How do you integrate $int 1/sqrt(7-3x^2)$ by trigonometric substitution?

Question

5405 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-10-30T07:35:00

$int1/sqrt(7 - 3x^2)dx = (sqrt(3)/3)sin^-1(sqrt(3/7)x) + C$

Given:

$int1/sqrt(7 - 3x^2)dx$

Divide by $sqrt(3)$ so that the integrand is in the form $1/sqrt(a^2 - x^2)$

$1/sqrt(3)int1/sqrt(7/3 - x^2)dx$

Please notice that $a^2 = 7/3$ , therefore, $a = sqrt(7/3)$

The reference for Trignometric substitutions says to substitute $x = asin(t)$

let $x = sqrt(7/3)sin(t)$ , then $dx = sqrt(7/3)cos(t)dt$

$1/sqrt(3)int(sqrt(7/3)cos(t))/sqrt(7/3 - 7/3sin^2(t))dt$

From the same reference, it says to substitute $a^2cos^2(t)$ for $a^2 - a^2sin^2(t)$ :

$1/sqrt(3)int(sqrt(7/3)cos(t))/sqrt(7/3cos^2(t))dt =$

$1/sqrt(3)int(sqrt(7/3)cos(t))/(sqrt(7/3)cos(t))dt =$

$1/sqrt(3)intdt = (sqrt(3)(t))/3 + C$

Solve the original substitution for t:

$x = sqrt(7/3)sin(t)$

$sin(t) = sqrt(3/7)x$

$t = sin^-1(sqrt(3/7)x)$

Substitute $sin^-1(sqrt(3/7)x)$ for $t$

$int1/sqrt(7 - 3x^2)dx = (sqrt(3)(sin^-1(sqrt(3/7)x)))/3 + C$

How do you integrate int 1/sqrt(7-3x^2)int 1/sqrt(7-3x^2) by trigonometric substitution?