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How do you find the antiderivative of #(2x)(sinx)(cosx)#?

Answer 1

Luke Alvarez

#int 2x*sin x*cos x d x#

#int x*2sin x cos x d x#

#2sin x cos x=sin 2x#

#int x* sin 2x *d x=-1/2x*cos 2x+1/4*sin 2x+C#

#int2x*sin x*cos x d x=-1/2x*cos 2x+1/4*sin 2x+C#

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Answer 2

Evelyn Agard

To find the antiderivative of ( (2x)(\sin(x))(\cos(x)) ), you can use integration by parts. Let ( u = 2x ) and ( dv = \sin(x)\cos(x) , dx ). Then, differentiate ( u ) to get ( du = 2 , dx ) and integrate ( dv ) to get ( v = \frac{1}{2}\sin^2(x) ). Now, apply the integration by parts formula: ( \int u , dv = uv - \int v , du ). Substituting the values, you'll get ( \int (2x)(\sin(x))(\cos(x)) , dx = (2x)(\frac{1}{2}\sin^2(x)) - \int (\frac{1}{2}\sin^2(x))(2) , dx ). Simplify and integrate the second term to obtain the final antiderivative.

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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.