Question

How can I integrate the problem below using integration by parts?

Answer 1

\[\int\limits\limits_{}^{}\frac{ e ^{sinx}cosxsinxdx }{ \sin ^{2}x+2sinx+1 }\]

Answer 2

can u tell the power of e???

Answer 3

Um, e is raised to sinx. Is this what you are asking?

Answer 4

yup

Answer 5

take sinx+1=t
can u do nxt???

Answer 6

The presence of so many instances of \(\large \sin \ x\) is daunting, but is tamed by the fact that in the numerator, a humble \(\large \cos \ x \) graces your problem with its presence....

Answer 7

denominator is (sinx+1)^2

Answer 8

That does make things simpler, @gorv :)

Answer 9

ah, you simplified it. yes.

Answer 10

Okay...
\[\huge \int\frac{e^{\sin(x)}\cos(x)\sin(x)}{[\sin(x)+1]^2}dx\]Can you sense what to do next, @taiga_aisaka ?

Answer 11

We've been taught that we need to use this formula:
\[uv-\int\limits_{}^{}vdu\]
So I guess that it is time to identify the u and the dV?

Answer 12

It's not yet readily apparent. Perhaps you can make one simple substitution before proceeding to Integration by Parts?

Answer 13

I'm sorry, but I don't really know what to replace. :/

Answer 14

Is it not tempting, for instance, to let
\[\Large t = \sin(x) + 1\]?

In that case, what is
\[\huge dt = ?\]

Answer 15

then it's cosxdx

Answer 16

yes... and also
\[\huge t - 1 = \sin(x)\]
So the integral becomes
\[\huge \int \frac{e^{t-1}(t-1)}{t^2}dt\]

Answer 17

t represents sinx+1
do I need to just replace the variable t with sinx+1??

Answer 18

Why go back when we switched to to to make integration easier?
Now you can begin the actual integration work :)

Answer 19

switched to t
*gah*
always the small typos :/

Answer 20

I'm confused (I'm sorry, I'm really a beginner in integrals). To get it straight, you used a variable to substitute a value to make the integral easier to evaluate? <--- I can't really explain it well.

Answer 21

Integration by Substitution?
They should have taught you that before going to Integration by Parts...
or else they're EEEEVVVVIIIIIIIILLLLL :D

Answer 22

Then they are EVIL. no, I'm not even aware that it is possible. That is why I'm confused as to where the "t" came from. So I'm going to evaluate the last integral you gave me the one with t's?

Answer 23

Yes.
(Seriously? Integration by parts without Integration by substitution? Bloody hell... o.O )

Answer 24

Oh Gosh. Thank you very much!

Answer 25

Wait... ONCE you integrate that integral with the t's, don't forget to switch back to x, namely with this relation:
\[\Huge t = \sin(x) +1\]
Okay? :)