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

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

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Answer 1 · 2016-12-22T16:59:39

$\sqrt{x^{2} + 1} + C$ $sqrt(x^2 + 1) + C$

Explanation:

Let $x = tan θ$ $x= tantheta$ . Then $d x = {sec}^{2} θ d θ$ $dx = sec^2thetad theta$

$\Rightarrow \int \frac{tan θ}{\sqrt{({tan}^{2} θ + 1)}} {sec}^{2} θ d θ$ $=> int tan theta/sqrt((tan^2theta + 1)) sec^2theta d theta$

$\Rightarrow \int \frac{tan θ}{\sqrt{{sec}^{2} θ}} {sec}^{2} θ d θ$ $=> int tantheta/sqrt(sec^2theta) sec^2theta d theta$

$\Rightarrow \int \frac{tan θ}{sec θ} {sec}^{2} θ d θ$ $=> int tantheta/sectheta sec^2theta d theta$

$\Rightarrow \int tan θ sec θ d θ$ $=> int tan theta sec theta d theta$

This is a common integral-- $\int (tan x sec x) d x = sec x + C$ $int(tanxsecx)dx = secx + C$ .

$\Rightarrow sec θ + C$ $=> sec theta + C$

We now draw an imaginary triangle.

enter image source here

The definition of $sec θ$ $sectheta$ is $\frac{hypotenuse}{side adjacent θ}$ $"hypotenuse"/("side adjacent" theta)$ because $sec x = \frac{1}{cos x}$ $secx = 1/cosx$ . In this image, $sec θ = \sqrt{x^{2} + 1}$ $sectheta = sqrt(x^2 + 1)$ .

Therefore, the integral can be simplified to $\sqrt{x^{2} + 1} + C$ $sqrt(x^2 + 1) + C$ .

Hopefully this helps!

Answer 2 · 2016-12-23T00:26:51

By inspection rather than trig substitution.

Explanation:

Notice that the $x$ $x$ in the top is proportional to the derivative of the function under the square root. So just write down ${(x^{2} + 1)}^{- \frac{1}{2} + 1}$ $(x^2+1)^(-1/2+1)$ as a trial (rather like integrating any power of $x$ $x$ ) and differentiate it. By the chain rule, the function under the square root will provide the $2 x$ $2x$ and the $2$ $2$ 's will cancel, so the trial was an immediate success.

Alternatively substitute $u = x^{2} + 1$ $u=x^2+1$ , $d x = \frac{d u}{2 \sqrt{u - 1}}$ $dx=(du)/(2sqrt(u-1))$ and the integral becomes $(\frac{1}{2}) \int u^{- \frac{1}{2}} d u$ $(1/2)int u^(-1/2)du$ = $u^{\frac{1}{2}} + C$ $u^(1/2)+C$ = $\sqrt{x^{2} + 1} + C$ $sqrt(x^2+1)+C$

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

2 Answers

Explanation:

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