Q.Choose the correct option: $2\int \frac{\cos^2 x-\sin^2 x}{\cos^2 x+\sin^2 x}dx =$
(A) $\sin 2x+c$ (B) $\cos 2x+c$ (C) $\tan 2x+c$ (D) $2\sin 2x+c$

Write $1+\sin2x$ as a perfect square $(\sin x+\cos x)^2$.

$1+\sin 2x=\sin^2x+\cos^2x+2\sin x\cos x=(\sin x+\cos x)^2$

$\sqrt{1+\sin2x}=\sqrt{(\sin x+\cos x)^2}=|\sin x+\cos x|$

Taking $\sin x+\cos x\ge0$ on the interval considered: …

Write $\cos^3x=\cos x(1-\sin^2x)$ and integrate, then convert to the $\sin3x$ form.

$\displaystyle\int\cos^3x\,dx=\int\cos x(1-\sin^2x)\,dx=\sin x-\dfrac{\sin^3x}{3}+c$

Using $\sin3x=3\sin x-4\sin^3x \Rightarrow \sin^3x=\dfrac{3\sin x-\sin3x}{4}$:

…