I am currently working on evaluating the integral:
$$ \int \frac{\sin x \cos x}{1 + \sin^4 x} \, dx. $$
In order to simplify the integrand, I've considered the substitution $ \tan(x) = t $.
In my textbook, I found the following trigonometric identities:
$$ \sin^2(x) = \frac{t^2}{1 + t^2}, $$
$$ \cos^2(x) = \frac{1}{1 + t^2}, $$
$$ \sin(x)\cos(x) = \frac{t}{1 + t^2}, $$ where $t=\tan(x)$.
While I can verify these identities by myself, my question pertains to how I would derive these identities without prior knowledge of their right-hand sides. Essentially, given only the idea to express them in terms of $ \tan(x) $, how would I approach deriving these identities from scratch?
I mean in this problem I would need $\sin(x)\cos(x)$ and $\sin^2(x)$. How do I start just knowing that I have to find them in terms of $\tan(x)$.
Thank you for any insights or explanations you can provide!
Best Answer
If $t=\tan x$, then\begin{align}t^2&=\tan^2x\\&=\frac{\sin^2x}{\cos^2x}\\&=\frac{\sin^2x}{1-\sin^2x},\end{align}and therefore$$\sin^2x=\frac{t^2}{1+t^2}\label{a}\tag1$$On the other hand, since $\cos^2x+\sin^2x=1$, it follows from \eqref{a} that$$\cos^2x=\frac1{1+t^2}.$$Finally\begin{align}\sin(x)\cos(x)&=\tan(x)\cos^2(x)\\&=\frac t{1+t^2}.\end{align}