[x²cosx²cosx dx, use integration by parts. What is the expression after the second integration by parts?x2sinx + 2x cost – 2sinxsinx dx + Cx2sinx – 2x cosx+2x2sinx – 2x cosx – 2x2sinx+2x cosx – 26. To evaluateA.B.C.D.2 f cost2 fo-2 fcOSXcosx dx + Ccosx dx + Ccosx dx + C

We will use integration by parts to solve this question.

[x²cos x²cosx dx, use integration by parts. What is the expression after the second integration by parts? x2sinx+2x cosx − 2 sinx dx + C efsir 2 f cos x2sinx − 2x cosx+2 x²sinx-2 – 2x cost – 2 cosx dx + C 2 / CONT -2 / COST x2sinx+2x cost – 2 6. To evaluate A. B. C. D. cosx dx + C cosx dx + C

[x²cos x²cosx dx, use integration by parts. What is the expression after the second integration by parts? x2sinx+2x cosx − 2 sinx dx + C efsir 2 f cos x2sinx − 2x cosx+2 x²sinx-2 – 2x cost – 2 cosx dx + C 2 / CONT -2 / COST x2sinx+2x cost – 2 6. To evaluate A. B. C. D. cosx dx + C cosx dx + C

Intermediate Algebra

10th Edition

ISBN:9781285195728

Author:Jerome E. Kaufmann, Karen L. Schwitters

Publisher:Jerome E. Kaufmann, Karen L. Schwitters

Chapter8: Conic Sections

Section8.2: More Parabolas And Some Circles

Problem 63.2PS: By expanding (xh)2+(yk)2=r2, we obtain x22hx+h22ky+k2r2=0. When we compare this result to the form...

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