Question

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

Answer 1

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

Answer 2

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

Answer 3

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.

Answer 4

@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?

Answer 5

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