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sha0403
Group Title
help me to integrate this :
integration of (3 sin^2 x cos x) dx
i confuse how to integrate this..
 one year ago
 one year ago
sha0403 Group Title
help me to integrate this : integration of (3 sin^2 x cos x) dx i confuse how to integrate this..
 one year ago
 one year ago

This Question is Closed

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.1
let sinx = t
 one year ago

Kainui Group TitleBest ResponseYou've already chosen the best response.2
Use "u"substitution since you see a function and its derivative next to it.
 one year ago

Kainui Group TitleBest ResponseYou've already chosen the best response.2
www.wolframalpha.com show me steps
 one year ago

Australopithecus Group TitleBest ResponseYou've already chosen the best response.1
you have to use identities to solve this \[\int\limits_{}^{}3 \sin^2 x \cos x\]
 one year ago

Australopithecus Group TitleBest ResponseYou've already chosen the best response.1
\[3\int\limits_{}^{}\sin^2 x \cos xdx\] \[3\int\limits_{}^{}(1\cos^{2}(x)) \cos xdx\]
 one year ago

Kainui Group TitleBest ResponseYou've already chosen the best response.2
No, sin^2(x)=u 1/2 du=cosxdx 3/2udu becomes your new integral.
 one year ago

Australopithecus Group TitleBest ResponseYou've already chosen the best response.1
f(x) = sin^2(x) f'(x) = cos(x)^2*sin(x)
 one year ago

Chlorophyll Group TitleBest ResponseYou've already chosen the best response.0
Why not u = sinx > du = cosxdx ??
 one year ago

Chlorophyll Group TitleBest ResponseYou've already chosen the best response.0
= 3 ∫ u² du = ....
 one year ago

Australopithecus Group TitleBest ResponseYou've already chosen the best response.1
\[3(\int\limits_{}^{}\cos(x)dx  \int\limits_{}^{}\cos^2(x)dx)\]
 one year ago

Chlorophyll Group TitleBest ResponseYou've already chosen the best response.0
It's crystal clear about the relation between sinx and cosxdx :)
 one year ago

Australopithecus Group TitleBest ResponseYou've already chosen the best response.1
remember for the future \[\cos^2(x) = \frac{1 + \cos(2x)}{2}\] so we have, \[3(\sin(x) + c  \int\limits_{}^{}\frac{1+\cos(2x)}{2}dx)\]
 one year ago

stgreen Group TitleBest ResponseYou've already chosen the best response.2
^done in 3rd reply
 one year ago

Australopithecus Group TitleBest ResponseYou've already chosen the best response.1
You can solve it my way too which is good practice for when you get integrals with trig functions to high powers
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.1
@Australopithecus its cos^3 x there and not square
 one year ago

Australopithecus Group TitleBest ResponseYou've already chosen the best response.1
oh sorry made a mistake lol
 one year ago

sha0403 Group TitleBest ResponseYou've already chosen the best response.0
where the best solution? i still confuse.. hu2
 one year ago

Australopithecus Group TitleBest ResponseYou've already chosen the best response.1
it can still be solved with that method
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.1
and sinx= t ,u whatever, thats the easiest thing which you can do..as i pointed out in the very beginning ..
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.1
cos^3 x will still involve substitution ..
 one year ago

Australopithecus Group TitleBest ResponseYou've already chosen the best response.1
\[3(\sin(x) + c  \int\limits_{}^{}\cos(x)\frac{1 + \cos(2x)}{2}dx\]
 one year ago

Australopithecus Group TitleBest ResponseYou've already chosen the best response.1
I'm not arguing that substitution will be involved in my solution but it still works for solving this integral it is just the long way
 one year ago

Australopithecus Group TitleBest ResponseYou've already chosen the best response.1
I think I made a mistake again though meh
 one year ago

Kainui Group TitleBest ResponseYou've already chosen the best response.2
@Australopithecus It's like suggesting someone make a tunnel through a mountain when you can just use a teleporter.
 one year ago

Australopithecus Group TitleBest ResponseYou've already chosen the best response.1
It is still applicable and stop hassling me I'm rusty ha
 one year ago

sha0403 Group TitleBest ResponseYou've already chosen the best response.0
so, the best solution? anyone?
 one year ago

Mathmuse Group TitleBest ResponseYou've already chosen the best response.0
@sha0403, to clarify, you should use any method provided except @Australopithecus'
 one year ago

sha0403 Group TitleBest ResponseYou've already chosen the best response.0
ok thanks u all for helping me..i appreciate it.. =)
 one year ago
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