Question

How do you find the integral of (sqrt(cot x)(csc^2 x) dx ?

Answer 1

\[\int\sqrt{\cot x}~\csc^2x~dx\]
Let \(u=\cot x\), then \(du=-\csc^2x~dx\), or \(-du=\csc^2x~dx\).
So you have
\[-\int\sqrt u~du\]

Answer 2

I'm still confused about how o find it could you explain a little more?

Answer 3

\[\large \int \sqrt{\color \green{cot(x)}} \color \red{csc^2(x)dx}\]

you'll see why I highlighted ina bit

We do a u-substitution letting u = cot(x)

We know the derivative of cot(x) = -csc^2(x) so
\[\large u = \color \green{cot(x)}\]\[\large du = \color \red{-csc^2(x)dx}\]

So what we are left with is (look at the relation between the highlights)

\[\large \int- \sqrt{u}du\]

We dont need that '-' sign right now...so lets just factor it out

\[\large - \int \sqrt{u} du\]

Now what is the integral of √u?

We know that \(\large \sqrt{x} = x^{1/2}\) so

\[\large - \int u^{1/2}du\]

Now just using basic integration steps we know that

\[\large \int x^a = \frac{x^{a + 1}}{a + 1}\]

so

\[\large - \int \frac{u^{1/2 + 1}}{1/2 + 1} = -\frac{u^{3/2}}{3/2} = -\frac{2u^{3/2}}{3}\]

And lastly, we must plug back in 'u' in terms of 'x'...we originally made u = cot(x) so

\[\large -\frac{2cot(x)^{3/2}}{3} + c\]

Answer 4

Thank you so much!!! I understand it now(:

Answer 5

Anytime :)