# Math Help - Trig Integral

1. ## Trig Integral

$\int \frac{sin(2x) - x}{x^2 + cos(2x)} dx$

No clue how to start this problem. Any help appreciated, I'll be looking through my notes to see if we've done something similar to this one in the mean time.

2. Try the substitution: $u = x^2 + \cos (2x)$

See where that takes you

3. Originally Posted by o_O
Try the substitution: $u = x^2 + \cos (2x)$

See where that takes you
$u=x^2+Cos(2x)$

$du=2x - Sin(2x)dx$

But I cant see how it applies to the Integral because the coefficient 2 in front of the x in the derivative. Am I missing to see something here?

4. You made an error when you took the derivative of $\cos (2x)$. Remember chain rule? A factor of 2 should then appear in your derivative which you can then factor out et voila. A standard integral.

5. Originally Posted by o_O
You made an error when you took the derivative of $\cos (2x)$. Remember chain rule? A factor of 2 should then appear in your derivative which you can then factor out et voila. A standard integral.
oh dang you're right I overlooked the 2. So the $\frac{d}{dx} Cos(2x) = -Sin(2x)(2x)' \rightarrow -2Sin(2x)$

This correct?

6. Yup. So: $du = \left(2x -2\sin (2x)\right)dx = -2 \left(\sin(2x) - x\right)dx$

which nicely appears in your numerator. Hence, a standard integral

7. Ok here's my work:

$u=x^2+Cos(2x)$
$du=3x-2Sin(2x)dx \rightarrow du=-2(-x+Sin(2x))dx\rightarrow -\frac{1}{2}du=-x+Sin(2x)dx$

$-\frac{1}{2}\int \frac{1}{u}du \rightarrow -\frac{1}{2}[ln|u|]$

$-\frac{1}{2}ln|x^2+Cos(2x)|+C$

Hows it look?

8. $du={\color{red}3}x-2Sin(2x)dx \rightarrow du=-2(-x+Sin(2x))dx\rightarrow -\frac{1}{2}du=-x+Sin(2x)dx$
Just a little typo which you didn't carry through. Good job!