1. ## Integration of 2/(3sin(2x)+4cos(2x))

Hello:
I am stuck at this sum:
$\displaystyle \int\frac{2}{3 sin 2x+4 cos 2x}$

Hello:
I am stuck at this sum:
$\displaystyle \int\frac{2}{3 sin 2x+4 cos 2x}$
$\displaystyle \int\frac{2}{3 \sin 2x+4 \cos 2x}\ \mathrm{d}x$

Let $\displaystyle u=2x \implies \mathrm{d}x= \frac{\mathrm{d}u}{2}$

$\displaystyle \therefore \int \frac{1}{3 \sin u+4 \cos u}\ \mathrm{d}u$

You have to use T-Substitution and solve: $\displaystyle t=\tan \left(\frac{u}{2}\right)$

Note: $\displaystyle \cos u = \frac{1-t^2}{1+t^2}$ and $\displaystyle \sin u = \frac{2t}{1+t^2}$.

3. Originally Posted by Air
$\displaystyle \int\frac{2}{3 \sin 2x+4 \cos 2x}\ \mathrm{d}x$

Let $\displaystyle u=2x \implies \mathrm{d}x= \frac{\mathrm{d}u}{2}$

$\displaystyle \therefore \int \frac{1}{3 \sin u+4 \cos u}\ \mathrm{d}u$

You have to use T-Substitution and solve: $\displaystyle t=\tan \left(\frac{u}{2}\right)$

Note: $\displaystyle \cos u = \frac{1-t^2}{1+t^2}$ and $\displaystyle \sin u = \frac{2t}{1+t^2}$.
The t substitution, it props up everywhere and yet no mention of its uses are in the book. Thanks a lot.
t=tan x, can you give me any site address where it and the places where it can apply is discussed, I have yet to get used to this substitution.

4. Originally Posted by Air
$\displaystyle \int\frac{2}{3 \sin 2x+4 \cos 2x}\ \mathrm{d}x$

Let $\displaystyle u=2x \implies \mathrm{d}x= \frac{\mathrm{d}u}{2}$

$\displaystyle \therefore \int \frac{1}{3 \sin u+4 \cos u}\ \mathrm{d}u$

You have to use T-Substitution and solve: $\displaystyle t=\tan \left(\frac{u}{2}\right)$

Note: $\displaystyle \cos u = \frac{1-t^2}{1+t^2}$ and $\displaystyle \sin u = \frac{2t}{1+t^2}$.
I noticed I gave you your first thanks

5. Actually, those $\displaystyle 2x$ work well, since

\displaystyle \begin{aligned} 4\cos 2x+3\sin 2x&=2\left( 2{{\cos }^{2}}x-2{{\sin }^{2}}x+3\sin x\cos x \right) \\ & =2\Big(\left( 2{{\cos }^{2}}x-\sin x\cos x \right)+\left( 4\sin x\cos x-2{{\sin }^{2}}x \right)\Big) \\ & =2(2\cos x-\sin x)(2\sin x+\cos x). \end{aligned}

Then the integral becomes,

$\displaystyle \int{\frac{dx}{(2\cos x-\sin x)(2\sin x+\cos x)}}=\int{\frac{{{(2\cos x-\sin x)}^{2}}+{{(2\sin x+\cos x)}^{2}}}{5(2\cos x-\sin x)(2\sin x+\cos x)}\,dx},$ which equals $\displaystyle \frac{1}{5}\left( \int{\frac{2\cos x-\sin x}{2\sin x+\cos x}\,dx}+\int{\frac{2\sin x+\cos x}{2\cos x-\sin x}\,dx} \right),$

and these are natural logs.

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### dx/3sin^2x 4cos^2x

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