Question

Evaluate this integral.

Answer 1

\[7\int\limits_{0}^{a} \frac{ dx }{ (a^2+x^2)^{\frac{ 3 }{ 2 }} }\] where a >0

Answer 2

\[\text{Let }x=a\tan u\\
dx=a\sec^2 u \;du\\
\text{The integral then becomes}\\
7\int_{0}^{a\tan a}\dfrac{a\sec^2 u}{(a^2+(a\tan u)^2)^{3/2}}du\]
Do you see where this is going?

Answer 3

Yeah i do but i am bad at trig substitution

Answer 4

Well, I've already made the substitution. All that's left is simplifying the integrand until you get something easier to work with.

Answer 5

why did you change the integral to atana

Answer 6

Because I changed the variable of integration. The limits 0 and a corresponded to x. Now we're integrating with respect to u.

I also made a mistake with those limits... They should actually be 0 and pi/4. Sorry!

Unsure how to get the limits?
Since x = a tanu, we also have
u = arctan(x/a).

Upper u = arctan(a/a) = arctan(1) = pi/4
Lower u = arctan(0/a) = arctan(0) = 0

Answer 7

So the integral should be
\[7\int_0^{\pi/4}... du\]
As for simplification,
\[\frac{a\sec^2 u}{(a^2+a^2\tan^2u)^{3/2}}\\
\frac{a\sec^2 u}{(a^2 (1+\tan^2u))^{3/2}}\\
\frac{a\sec^2 u}{(a^2)^{3/2} (1+\tan^2u)^{3/2}}\\
\frac{a\sec^2 u}{a^3 (\sec^2u)^{3/2}}\\
\frac{\sec^2 u}{a^2 \sec^3u}\\
\frac{1}{a^2 \sec u}\\\]
So now you're down to
\[7\int_0^{\pi/4}\frac{1}{a^2\sec u} du\]

Answer 8

Thank you soo much for showing me that! I kept messing it up. I think i got it from here! :)

Answer 9

You're welcome!