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How do you find the integral of #int cos^3x sin^2x dx#?

Answer 1

Evelyn Agard

#intcos^3xsin^2xdx=(1/3)sin^3x-(1/5)*sin^5x+c#

#intcos^3xsin^2xdx#

#=intcos^2x*sin^2x*(cosx)dx#

#=int (1-sin^2)sin^2x*d(sinx)#

Hints:

#(d(sinx))/dx=cosx#

#d(sinx)=cosxdx#

Consequently,

#=int(sin^2x-sin^4x)*d(sinx)# #==(1/3)sin^3x-(1/5)*sin^5x+c#

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Answer 2

Christopher Agresta

To find the integral of (\int \cos^3(x) \sin^2(x) ,dx), you can use trigonometric identities to rewrite the integral in a more manageable form. Specifically, you can use the identity (\sin^2(x) = 1 - \cos^2(x)) and then apply substitution or integration by parts. The process may involve several steps, including applying trigonometric identities, simplifying expressions, and performing integration techniques. The final result will be the antiderivative of the given expression.

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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.