\int\limits \frac{ (1-x^2)^\frac{ 3 }{ 2 } }{ x^6 }dx
I should get - (1-x^2)^(5/2) over 5x^5

Question

\int\limits \frac{ (1-x^2)^\frac{ 3 }{ 2 } }{ x^6 }dx
I should get - (1-x^2)^(5/2) over 5x^5

rock_mit182 · Answer

$$\int\limits \frac{ (1-x^2)^\frac{ 3 }{ 2 } }{ x^6 }$$

rock_mit182 · Answer

@campbell_st @mathmate

rock_mit182 · Answer

please give me a hand

anonymous · Answer

Quick tip: wrap your code in `` to have it display properly.

I would try a trig sub, setting $x=\sin u$, then $dx=\cos u\,du$.
$$\int\frac{(1-x^2)^{3/2}}{x^6}\,dx=\int\frac{(1-\sin^2u)^{3/2}\cos u}{\sin^6u}\,du=\int\frac{\cos^4u}{\sin^6u}\,du=\int\cot^4u\csc^2u\,du$$
Substituting $t=\cot u$ would be the final step, which gives $-dt=\csc^2u\,du$.

anonymous · Answer

Cut-off part is just $du$.

rock_mit182 · Answer

@SithsAndGiggles according to wolfram the answer is $$\frac{ - (1-x^2)^\frac{ 5 }{ 2 } }{ 5x^5}$$

anonymous · Answer

What's the problem? Implementing the $t$ sub gives
$$-\int t^4\,dt=-\frac{1}{5}t^5+C=-\frac{1}{5}\cot^5u+C=-\frac{1}{5}\cot^5(\arcsin x)+C$$
Are you having trouble showing that this is equal to the answer?