Question

Please check my work.

Answer 1

Our teacher got the same answer in class. But for some reason, WebAssign isn't accepting my answer, Can someone see if they see any errors?

Answer 2

I havent the time now. I'll take a look in about 1/2 hour if you havent had it solved by then

Answer 3

there can be many solutions to these sort of problem. you can solve it in sin and cos as well. try doing it. I hope it accepts. good luck.

Answer 4

If you're allowed to do so, look up the integrals of (sec x)^2 and of (sec x).

It's great that you're trying to integrate these from scratch.

However, I don't see why you're apparently using integration by parts (which I recognize because you've written down u, du, v and dv.

Focus on the integral of (sec x)^3 first, then on that of (sec x), separately.

Perhaps if you try to defend your use of integration by parts here, you'll realize that some other method may be better suited here.

Answer 5

look at the last line

i think there's a very unfortunate arithmetic error in these

Answer 6

tbc, the 9 is part of the integrand.

Answer 7

\[I=9 \int\limits \tan ^2x \sec x dx=9 \int\limits \left( \sec ^2x-1 \right)\sec x ~dx=9 \int\limits \sec ^3x ~dx-9 \int\limits \sec x~dx\]
\[=9 \int\limits \sec x \sec ^2x~dx-9 \int\limits \sec x~dx\]
\[=9\left( \sec x \tan x \right)-9\int\limits \sec x \tan x*\tan x~dx-9\ln \left| \sec x+\tan x \right|+c\]
\[=9 \sec x \tan x-9 \int\limits \tan ^2x \sec x~dx-9\ln \left| \sec x+\tan x \right|+c\]
\[=9 \sec x \tan x-I-9\ln \left| \sec x+\tan x \right|+c\]
\[I+I=2I=9\sec x \tan x-9\ln \left| \sec x+\tan x \right|+c\]
I=?