How do you integrate $\int \frac{x}{\sqrt{x^{2} - 7}}$ $int x/sqrt(x^2-7)$ by trigonometric substitution?

Question

How do you integrate $\int \frac{x}{\sqrt{x^{2} - 7}}$ $int x/sqrt(x^2-7)$ by trigonometric substitution?

2 Answers

Impact of this question

15440 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-11T15:29:46

$\int \frac{x d x}{\sqrt{x^{2} - 7}} = \sqrt{x^{2} - 7} + C$ $int (xdx)/sqrt(x^2-7) = sqrt(x^2-7)+C$

Explanation:

You do not really need a trigonometric substitution here.

If you pose:

$t = x^{2} - 7$ $t= x^2-7$
$d t = 2 x d x$ $dt = 2xdx$

the integral becomes:

$\int \frac{x d x}{\sqrt{x^{2} - 7}} = \frac{1}{2} \int \frac{d t}{\sqrt{t}} = \sqrt{t} + C = \sqrt{x^{2} - 7} + C$ $int (xdx)/sqrt(x^2-7) = 1/2 int (dt)/sqrt(t) = sqrt(t) + C = sqrt(x^2-7)+C$

Answer 2 · 2017-01-11T18:01:30

Although you do not need trigonometric substitution, it can be used for this integral.

Explanation:

Recall that ${sec}^{2} θ - 1 = {tan}^{2} θ$ $sec^2theta-1 = tan^2 theta$ , so that's the basic substitution we'll use. (We could use hyperbolic functions instead, but not everyone is familiar with them when first learning trig sub.)

We need, not just ${sec}^{2} θ - 7$ $sec^2 theta - 7$ , but $7 {sec}^{2} θ - 7$ $7sec^2 theta - 7$ so that we can factor out the $7$ $7$

$\int \frac{x}{\sqrt{x^{2} - 7}} d x$ $int x/sqrt(x^2-7) dx$

Let $x = \sqrt{7} sec θ$ $x = sqrt7 sec theta$ , so that $d x = \sqrt{7} sec θ tan θ d θ$ $dx = sqrt7 sec theta tan theta d theta$

and $\sqrt{x^{2} - 7} = \sqrt{7 {sec}^{2} θ - 7} = \sqrt{7} tan θ$ $sqrt(x^2-7) = sqrt (7sec^2 theta - 7) = sqrt7 tan theta$

$\int \frac{x}{\sqrt{x^{2} - 7}} d x = \int \frac{\sqrt{7} sec θ}{\sqrt{7} tan θ} \sqrt{7} sec θ tan θ d θ$ $int x/sqrt(x^2-7) dx = int (sqrt7sec theta)/(sqrt7 tan theta) sqrt7 sec theta tan theta d theta$

$= \int \sqrt{7} {sec}^{2} θ d θ = \sqrt{7} tan θ + C$ $= int sqrt7 sec^2 theta d theta = sqrt7 tan theta +C$

With $x = \sqrt{7} sec θ$ $x = sqrt7 sec theta$ , we get $tan θ = \frac{\sqrt{x^{2} - 7}}{\sqrt{7}}$ $tan theta = sqrt(x^2-7)/sqrt7$

Therefore,

$\int \frac{x}{\sqrt{x^{2} - 7}} d x = \sqrt{x^{2} - 7} + C$ $int x/sqrt(x^2-7) dx = sqrt(x^2-7) +C$

Science

Math

Humanities

... and beyond

How do you integrate $\int \frac{x}{\sqrt{x^{2} - 7}}$ $int x/sqrt(x^2-7)$ by trigonometric substitution?

2 Answers

Explanation:

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you integrate ∫x√x2−7int x/sqrt(x^2-7) by trigonometric substitution?

2 Answers

Explanation:

Explanation:

Related questions

Impact of this question

How do you integrate $\int \frac{x}{\sqrt{x^{2} - 7}}$ $int x/sqrt(x^2-7)$ by trigonometric substitution?