evaluating j 6 8w2w2 3w 7dw we make the substitution p therefore dp the integral later transform

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Darshan Maheshwari · Answer

Let&#x27;s talk about question number 33. So in this case, we are going to substitute E raised to the

K B · Answer

So in this question we&#x27;re asked to use cast to evaluate this interval. So we&#x27;re going to rewrite

Ramzi Deek · Answer

So we have integral of 2w plus 4 times cosine 2 w dw, cosine 2w dw. So let&#x27;s take U as 2w plus 4

Gregory Higby · Answer

The theme in this problem is the integral of some power is equal to adding one to the exponent.

Wendi Zhao · Answer

The derivative of this form equals 5E to the 5x power times C1 cosine 7x plus C2 sine 7x and plu

Evaluating J (6 - 8w)(2w2 - 3w + 7)dw we make the substitution p : therefore dp The integral later transforms to --2 dp. Hence f(6 - 8w)(2w2 -3w+ 7)dw= +C.