• May 17th 2006, 11:53 AM
bobby77
Integrate sqrt(x)csc^2[x^(3/2)]dx
• May 17th 2006, 12:38 PM
topsquark
Quote:

Originally Posted by bobby77
Integrate sqrt(x)csc^2[x^(3/2)]dx

$\int dx \, \sqrt{x} \, csc^2 \left ( x^{3/2} \right )$

Let $y = x^{3/2}$
Thus $dy = \frac{3}{2} \sqrt{x}\, dx$

$\int dx \, \sqrt{x} \, csc^2 \left ( x^{3/2} \right ) = \frac{2}{3} \int dy \, csc^2(y)$

$= -\frac{2}{3}cot(y) = -\frac{2}{3} cot \left ( x^{3/2} \right )$

-Dan
• May 17th 2006, 12:48 PM
CaptainBlack
Quote:

Originally Posted by bobby77
Integrate sqrt(x)csc^2[x^(3/2)]dx

$\frac{d}{dx} \left(\frac{2}{3} x^{3/2}\right) =x^{1/2}$, so the
integrand is the derivative of a multiple of $\cot(x^{3/2})$.

From here the fundamental theorem of Calculus should give you the answer.

RonL
• May 17th 2006, 12:54 PM
topsquark
Quote:

Originally Posted by topsquark
$\int dx \, \sqrt{x} \, csc^2 \left ( x^{3/2} \right )$

Let $y = x^{3/2}$
Thus $dy = \frac{3}{2} \sqrt{x}\, dx$

$\int dx \, \sqrt{x} \, csc^2 \left ( x^{3/2} \right ) = \frac{2}{3} \int dy \, csc^2(y)$

$= -\frac{2}{3}cot(y) = -\frac{2}{3} cot \left ( x^{3/2} \right )$

-Dan

Actually the answer is: $-\frac{2}{3} cot \left ( x^{3/2} \right ) + C$. Don't get sloppy like me and forget the constant! :)

-Dan