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How do you evaluate the definite integral #int sin^(5)x * cos^(20)x dx# from [0,pi/2]?

Answer 1

#1/21-2/23+1/25#

#int_0^(pi/2) sin^(5)x * cos^(20)x dx =int_0^(pi/2) (1-cos^2x)^2cos^20 x sin x dx# #=int_0^(pi/2) (1-2cos^2x+cos^4x)cos^20 x sin xdx =# #=int_0^(pi/2)(cos^20x-2cos^22 x+cos^24x)sinx dx =# #=-1/21(0-1)+2/23(0-1)-1/25(0-1)=# #=1/21-2/23+1/25#

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Answer 2

Cameron Alexander

To evaluate the definite integral ∫ sin^5(x) * cos^20(x) dx from 0 to π/2, you can use the substitution u = sin(x), then du = cos(x) dx. This transforms the integral into ∫ u^5 * (1 - u^2)^10 du. You can then expand (1 - u^2)^10 using the binomial theorem, integrate the resulting polynomial, and evaluate it from 0 to 1.

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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.