integral of $\int5\sec(4x)\tan(4x)\,dx$

Question

I was thinking about using $u$-substitution, so I did here's what I got $$ \int5\sec(4x)\tan(4x)\,dx =\int\sec(u)\tan(u)(4)\,du= \frac{5}{4} \int \sec(u)\tan(u)\,du= \frac{5}{4} \sec(4x) + C.$$

Thanks

Answer 1

Your substitution was wrong. Letting $4x=u$ then $4dx=du$ or $dx=du/4$.

Answer 2

If you do $4x=u$, you have to do $4\,\mathrm dx=\mathrm du$ too. So, you get$$\int5\sec(4x)\tan(4x)\,\mathrm dx=\frac54\int\sec(4x)\tan(4x)4\,\mathrm dx)=\frac54\int\sec(u)\tan(u)\,\mathrm du.$$

Answer 3

Note that since $\frac d{dx}(4x)=4$, $$\int5\sec(4x)\tan(4x)\,dx=5\cdot\frac14\int\sec(4x)\tan(4x)\cdot\frac{d(4x)}{dx}\,dx$$ so if $u=4x$ then $$\int5\sec(4x)\tan(4x)\,dx=\frac54\int\sec u\tan u\,du.$$