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Mathematics 6 Online
OpenStudy (anonymous):

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

OpenStudy (shubhamsrg):

let sinx = t

OpenStudy (kainui):

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

OpenStudy (anonymous):

OpenStudy (kainui):

www.wolframalpha.com show me steps

OpenStudy (anonymous):

^^shown

OpenStudy (australopithecus):

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

OpenStudy (kainui):

...

OpenStudy (australopithecus):

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

OpenStudy (kainui):

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

OpenStudy (australopithecus):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

= 3 ∫ u² du = ....

OpenStudy (australopithecus):

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

OpenStudy (anonymous):

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

OpenStudy (australopithecus):

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

OpenStudy (anonymous):

^done in 3rd reply

OpenStudy (australopithecus):

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

OpenStudy (shubhamsrg):

@Australopithecus its cos^3 x there and not square

OpenStudy (australopithecus):

oh sorry made a mistake lol

OpenStudy (anonymous):

where the best solution? i still confuse.. hu2

OpenStudy (australopithecus):

it can still be solved with that method

OpenStudy (shubhamsrg):

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

OpenStudy (shubhamsrg):

cos^3 x will still involve substitution ..

OpenStudy (australopithecus):

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

OpenStudy (australopithecus):

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

OpenStudy (australopithecus):

I think I made a mistake again though meh

OpenStudy (kainui):

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

OpenStudy (australopithecus):

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

OpenStudy (anonymous):

so, the best solution? anyone?

OpenStudy (anonymous):

3rd

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

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