x$ $du = sec^2xdx$ and $d x = \frac{d u}{{sec}^{2} x}$ $dx= (du)/sec^2x$

$= \int \frac{{sec}^{2} x}{u} \cdot \frac{d u}{{sec}^{2} x}$ $=intsec^2x/u * (du)/sec^2x$

$= \int (\frac{1}{u}) d u$ $= int(1/u) du$

This can be integrated as $\int (\frac{1}{x}) d x = ln | x | + C$ $int(1/x)dx = ln|x| + C$ .

$= ln | u | + C$ $= ln|u| + C$

$= ln | 1 + tan x | + C$ $= ln|1 + tanx| + C$ , where $C$ $C$ is a constant

Hopefully this helps!

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How do you evaluate the integral $\int \frac{{sec}^{2} x}{1 + tan x} d x$ $int sec^2x/(1+tanx)dx$ ?

1 Answer

Explanation:

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How do you evaluate the integral $\int \frac{{sec}^{2} x}{1 + tan x} d x$ $int sec^2x/(1+tanx)dx$ ?