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Integrating tan2x

Hi all
How do you integrate tan2x, since the answer is meant to be 1/2 sec2x + c, but im sure it must involve ln(...) no?

tan2x=\frac{2tanx}{1-tan^2x}

Hmmm, I don't know, haven't done much trig calculus if any at all.

Would you integrate this in parts?

With

\frac{1}{1-tan^2x}

and

2tanx

You would use a substitution u = 2x.

OP

nope, the simple way is to rewrite tan2x as sin2x/cos2x - then integrate by substitution. I get the answer as -1/2 ln(cos2x) + c. How can you manipulate this to equal 1/2 ln(secx) +c ????
Thanks

The Philosopher

nope, the simple way is to rewrite tan2x as sin2x/cos2x - then integrate by substitution. I get the answer as -1/2 ln(cos2x) + c. How can you manipulate this to equal 1/2 ln(secx) +c ????
Thanks

Log laws. But it would be 2x, not x.

OP

Hmm in the text book it says x - thats what was confusing me!
Please could you run through it step by step (the log laws bit)?

The Philosopher

Hmm in the text book it says x - thats what was confusing me!
Please could you run through it step by step (the log laws bit)?

a \log(x) = \log(x^{a})

Or even

\log x = - \log(1/x)

.

I can't see how the "answer" is right then. Integrating tanx is ln secx.

The Edexcel formula book shows the integral of tan kx. Unfortunately, I've lost mine.

Derive it?

\int \tan(kx)\text{d}x = \frac{1}{k}ln|\sec(kx)| + c

I'm pretty sure that's in the formula book...I've forgotten everything mathsy, but remember that being one of the simplest integrations.

I'd call

\int \cos{x} \text{d}x = \sin{x} + c

easy, but, you know... :p:

:p:

-1/2ln(cos2x)+c = 1/2ln((cos2x)^-1)+c = 1/2ln(1/cos2x)+c = 1/2lnsec2x +c

Original post by Supertangi

-1/2ln(cos2x)+c = 1/2ln((cos2x)^-1)+c = 1/2ln(1/cos2x)+c = 1/2lnsec2x +c

You're replying to an 8 year old post...

it seems nobody really made it clear

Original post by Supertangi

it seems nobody really made it clear

you missed out those funny vertical brackets

Original post by the bear

you missed out those funny vertical brackets

yep
seem so

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