Trig Integral

**VitaX** · February 21st 2010, 05:02 PM

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

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.

**o_O** · February 21st 2010, 05:08 PM

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

See where that takes you

**VitaX** · February 21st 2010, 05:23 PM

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?

**o_O** · February 21st 2010, 05:25 PM

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.

**VitaX** · February 21st 2010, 05:30 PM

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?

**o_O** · February 21st 2010, 05:47 PM

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

**VitaX** · February 21st 2010, 05:49 PM

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?

**o_O** · February 21st 2010, 06:14 PM

$du={\color{red}3}x-2Sin(2x)dx \rightarrow du=-2(-x+Sin(2x))dx\rightarrow -\frac{1}{2}du=-x+Sin(2x)dx$

Just a little typo which you didn't carry through. Good job!

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