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

1 Answer
Mind
Jan 27, 2015

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

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