# Thread: Finding the indefinite integral

1. ## Finding the indefinite integral

⌡ cot^2(x) dx

I approached this question by replacing cot^2 (x) w/ [cos^2(x)]/[sin^2(x)] => substituting sin^2(x) w/ u^2 => and ended up w/ -(1/3)cot(x) csc^2(x) + C

The answer key just gives -cot (x) - x + C

I have no idea of drawing any correlation so far except I didnt approach the trig sign from the correct position to begin with. Help with detailed algebraic work to this answer is what I ask please.

2. Originally Posted by ugkwan

⌡ cot^2(x) dx

I approached this question by replacing cot^2 (x) w/ [cos^2(x)]/[sin^2(x)] => substituting sin^2(x) w/ u^2 => and ended up w/ -(1/3)cot(x) csc^2(x) + C

The answer key just gives -cot (x) - x + C

I have no idea of drawing any correlation so far except I didnt approach the trig sign from the correct position to begin with. Help with detailed algebraic work to this answer is what I ask please.
From the Pythagorean Identity: $\displaystyle \cot^2 (x) = \text{cosec}^2 (x) - 1$.

And since the derivative of $\displaystyle \cot (x)$ is $\displaystyle - \text{cosec}^2 (x)$ ....

3. Originally Posted by ugkwan

⌡ cot^2(x) dx

I approached this question by replacing cot^2 (x) w/ [cos^2(x)]/[sin^2(x)] => substituting sin^2(x) w/ u^2 => and ended up w/ -(1/3)cot(x) csc^2(x) + C
If you let $\displaystyle u^2= sin^2(x)$, then $\displaystyle u= sin(x)$ and $\displaystyle du= cos(x) dx$. Now you have two "cosines" in the problem so you will have $\displaystyle \int\frac{cos(x)}{sin^2(x)} (cos(x)dx)= \int \frac{cos(x)}{u^2} du$ and you still have that "cos(x)" in the integral. You could then argue that $\displaystyle cos(x)= \sqrt{1- sin^2(x)}$ and so change the integral to $\displaystyle \int \frac{\sqrt{1- u^2}}{u^2} du$ but I don't that that's any easier!

The answer key just gives -cot (x) - x + C

I have no idea of drawing any correlation so far except I didnt approach the trig sign from the correct position to begin with. Help with detailed algebraic work to this answer is what I ask please.[/QUOTE]