How to solve and indefinite integral: $$\int{\frac{dx}{\sin^2x+5\cos^2x}}.$$ The only method that can be used is $u$-substitution.
The only thing I've done so far is that I've tried modifying it so there is either only $\sin^2x$ or $\cos^2x$ left, so we get the following two integrals, respectively: $$\int{\frac{dx}{5-4\sin^2x}}$$ $$\int{\frac{dx}{1+4\cos^2x}}.$$
How would you go about finishing this problem using any of these two paths.
My question was deleted for being a duplicate of a question of solving and indefinite integral: $$\int{\frac{dx}{1+\sin^4x}}$$ which it obviously isn't.
The go-to is often the Weierstrass substitution, $u=\tan(x/2)$. This can be clunky, but it is effective for rational functions of sine and cosine.
A slightly more pleasant starting route can be achieved with
$$I := \int \frac{\mathrm{d}x}{\sin^2 x + 5 \cos^2 x} =\int \frac{\mathrm{d}x}{\sin^2 x + \cos^2 x + 4 \cos^2 x} =\int \frac{\mathrm{d}x}{1 + 4 \cos^2 x}$$
From here, divide the top and bottom by $\cos^2 x$:
$$I = \int \frac{\sec^2 x \, \mathrm{d} x}{\sec^2 x + 4}= \int \frac{\sec^2 x \, \mathrm{d} x}{\sec^2 x -1+5}= \int \frac{\sec^2 x \, \mathrm{d} x}{\tan^2 x + 5}$$
Then $u=\tan x$ gives $\mathrm{d}u = \sec^2 x \, \mathrm{d} x$ and
$$I = \int \frac{\mathrm{d}u}{u^2 + 5}$$
This is a fairly standard integral to solve with the substitution $u = \tan \theta$, or with a standard antiderivative,
$$\int \frac{\mathrm{d}\xi}{1+\xi^2} = \arctan \xi + C$$