$\displaystyle \int x(1-2e^{cotx^{2}})csc^{2}(x^{2})dx$

$\displaystyle \int5 e^{\frac{1}{2}lnx}dx$

$\displaystyle \int ln e^{\frac{x}{2}}dx$

Can you show how it's done?

Thank you

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- Aug 15th 2009, 08:13 PMhonestliarHow do you solve this calculus problem?
$\displaystyle \int x(1-2e^{cotx^{2}})csc^{2}(x^{2})dx$

$\displaystyle \int5 e^{\frac{1}{2}lnx}dx$

$\displaystyle \int ln e^{\frac{x}{2}}dx$

Can you show how it's done?

Thank you - Aug 15th 2009, 08:35 PMsongoku
Hi honestliar

Do you mean integration ?

1. let : $\displaystyle u=1-2e^{(\cot x^2)}$

2. hint : $\displaystyle e^{\ln a}=a$

3. hint : $\displaystyle \log a^b = b*\log a$ - Aug 17th 2009, 05:01 AMtom@ballooncalculus
Expand the brackets and tackle 2 separate integrals.

Just in case a picture helps...

http://www.ballooncalculus.org/asy/intChain/split1.png

As usual, straight continuous lines differentiate downwards (integrate up) with respect to x, and the straight dashed lines similarly but with respect to the dashed balloon expression. So the triangular network...

http://www.ballooncalculus.org/asy/chain.png

... is the chain rule.

________________________

Don't integrate - balloontegrate!

Balloon Calculus Forum - Aug 17th 2009, 09:10 PMsongoku
Hi honestliar

1.

let : $\displaystyle u=1-2e^{(\cot x^2)}$

$\displaystyle du=-2e^{(\cot x^2)}\; (-\csc^{2}(x^{2})) \; (2x) \; dx= 4x\; e^{(\cot x^2)}\; \csc^{2}(x^{2})\; dx$

Can you continue ? :)

P.S: I just realized that Mr. F edited my post. Thx Mr. F - Aug 18th 2009, 01:07 AMtom@ballooncalculus
Hi, songoku - I couldn't (yet) make your sub (with parts, maybe?) work for the big one. Hope mine was ok...

- Aug 18th 2009, 05:07 AMsongoku
Hi tom@ballooncalculus

Your method is ok. Hope honestliar understands it :)

As for mine, subs. dx into the question and state $\displaystyle e^{(\cot x^2)}$ in term of u. - Aug 18th 2009, 02:01 PMtom@ballooncalculus
D'oh! Cheers for that... and if you don't mind another picture...

http://www.ballooncalculus.org/asy/intChain/reform.png

... where we're doing just as you say, expressing $\displaystyle e^{\cot(x^2)}$ in terms of u (i.e. the dashed balloon expression)...

http://www.ballooncalculus.org/asy/intChain/reform1.png

... in order to divide by $\displaystyle e^{\cot(x^2)}$ and thus cancel the extra one appearing in the derivative by-product. I'm using F and G just to point up how the integration with respect to u splits into two.

Obviously, the lower equals sign depends on several lines of algebra that aren't shown, so the picture's only an overview.

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