How do I evaluate the integral: $int_0^(7sqrt(3/2))dx / sqrt(49-x^2)$ ?

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Answer 1 · 2015-01-27T18:08:56

First, recall the antiderivative formula:
$int dx/sqrt(a^2-u^2) = arcsin(u/a) + C$

Thus, $int dx/sqrt(49-x^2) = int dx/sqrt(7^2-x^2)$

We have $int dx/sqrt(7^2-x^2) = arcsin(x/7) + C$

Substitute in the constraints we

$int_0^(7sqrt(3/2)) dx/sqrt(7^2-x^2) = arcsin((7sqrt(3/2))/7) - arcsin(0) = arcsin((7sqrt(3/2))/7)$

Please find the formula sheet below for reference:
$www.eeweb.com$

How do I evaluate the integral: int_0^(7sqrt(3/2))dx / sqrt(49-x^2)int_0^(7sqrt(3/2))dx / sqrt(49-x^2)?