Integrating $\int_{0}^{π/4} \frac{\sin x+\cos x}{\sin^4 x+\cos^2 x} dx$ Here is the problem
$$\int_{0}^{π/4} \frac{\sin x+\cos x}{\sin^4 x+\cos^2 x} dx$$
I have tried diving by $\cos^2 x$ and using partial fractions. Also substituting $\tan$ formulas or separating and partial fraction mess up the limits for me. 
So can I get a full solution with answer please?
 A: Hints: 
1) Divide it into two integrals : $$\int \frac{-d\cos{x}}{(1-\cos^{2}{x})^2+\cos^{2}{x}} + \int \frac{d\sin{x}}{\sin^{4}{x} + 1 - \sin^{2}{x}}$$
Then use rational functions 
2) Use $\tan{\frac x2} = t$ then it's easy to get rational functions in numerator and denominator. 
A: Use your trig identities to create a $u$-substitution. We have
$$
[\sin(x) + \cos(x)]dx = -d[\cos(x) - \sin(x)]
$$
and
$$
\sin(x)^4 + \cos(x)^2 = 1 - \sin(x)^2\cos(x)^2 = 1 - \frac{1}{4}\left(1 - [\cos(x) - \sin(x)]^2\right)^2,
$$
so
$$
\int_0^{\pi/4}\frac{\sin(x)+\cos(x)}{\sin(x)^4 + \cos(x)^2}dx = 4\int_0^1\frac{du}{4 - (1-u^2)^2} = 4\int_0^1\frac{du}{(3-u^2)(1+u^2)}
$$
and I'm guessing you know where to go from here.
Note that if you didn't know those trig identities for the denominator, you could still use $u = \cos(x) -\sin(x) = \sqrt{2}\cos\left(x+\frac{\pi}{4}\right)$, then plug in $x = \cos^{-1}\left(\frac{u}{\sqrt{2}}\right) - \frac{\pi}{4}$ and do the algebra. It's not fun, but it gets there in the end.
