# Find the integral `int 1/sqrt(49-(36x^2))dx`

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### 1 Answer

The integral `int 1/sqrt(49-(36x^2))dx` has to be determined.

`int 1/sqrt(49-(36x^2))dx`

=> `int 1/(7*sqrt(1 - (6*x/7)^2)) dx`

Let `y = (6x/7) => dy = 6/7`

=> `(7/6)*int 1/(7*sqrt (1 - y^2))*dy`

The integral `int 1/sqrt (1 - x^2) dx` = `sin^-1(x)`

=> `(1/6)*sin^-1((6x)/7) + C`

**The integral `int 1/sqrt(49-(36x^2))dx = (1/6)*sin^-1((6x)/7) + C` **