Find the general antiderivative...

[thatguy47]

Find the general antiderivative. Use theta for ?.
g(?) = sec(7?)tan(7?)

I can't get the right answer on this. So far I've gotten: 1/(7cos7?) + C which is incorrect

Here's my work:
[integral] sec(7?)tan(7?)
= [integral] (sin(7?))/(cos(7?))^2
u=cos7?
du = -7sin7? d?

=[integral] (sin7?/u^2)(1/(-7sin7?))
= (-1/7) [integral] (1/u^2)du
= (-1/7) (-u^(-1)) + C
= (1/7u) + C

= 1/7cos7? + C which is wrong... please help

 

[BigGlenntheHeavy]

\(\displaystyle \int sec(7\theta)tan(7\theta)d\theta \ = \ \int \frac{sin(7\theta)d\theta}{cos^{2}(7\theta)}\)

\(\displaystyle Let \ u \ = \ cos(7\theta), \ then \ du \ = \ -7sin(7\theta)d\theta\)

\(\displaystyle Ergo, \ we \ have \ \frac{-1}{7}\int\frac{du}{u^{2}}, \ = \ \frac{1}{7}sec(7\theta)+C, \ why \ is \ this \ wrong?\)

 

[thatguy47]

edit:

thanks,

thats right

 

[daon]

Why so much work?


\(\displaystyle \frac{d}{dx}sec(x) = sec(x)tan(x)\)

So, suppose \(\displaystyle u=7\theta\)

\(\displaystyle \int sec(7\theta)tan(7 \theta) d \theta = \frac{1}{7}\int sec(u)tan(u)du = ...\)