Evaluate ∫(0,2) ∫(y^2,4) (sqrt(x))sinxdxdy...by using integration by parts
normally is fine too, without integration by parts. But integration by parts would be more helpful!
...HMM well that's an odd notation... is it: \[\int^2_0\int^4_{y^2}\sin(x)\sqrt{x}\phantom{.}dx\phantom{.}dy\]
yes!
ALRIGHT! Well I would first find the primitive of \(\sin(x)\sqrt{x}\) first before doing anything! Just to point out, the integral is not very nice according to this: http://www.wolframalpha.com/input/?i=integral+of+sin%28x%29sqrt%28x%29 BUT I'LL DO MY BEST!
Ok, by the way. If you are doing it the normal way it is much nicer switching the order. But if not it will be very difficult. But I will get extra credit if I use parts and I really need the extra credit haha.
Haha I give up...I tried but I think im not getting anywhere... here is what I had so far... but I don't think that's going anywhere...SORRY. \[\eqalign{ &\int\sin(x)\sqrt{x}\phantom{.}dx \\ =&\int \sin(x)u\phantom{.}dx\to\phantom{space}u=\sqrt{x}\to u^2=x\to dx=2u\phantom{.}du\\ =&\int\sin(u^2)(u)(2u)du\\ =&2\int u^2\sin(u^2)du \\ =&2\int a\phantom{.}db \\ &\\ a&=u^2\phantom{...}db=\sin(u^2) \\ da&=2u\phantom{....}b=\small{\int}\sin(u^2) \\ &\\ =&2\int a\phantom{.}db \\ =&2\left[ab-\int b\phantom{.}da\right] \\ =&2\left[u^2\int\sin(u^2)-\int2u\left(\int\sin(u^2)\right)\right] \\ }\]
its ok thanx though!!
Could you help me do the integral with by switching the order?
@zpupster hey, where are you in the problem
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