How do I evaluiate $\int sec (x) (sec (x) + tan (x)) d x$ $intsec(x)(sec(x) + tan(x)) dx$ ?

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How do I evaluiate $\int sec (x) (sec (x) + tan (x)) d x$ $intsec(x)(sec(x) + tan(x)) dx$ ?

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Answer 1 · 2018-05-01T06:44:42

$\int sec x (sec x + tan x) d x = sec x + tan x + c$ $int secx(secx+tanx) dx = secx + tanx + c$

Explanation:

I don't believe that there is any trigonometric substitution involved here actually.

The integrand $sec x (sec x + tan x)$ $secx(secx+tanx)$ can be expanded out to get ${sec}^{2} x + sec x tan x$ $sec^2x +secxtanx$ which are actually two simple functions to be antidifferentiated.

$\int {sec}^{2} x d x = tan x + c$ $int sec^2x dx = tanx + c$ because $\frac{d}{d x} (tan x) = {sec}^{2} x$ $d/dx(tanx) = sec^2x$ .

$\int sec x tan x d x = sec x$ $int secxtanx dx = secx$ which we will prove using substitution.

Expressing that in terms of sine and cosine, we get $sec x tan x = \frac{sin x}{{cos}^{2} x}$ $secxtanx = sinx/cos^2x$ so the integral becomes:

$\int \frac{sin x}{{cos}^{2} x} d x$ $int sinx/cos^2x dx$ , to which we apply the substitution $u = cos x$ $u=cosx$ .

$u = cos x$ $u=cosx$
$:. (du)/dx=-sinx$
$:. -du = sinxdx$

Now, using the change of variable rule, we get:

$- int 1/u^2 du$
$=1/u + c$
$= secx + c$

$:. int secx(secx+tanx) dx = secx + tanx + c$

And there you have it!

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How do I evaluiate $\int sec (x) (sec (x) + tan (x)) d x$ $intsec(x)(sec(x) + tan(x)) dx$ ?

1 Answer

Explanation:

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How do I evaluiate intsec(x)(sec(x) + tan(x)) dx∫sec(x)(sec(x)+tan(x))dxintsec(x)(sec(x) + tan(x)) dx?

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How do I evaluiate $\int sec (x) (sec (x) + tan (x)) d x$ $intsec(x)(sec(x) + tan(x)) dx$ ?