How to find Integration of Trigonometric function with this rules $R(-\sin x,-\cos x)=R(\sin x,\cos x).$..

Question

$$\int \dfrac{dx}{1+\sin^2 x}$$
If this formula is true, $R(- \sin x,-\cos x)=R(\sin x,\cos x)$, this thing can be used $t=tgx$, $\sin x=\dfrac{t}{\sqrt{t^2+1}}$, $\cos x=\dfrac{1}{\sqrt{t^2+1}}$, $dx=\dfrac{dt}{t^2+1}$, so we apply this rules if the formula is true. I don't know how this work and there are 2-3 more like this. If anyone can explain how this formulas need be used it will be very helpful for me. :)

Answer 1

You Can Substitute $$\sin(x)=\frac{2t}{1+t^2}$$ and $$dx=\frac{dt}{1+t^2}$$ so our integral is given by $$\int\frac{1}{1+\left(\frac{2t}{1+t^2}\right)^2}dt$$

Answer 2

Hint: $$\int\dfrac{a\sin^2x+b\sin x\cos x+c\cos^2x}{p\sin^2x+q\sin x\cos x+r\cos^2x}dx$$

Divide numerator & denominator by $\cos^2x$ and choose $\tan x=t$