$\displaystyle \int_0^1 {e^{sin2x}} {cos2x\,dx}$

$\displaystyle \int \frac{e^{2x}}{(1+e^x)^{1/2}}{\,dx}$

can someone help me?

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- Sep 18th 2009, 10:25 AMkokolilyIntegration by substitution
$\displaystyle \int_0^1 {e^{sin2x}} {cos2x\,dx}$

$\displaystyle \int \frac{e^{2x}}{(1+e^x)^{1/2}}{\,dx}$

can someone help me? - Sep 18th 2009, 11:22 AMtom@ballooncalculus
Hi Kokolily, I'll add a picture that hopefully shows the thought process.

In the second problem, the inner function worth identifying as such (by some method or other) is 1 + e^x.

Edit:

Just in case a picture helps...

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

... where

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

... is the chain rule. Straight continuous lines differentiate downwards (integrate up) with respect to x, and the straight dashed line similarly but with respect to the dashed balloon expression (which is the inner function of the composite and hence subject to the chain rule).

So, to determine the function F, focus on getting the anti-derivative of 1/2e to the power of the dashed balloon, just like you would if it were 1/2 e to the power of some variable like u.

Call it u, if that helps!

http://www.ballooncalculus.org/asy/i...n/expTrigU.png

In the second problem, we have a choice between taking e or else $\displaystyle 1 + e^x$ or else $\displaystyle \sqrt{1 + e^x}$ as the most likely inner function... but either way, that makes e^x the 'by-product' derivative of the chain-rule differentiation. Remember a chain-rule differentiation is what we're trying to work backwards through, when we do - or avoid doing - u-substitution.

http://www.ballooncalculus.org/asy/i...actionExp1.png

OK, what happens if we choose e^x...

http://www.ballooncalculus.org/asy/i...actionExp2.png

... no obvious leads, so let's try 1 + e^x...

http://www.ballooncalculus.org/asy/i...ractionExp.png

... which is progress, because if we just re-write e^x in terms of u, I mean the dashed balloon...

http://www.ballooncalculus.org/asy/i...actionExp3.png

... then we can see a way ahead, because the fraction breaks nicely into two...

http://www.ballooncalculus.org/asy/i...actionExp4.png

... i.e...

http://www.ballooncalculus.org/asy/i...actionExp5.png

__________________________________________

Don't integrate - balloontegrate!

http://www.ballooncalculus.org/forum/top.php

Draw balloons with LaTeX: http://www.ballooncalculus.org/asy/doc.html - Sep 19th 2009, 06:10 AMkokolily
thank you so much.i'll work it out(Happy)

Edit:

$\displaystyle \int \frac{e^{2x}}{(1+e^x)^{1/2}}{\,dx}$

can anyone show me step by step in solving this question?cz i'm not a fast learner(Doh) - Sep 19th 2009, 08:54 AMskeeter
$\displaystyle u = 1+e^x$

$\displaystyle e^x = u-1$

$\displaystyle du = e^x \, dx$

$\displaystyle \int \frac{e^{x}}{(1+e^x)^{1/2}} \, \cdot e^x \, dx$

substitute ...

$\displaystyle \int \frac{u-1}{u^{1/2}} \, du$

$\displaystyle \int u^{1/2} - u^{-1/2} \, du$

integrate, then back substitute.