Integrate: (cos^(3)x-sin^(2)x)/(cos^(2)x) dx

Question

anonymous · Answer

$$(\cos ^{3}x-\sin ^{2}x)/\cos ^{2}x $$

anonymous · Answer

$$\begin{align*}\int\frac{\cos^3x-\sin^2x}{\cos^2x}\mathrm{d}x&=\int\left(\frac{\cos^3x}{\cos^2x}-\frac{\sin^2x}{\cos^2x}ight)\mathrm{d}x\&=\int\left(\cos x-	an^2 xight)\mathrm{d}x\end{align*}$$Can you take it from here?

anonymous · Answer

Actually, that was where I got to before I got stuck. I didn't know what to do with the exponent of tan^(2)x when integrating.

anonymous · Answer

@haha31896 do you recall the Pythagorean identity $\cos^2x+\sin^2x=1$? Dividing through by $\cos^2x$ yields $1+\tan^2x=\sec^2x$, which may be rearranged to yield $\tan^2x=\sec^2x-1$... can you do it now?

anonymous · Answer

So is the final answer x+sinx-tanx+c?