Question

Evaluate the integral

integral of sec^3(2x)dx

I have tried it so many times and I keep getting it wrong.

My answer is:

1/4(tan(2x)sec(2x)-log(cos(x)-sin(x))+log(sin(x)+cos(x)))

Answer 1

your answer is correct.
just add +c to it.

Answer 2

oh i forgot to add that in the open question. but, it still says wrong.

Answer 3

Well, start with integration by parts,
\[ \int \sec^2(2x) \sec(2x) \ dx = \frac{1}{2} \tan(2x) \sec(2x) - \int \tan^2(2x)\sec(2x) \ dx \]
Now, that integral on the right hand side equals
\[ \int \sec^3(2x) \ dx - \int \sec(2x) \ dx \]
Thus rearranging
\[ 2 \int \sec^3(2x) \ dx = \frac{1}{2} \tan(2x)\sec(2x) + \int \sec(2x) \ dx \]

So far so good?

Answer 4

then maybe simplify the 2 log terms using logA-log B = log (A/B)

Answer 5

yupp good.

Answer 6

log [ (cos + sin)/(cos -sin) ]

Answer 7

we can avoid integration by parts(if wanted to), if we can use some standard results....

Answer 8

Now, you know how to integrate sec(2x)? Multiply by (sec 2x + tan 2x) / (sec 2x + tan 2x)
and we have
\[ \int \frac{ \sec^2(2x) + \sec(2x)\tan(2x)}{\tan 2x + \sec 2x} dx = \frac{1}{2}\ln( \tan 2x + \sec 2x) \]

Now you pull everything together

Answer 9

I got it! It was a simple problem. I forgot a multiplication sign lol!

Answer 10

ok