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Original Question
What is the integral of sin4xcos4x dx?Asked by Bill
What is the integral of sin4xcos4x dx?
Answers
Answered by
John
In this problem, it is easiest to substitute twice.
First, substitute 4x for U and 4dx for dU (remember to balance it out by adding a 1/4 to the resulting integral).
Afterwards, substitute sin(U) for W and cos(U)dU for dW and you should get the integral of 1/4*W*dW, after which solve like a normal integral, which should be
(1/8)*(W^2) + C and replace the W's and U's for the substitutions made above and you should end up with
(1/8)*((sin4x)^2) + C.
First, substitute 4x for U and 4dx for dU (remember to balance it out by adding a 1/4 to the resulting integral).
Afterwards, substitute sin(U) for W and cos(U)dU for dW and you should get the integral of 1/4*W*dW, after which solve like a normal integral, which should be
(1/8)*(W^2) + C and replace the W's and U's for the substitutions made above and you should end up with
(1/8)*((sin4x)^2) + C.
Answered by
Reiny
simpler way:
recognize that sin4xcos4x = (1/2)sin8x
and the integral of that is
(-1/16)cos(8x) + c
recognize that sin4xcos4x = (1/2)sin8x
and the integral of that is
(-1/16)cos(8x) + c