1. trigonometry 3

$$\int\frac{1}{sin^2(x)+2cos^2(x)}dx$$
My attempt
$$=\int\frac{1}{2-sin^2(x)}dx = \int\frac{1}{1+cos^2(x)}dx$$
Not sure how to proceed here so tried substitution t=tan(x/2).
$$t=tan(x); cos(x)=\frac{1-t^2}{1+t^2}; dx=\frac{2}{1+t^2}dt$$
But it didn't go well... It seems that there should be a better way. It's also possible that i screwed up.

$$=\int\frac{1}{1+(\frac{1-t^2}{1+t^2})^2}*\frac{2}{1+t^2}dt$$
$$=\int\frac{1+t^2}{1+t^4}dt$$

Does it look ok so far?

2. Re: trigonometry 3

No, that does not look correct so far.

$t = \tan x$, $dx = \dfrac{1}{1+t^2}dt$?

$\cos^2 x = \dfrac{1}{1+t^2}$

Then, you have:

$\displaystyle \int \dfrac{1}{1+ \tfrac{1}{1+t^2} }\cdot \dfrac{dt}{1+t^2} = \int \dfrac{dt}{2+t^2} = \dfrac{1}{\sqrt{2}}\tan^{-1}\left( \dfrac{t}{\sqrt{2}} \right) + C$

3. Re: trigonometry 3

Thanks I've learned a lot, especially when you explained how cos^2(x) relates to tan(x) and t in another thread here.

It's interesting how you prefer using t=tanx but our book insist on t=tan(x/2). Says it a general method of integrating all rational functions that have sin and cosx.