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help me to integrate this : integration of (3 sin^2 x cos x) dx i confuse how to integrate this..

Mathematics
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let sinx = t
Use "u"-substitution since you see a function and its derivative next to it.
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Other answers:

www.wolframalpha.com show me steps
^^shown
you have to use identities to solve this \[\int\limits_{}^{}3 \sin^2 x \cos x\]
...
\[3\int\limits_{}^{}\sin^2 x \cos xdx\] \[3\int\limits_{}^{}(1-\cos^{2}(x)) \cos xdx\]
No, sin^2(x)=u 1/2 du=cosxdx 3/2udu becomes your new integral.
f(x) = sin^2(x) f'(x) = cos(x)^2*sin(x)
Why not u = sinx --> du = cosxdx ??
= 3 ∫ u² du = ....
\[3(\int\limits_{}^{}\cos(x)dx - \int\limits_{}^{}\cos^2(x)dx)\]
It's crystal clear about the relation between sinx and cosxdx :)
remember for the future \[\cos^2(x) = \frac{1 + \cos(2x)}{2}\] so we have, \[3(\sin(x) + c - \int\limits_{}^{}\frac{1+\cos(2x)}{2}dx)\]
^done in 3rd reply
You can solve it my way too which is good practice for when you get integrals with trig functions to high powers
@Australopithecus its cos^3 x there and not square
oh sorry made a mistake lol
where the best solution? i still confuse.. hu2
it can still be solved with that method
and sinx= t ,u whatever, thats the easiest thing which you can do..as i pointed out in the very beginning ..
cos^3 x will still involve substitution ..
\[3(\sin(x) + c - \int\limits_{}^{}\cos(x)\frac{1 + \cos(2x)}{2}dx\]
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
I think I made a mistake again though meh
@Australopithecus It's like suggesting someone make a tunnel through a mountain when you can just use a teleporter.
It is still applicable and stop hassling me I'm rusty ha
so, the best solution? anyone?
3rd
@sha0403, to clarify, you should use any method provided except @Australopithecus'
ok thanks u all for helping me..i appreciate it.. =)

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