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



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shubhamsrg
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let sinx = t

Kainui
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Use "u"substitution since you see a function and its derivative next to it.

stgreen
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Kainui
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www.wolframalpha.com show me steps

stgreen
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^^shown

Australopithecus
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you have to use identities to solve this
\[\int\limits_{}^{}3 \sin^2 x \cos x\]

Kainui
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...

Australopithecus
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\[3\int\limits_{}^{}\sin^2 x \cos xdx\]
\[3\int\limits_{}^{}(1\cos^{2}(x)) \cos xdx\]

Kainui
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No, sin^2(x)=u
1/2 du=cosxdx
3/2udu becomes your new integral.

Australopithecus
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f(x) = sin^2(x)
f'(x) = cos(x)^2*sin(x)

Chlorophyll
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Why not
u = sinx > du = cosxdx ??

Chlorophyll
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= 3 ∫ u² du = ....

Australopithecus
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\[3(\int\limits_{}^{}\cos(x)dx  \int\limits_{}^{}\cos^2(x)dx)\]

Chlorophyll
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It's crystal clear about the relation between sinx and cosxdx :)

Australopithecus
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remember for the future
\[\cos^2(x) = \frac{1 + \cos(2x)}{2}\]
so we have,
\[3(\sin(x) + c  \int\limits_{}^{}\frac{1+\cos(2x)}{2}dx)\]

stgreen
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^done in 3rd reply

Australopithecus
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You can solve it my way too which is good practice for when you get integrals with trig functions to high powers

shubhamsrg
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@Australopithecus its cos^3 x there and not square

Australopithecus
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oh sorry made a mistake lol

sha0403
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where the best solution? i still confuse.. hu2

Australopithecus
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it can still be solved with that method

shubhamsrg
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and sinx= t ,u whatever, thats the easiest thing which you can do..as i pointed out in the very beginning ..

shubhamsrg
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cos^3 x will still involve substitution ..

Australopithecus
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\[3(\sin(x) + c  \int\limits_{}^{}\cos(x)\frac{1 + \cos(2x)}{2}dx\]

Australopithecus
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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

Australopithecus
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I think I made a mistake again though meh

Kainui
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@Australopithecus It's like suggesting someone make a tunnel through a mountain when you can just use a teleporter.

Australopithecus
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It is still applicable and stop hassling me I'm rusty ha

sha0403
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so, the best solution? anyone?

stgreen
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3rd

Mathmuse
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@sha0403, to clarify, you should use any method provided except @Australopithecus'

sha0403
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ok thanks u all for helping me..i appreciate it.. =)