## Calculus 8th Edition

$$\int \frac{sin(\phi)}{cos^{3}(\phi)}d\phi=\frac{1}{2}tan^{2}\phi+C$$
$$\int \frac{sin(\phi)}{cos^{3}(\phi)}d\phi=\int tan\phi\frac{1}{cos^{2}\phi}d\phi$$ $$=\int tan\phi\,sec^{2}\phi\,d\phi$$ $$=\int tan\phi\,d(tan\phi)$$ $$=\frac{1}{2}tan^{2}\phi+C$$ $\frac{1}{2}tan^{2}\phi +C= \frac{1}{2}sec^{2}\phi-\frac{1}{2}+C=\frac{1}{2}sec^{2}\phi+C_{1}$ The two answers are the same.