# Thread: Trig identies and integrals

1. ## Trig identies and integrals

I have a step in my book that I cannot figure $\int cot^2\theta csc^2 \theta d \theta$

and next step is $-\frac{1}{3}cot^3\theta +c$

I used $1+cot^2\theta = csc^2 \theta$ but that does not work -

2. Hello !
Originally Posted by weezie23
I have a step in my book that I cannot figure $\int cot^2\theta csc^2 \theta d \theta$

and next step is $-\frac{1}{3}cot^3\theta +c$

I used $1+cot^2\theta = csc^2 \theta$ but that does not work -
What is the derivative of $\cot \theta$ ?

3. Originally Posted by weezie23
I have a step in my book that I cannot figure $\int cot^2\theta csc^2 \theta d \theta$

and next step is $-\frac{1}{3}cot^3\theta +c$

I used $1+cot^2\theta = csc^2 \theta$ but that does not work -
that's because an identity is not what's needed here. it is a straight substitution problem. let $u = \cot \theta$

4. Originally Posted by Moo
Hello !

What is the derivative of $\cot \theta$ ?
geez! you just keep beating me today!

$(m \! \circ \! o)(x)=\mu$

5. Originally Posted by weezie23
I have a step in my book that I cannot figure $\int cot^2\theta csc^2 \theta d \theta$

and next step is $-\frac{1}{3}cot^3\theta +c$

I used $1+cot^2\theta = csc^2 \theta$ but that does not work -
suppose $t = cot \theta$

so, $dt=-csc^2 \theta \Rightarrow -dt=csc^2 \theta$

now, $\int cot^2\theta csc^2 \theta d \theta$

$= -\int t^2 dt$

$=-\frac{t^3}{3} +c$

$=-\frac{1}{3}cot^3\theta +c$

6. ## I got it

u = cot x

du = -csc^2x

integral u^2du = U^3/3+C

thanks for refocusing me!!

7. Originally Posted by weezie23
u = cot x

du = csc^2x

integral u^2du = U^3/3+C

thanks for refocusing me!!
ahem, $du = {\color{red}-} \csc^2 x~{\color{red}dx}$