Integration (sin^2 x, cos^2 x) (1 Viewer)

ashokkumar · Mar 29, 2010

Can someone please show me how to integrate:

sin^2 x.Cos^2 x

answer:

1/8 (x - 0.25sin4x)

Thanks

Drongoski · Mar 29, 2010

ashokkumar · Mar 29, 2010

Thank you, thank you! +Rep

fullonoob · Mar 29, 2010

Drongoski said:
$\int \ sin^{2}x\cdot cos^{2}x\ dx\ \ =\ \ \int (\frac{1}{2}sin2x)^{2}\ dx\\ \\ =\ \ \frac{1}{8}\int (1-cos4x)\ dx\ \ =\ \ \frac{1}{8}(x-\frac{1}{4}sin4x)\ +\ C\\ \\ =\ \ \frac{1}{8}x\ -\ \frac{1}{32}sin4x \ +\ C$

does I(1/2sin2x)^2 = 1/8 I (1-cos4x) formula?? (haven't learnt this yet

)
i would try do Isinaxdx = -1/a cosax somehow
just wna know if its a new formula

Drongoski · Mar 29, 2010

As I have mentioned before, carrying the following trig identities in your head will serve you well.

$2sin^{2}\theta \ =\ 1 \ -\ cos2\theta\\ \\ 2cos^{2}\theta \ =\ 1\ +\ cos2\theta \\ \\ tan^{2}\theta \ =\ \frac{2sin^{2}\theta }{2cos^{2}\theta }\ \ =\ \ \frac{1-cos2\theta }{1+cos2\theta }$

thongetsu · Mar 30, 2010

lol i hope i beat you in the m.ext1 test tomorrow whoever you are in kumar's class xD

hello-there · Apr 6, 2010

nice +1 rep

Nahidul · Apr 11, 2010

fullonoob said:
does I(1/2sin2x)^2 = 1/8 I (1-cos4x) formula?? (haven't learnt this yet )
i would try do Isinaxdx = -1/a cosax somehow
just wna know if its a new formula

actually what he did bro, was quite smart: $\sin 2x = 2\sin x\cos x \left \therefore (\sin 2x)^{2}=(2\sin x\cos x)^{2} \therefore sin^{2}2x = 4\sin ^{2}x\cos ^{2}x\therefore \frac{1}{4}\sin ^{2}2x=\sin ^{2}x\cos ^{2}x \therefore (\frac{1}{2}\sin 2x)^{2}=\sin ^{2}x\cos ^{2} x \therefore \int \sin ^{2}x\cos ^{2}xdx=\int( \frac{1}{2}\sin 2x)^{2}$

its not exactly a formula you have to remember.
hope that simplifies things

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Integration (sin^2 x, cos^2 x) (1 Viewer)

ashokkumar

Mr

Drongoski

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ashokkumar

Mr

fullonoob

fail engrish? unpossible!

Drongoski

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thongetsu

Where aren't I?

hello-there

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Nahidul

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