$\displaystyle \int_0^1 {e^{sin2x}} {cos2x\,dx}$
$\displaystyle \int \frac{e^{2x}}{(1+e^x)^{1/2}}{\,dx}$
can someone help me?
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...
... where
... 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!
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.
OK, what happens if we choose e^x...
... no obvious leads, so let's try 1 + e^x...
... which is progress, because if we just re-write e^x in terms of u, I mean the dashed balloon...
... then we can see a way ahead, because the fraction breaks nicely into two...
... i.e...
__________________________________________
Don't integrate - balloontegrate!
http://www.ballooncalculus.org/forum/top.php
Draw balloons with LaTeX: http://www.ballooncalculus.org/asy/doc.html
thank you so much.i'll work it out
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
$\displaystyle u = 1+e^x$
$\displaystyle e^x = u-1$
$\displaystyle du = e^x \, dx$
$\displaystyle \int \frac{e^{x}}{(1+e^x)^{1/2}} \, \cdot e^x \, dx$
substitute ...
$\displaystyle \int \frac{u-1}{u^{1/2}} \, du$
$\displaystyle \int u^{1/2} - u^{-1/2} \, du$
integrate, then back substitute.