How do you integrate $\int \frac{1}{t^{3} \sqrt{t^{2} - 1}}$ $int 1/(t^3sqrt(t^2-1))$ by trigonometric substitution?

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How do you integrate $\int \frac{1}{t^{3} \sqrt{t^{2} - 1}}$ $int 1/(t^3sqrt(t^2-1))$ by trigonometric substitution?

How do you integrate $\int \frac{1}{t^{3} \sqrt{t^{2} - 1}} d t$ $int 1/(t^3sqrt(t^2-1))dt$ by trigonometric substitution?

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Answer 1 · 2016-08-31T10:59:03

$\frac{1}{2} (a r c sec t + \frac{\sqrt{t^{2} - 1}}{t^{2}}) + C$ $1/2(arc sect+sqrt(t^2-1)/t^2)+C$ .

Explanation:

Let $I = \int \frac{1}{t^{3} \sqrt{t^{2} - 1}} d t$ $I=int1/(t^3sqrt(t^2-1))dt$ .

We subst. $t = sec x, so that, d t = sec x tan x d x$ $t=sec x, "so that," dt=sec xtan xdx$ . Hence,

$I = \int \frac{sec x tan x}{{sec}^{3} x tan x} d x = \int {cos}^{2} x d x = \int \frac{1 + cos 2 x}{2} d x$ $I=int(secxtanx)/(sec^3xtanx)dx=intcos^2xdx=int(1+cos2x)/2dx$

$= \frac{1}{2} {x + \frac{sin 2 x}{2}} = \frac{1}{2} (x + sin x cos)$ $=1/2{x+(sin2x)/2}=1/2(x+sinxcos)$

Now, $sec x = t \Rightarrow x = a r c sec t, cos x = \frac{1}{t}, sin x = \sqrt{1 - \frac{1}{t^{2}}}$ $secx=t rArr x=arc sect, cos x=1/t, sinx=sqrt(1-1/t^2)$ .

Therefore, $I = \frac{1}{2} {a r c sec t + (\frac{1}{t}) (\frac{\sqrt{t^{2} - 1}}{t}}$ $I=1/2{arc sect+(1/t)(sqrt(t^2-1)/t}$ , or,

$I = \frac{1}{2} (a r c sec t + \frac{\sqrt{t^{2} - 1}}{t^{2}}) + C$ $I=1/2(arc sect+sqrt(t^2-1)/t^2)+C$ .

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How do you integrate $\int \frac{1}{t^{3} \sqrt{t^{2} - 1}}$ $int 1/(t^3sqrt(t^2-1))$ by trigonometric substitution?

How do you integrate $\int \frac{1}{t^{3} \sqrt{t^{2} - 1}} d t$ $int 1/(t^3sqrt(t^2-1))dt$ by trigonometric substitution?

1 Answer

Explanation:

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Explanation:

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How do you integrate $\int \frac{1}{t^{3} \sqrt{t^{2} - 1}}$ $int 1/(t^3sqrt(t^2-1))$ by trigonometric substitution?

How do you integrate $\int \frac{1}{t^{3} \sqrt{t^{2} - 1}} d t$ $int 1/(t^3sqrt(t^2-1))dt$ by trigonometric substitution?