How do I evaluate the integral: int_0^(7sqrt(3/2))dx / sqrt(49-x^2)∫7√320dx√49−x2?

1 Answer
Mind
Jan 27, 2015

First, recall the antiderivative formula:
int dx/sqrt(a^2-u^2) = arcsin(u/a) + C∫dx√a2−u2=arcsin(ua)+C

Thus, int dx/sqrt(49-x^2) = int dx/sqrt(7^2-x^2)∫dx√49−x2=∫dx√72−x2

We have int dx/sqrt(7^2-x^2) = arcsin(x/7) + C∫dx√72−x2=arcsin(x7)+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) ∫7√320dx√72−x2=arcsin⎛⎜ ⎜⎝7√327⎞⎟ ⎟⎠−arcsin(0)=arcsin⎛⎜ ⎜⎝7√327⎞⎟ ⎟⎠

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

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