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How do you integrate #(sinx)(cosx)(cos2x)dx#?

Answer 1

Charles Arocha

You can try this:

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Answer 2

Cameron Alexander

First use a double-angle formula to replace #cos(2x)# by #2cos^{2}(x)-1#. Then distribute #cos(x)# through to rewrite your integrand as #(2cos^{3}(x)-cos(x))sin(x)#. Now do a substitution: #u=cos(x), du=-sin(x)dx#, making your integral transform to #\int(u-2u^{3})du=u^{2}/2-u^{4}/2+C=\frac{1}{2}\cos^{2}(x)-\frac{1}{2}\cos^{4}(x)+C.# There are lots of alternative ways of writing this answer because of all the trigonometric identities out there. You could check, for instance, that it is equivalent to #-\frac{1}{16}\cos(4x)+C#.

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Answer 3

Elijah Abram

To integrate ( (\sin{x})(\cos{x})(\cos{2x}) ) with respect to ( x ), you can use integration by parts. Let ( u = \sin{x} ) and ( dv = (\cos{x})(\cos{2x})dx ). Then, ( du = \cos{x}dx ) and ( v = \int (\cos{x})(\cos{2x})dx ).

Integrating ( v ) requires applying the angle addition formula: ( \cos{A}\cos{B} = \frac{1}{2}[\cos{(A-B)} + \cos{(A+B)}] ).

After integrating ( v ), you'll need to apply the integration by parts formula:

[ \int u , dv = uv - \int v , du ]

This will give you the result of the integration.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.