Question

compute the indefinite integral.
\[\int\limits (dx) /(x^2+4)^{5/2} \]

Answer 1

I don't have heaps of time, but I can get you started.

Make a substitution,\[x=2\tan \theta\]Then\[dx=2\sec ^2 \theta d \theta\]and your integral becomes,\[\int\limits_{}{}\frac{dx}{(x^2+4)^{5/2}}=\int\limits_{}{}\frac{2\sec^2 \theta }{(4+4\tan^2 \theta)^{5/2}}d \theta =\int\limits_{}{}\frac{2\sec^2 \theta }{2^5(1+\tan^2 \theta)}d \theta\]\[=\int\limits_{}{}\frac{2\sec^2 \theta }{2^5(\sec^2 \theta)^{5/2}}d \theta=\frac{1}{2^4}\int\limits_{}{}\frac{d \theta }{\sec^3 \theta}=\frac{1}{2^4}\int\limits_{}{}\cos^3 \theta d \theta\]

Answer 2

The denominator on the third integral in the firs line should be raised to the power of 5/2

Answer 3

You can solve this now using a reduction formula on cos^3(theta), or by using integration by parts on cos^3(theta) a couple of times.

Once you have your answer, remember to undo your substitution; that is\[\theta = \tan^{-1}\frac{x}{2}\]and add a constant.

If I didn't have to rush off, I'd finish it. I'll look in later to see how you went.

Good luck :)

Answer 4

\[{1\over12 }\left( \dfrac {x}{(x^2+4)^{3/2}}+\tan^{-1}(\dfrac{x}{2})\right)\]

Answer 5

thank you so much guys!