Integrate sqrt(x)csc^2[x^(3/2)]dx

2. Originally Posted by bobby77
Integrate sqrt(x)csc^2[x^(3/2)]dx
$\displaystyle \int dx \, \sqrt{x} \, csc^2 \left ( x^{3/2} \right )$

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

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

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

-Dan

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

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

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

RonL

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

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

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

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

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

-Dan