Indefinite integration, trig integral manipulation
What's the thought process?
25 Replies
@Apu
Note for OP
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Transcription requested by Opt
The hint is given in the. Answer itself in the the answer of the integration so. If you start with. Take it to be one take M and N to be one and see if that works or take one of them to be minus one, which means. Eco do says it divided and then differentiate to see if you get a similar form. I think that should work.
One thing i can clearly see is that the derivative of cosx+xsinx is xcosx
Maybe u can use some substitution
I won't say differentiate both sides or @Saul Goodman will hunt me down
ðŸ˜
I dont have my phone to send a pic, so let me type out this mess in LaTex
Multiply and divide by -xcosx
Let a = xsin(x)+cos(x) and b = sin(x)-xcos(x)
$$\implies da = xcos(x) dx and db = xsin(x) dx$$
$$(a)db = \left(x^{2}sin^{2}(x)+xcos(x)sin(x)\right)dx$$
$$(b)da = \left(x^{2}cos^{2}(x)+xcos(x)sin(x)\right)dx$$
$$a(db)-b(da) = x^{2}dx$$
$$I = \int\frac{a(db)-b(da)}{a^{2}}$$
$$I = \int d\left(\frac{a}{b}\right)$$
$$I = \frac{a}{b}+C$$
Opt
@Nimboi @Nimboi [ping if answering]
,rotate
ah brilliant
i'm really not sure you thought of that
what's going on in your solution, can you clarify?
Bc in the denominator it is square of something so most likely it is u/v rule, so u diff that and u get -xcosx
With that u multiply and divide so u can split it and do by parts
That is it
What?
how you thought of that* horrible typo 💀
Yeah, the meaning is very different.
lmfao
oh huh
that's pretty cool
no really, can you describe the thought process that lead to that
led*
why cant i type???
I differentiated the expressions. Realised there was some symmetry stuff going on. Multiplied them with each other. Cancelling terms appeared.
Also there's a mistake there
see when you put it like that it sounds simple
It's d(b/a)
I'm crazy
It's b/a +C
Not a/b
gotchu gotchu
+solved @Opt @stormycloud
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