Question

f(x)=(9+x^2)^(-1/2) on the interval from [0,4]
throwing a blank on the integral

Answer 1

\[\int_0^4\frac{1}{\sqrt{9+x^2}}~dx\]
Try a trigonometric substitution. Let \(x=3\tan u\), so you have \(dx=3\sec^2u~du\). The limits of integration change as well; since \(u=\arctan \dfrac{x}{3}\),
\[\text{upper: }u=\arctan\frac{4}{3}\\
\text{lower: }u=\arctan\frac{0}{3}=0\]
\[\int_0^{\arctan(4/3)}\frac{3\sec^2u}{\sqrt{9+(3\tan u)^2}}~du\\
\int_0^{\arctan(4/3)}\frac{3\sec^2u}{\sqrt{9+9\tan^2u}}~du\\
\int_0^{\arctan(4/3)}\frac{\sec^2u}{\sqrt{1+\tan^2u}}~du\\
\int_0^{\arctan(4/3)}\frac{\sec^2u}{\sqrt{\sec^2u}}~du\\
\int_0^{\arctan(4/3)}\sec u~du\]

Answer 2

thanks

Answer 3

You're welcome.