Question

please help..integrate (16-19x^2)^3/2 /x^6 dx..
thank you.

Answer 1

Is that really 19x^2 or is it just 9x^2?

Answer 2

sorry it is 9x^2

Answer 3

Let \(\large x=\frac{4}{3}\sin\theta\). Can you change all of the x's to theta?

Answer 4

i think

Answer 5

this topic was integration by parts

Answer 6

Can't do it by integration by parts. :(

Answer 7

it is integration by substitution using trigonometric substitution..please help

Answer 8

\[\begin{align}
\text{Let }&x=\frac{4}{3}\sin\theta\\
&dx=\frac{4}{3}\cos\theta d\theta\\
&x^6=\frac{2^{12}}{3^6}\sin^6\theta\\
&(16-9x^2)^{3/2}=8\cos^3\theta
\end{align}\]
Substitute the last three expressions into the integrand and simplify until you can evaluate the integral.

Answer 9

thank u..i'm trying now..

Answer 10

@blockcolder is there other way to solve this? because i was confuse where did u get the 4/3?thank u

Answer 11

how and why does trig come into this?