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How do you find the antiderivative of #int sinx(cosx)^(3/2) dx#?

Answer 1

Christopher Allers

#-2/5(cosx)^(5/2)+C#

Since we don't want to have to deal with the #3//2# power, let #u=cosx#. This implies that #du=-sinxcolor(white).dx#.

So, we need to modify our integral just a little:

#intsinx(cosx)^(3/2)dx=-int(cosx)^(3/2)(-sinxcolor(white).dx)=-intu^(3/2)color(white).du#

Integrate this using the rule #intu^ncolor(white).du=u^(n+1)/(n+1)#, where #n!=-1#.

#=-u^(5/2)/(5/2)=-2/5u^(5/2)=-2/5(cosx)^(5/2)+C#

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Answer 2

Christopher Agresta

To find the antiderivative of ∫sin(x)(cos(x))^(3/2) dx, you can use the substitution method. Let u = cos(x), then du = -sin(x) dx. Substitute these into the integral:

∫sin(x)(cos(x))^(3/2) dx = ∫-u^(3/2) du

Now integrate -u^(3/2) with respect to u:

= -(2/5)u^(5/2) + C

Substitute back u = cos(x):

= -(2/5)cos^(5/2)(x) + C

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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.