What is the integral of $\int sec x ln (sec x + tan x) d x$ $int secxln(secx + tanx) dx$ ?

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What is the integral of $\int sec x ln (sec x + tan x) d x$ $int secxln(secx + tanx) dx$ ?

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Answer 1 · 2017-04-08T15:01:19

The integral equals $\frac{1}{2} {ln}^{2} (sec x + tan x) + C$ $1/2ln^2(secx + tanx) + C$

Explanation:

Let $u = ln (sec x + tan x)$ $u = ln(secx + tanx)$ . Then by the chain rule we have

$d u = \frac{sec x tan x + {sec}^{2} x}{sec x + tan x} d x$ $du = (secxtanx + sec^2x)/(secx + tanx)dx$ and $d x = \frac{sec x + tan x}{sec x tan x + {sec}^{2} x} d u$ $dx = (secx + tanx)/(secxtanx + sec^2x)du$

We now have, calling the integral $I$ $I$ :

$I = \int sec x \cdot u \cdot \frac{sec x + tan x}{sec x tan x + {sec}^{2} x} d u$ $I = int secx * u * (secx + tanx)/(secxtanx + sec^2x)du$

If we do a little factoring we get:

$I = \int sec x \cdot u \cdot \frac{sec x + tan x}{sec x (tan x + sec x)} d u$ $I = int secx * u * (secx + tanx)/(secx(tanx + secx)) du$

$I = \int u d u$ $I = int u du$

$I = \frac{1}{2} u^{2} + C$ $I = 1/2u^2 + C$

$I = \frac{1}{2} {ln}^{2} (sec x + tan x) + C$ $I = 1/2ln^2(secx + tanx) + C$

Hopefully this helps!

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What is the integral of $\int sec x ln (sec x + tan x) d x$ $int secxln(secx + tanx) dx$ ?

1 Answer

Explanation:

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What is the integral of int secxln(secx + tanx) dx∫secxln(secx+tanx)dxint secxln(secx + tanx) dx?

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What is the integral of $\int sec x ln (sec x + tan x) d x$ $int secxln(secx + tanx) dx$ ?