How is $\frac{\sin^2x}{\cos^2x+1}=\frac{\tan^2x}{1+\sec^2x}$?

Question

I am reading my Calculus material and they present me this:

$$ \int\frac{\sin^2x}{\cos^2x+1}dx=\int\frac{\tan^2x}{1+\sec^2x}dx $$

I tried around playing with trig identities but I can't reach this equality. So, How is $$\frac{\sin^2x}{\cos^2x+1}=\frac{\tan^2x}{1+\sec^2x}$$? How can I manipulate the initial expression to get to the final one?

Thanks in advance

Answer 1

For $\cos x \ne 0$, $$\frac{\sin^2x}{\cos^2x+1}=\frac{\dfrac{\sin^2x}{\cos^2x}}{\dfrac{\cos^2x}{\cos^2x}+\dfrac1{\cos^2x}}=\dfrac{\tan^2x}{1+\sec^2x}$$