If int[pi/3,pi/2][sin(x)/(1+cos(x))]dx = int[3/2,1][f(u)]du, then what are "u", "du"?
\(\displaystyle \displaystyle \mbox{If }\, \int_{\pi/3}^{\pi/2}\, \dfrac{\sin(x)}{1\, +\, \cos(x)}\, dx\, =\, \int_{3/2}^1\, f(u)\, du,\)
. . .\(\displaystyle \mbox{then which of the following statements is true?}\)
\(\displaystyle \mbox{A. }\, u\, =\, 1\, +\, \cos(x),\, du\, =\, \sin(x)\, dx\)
\(\displaystyle \mbox{B. }\, u\, =\, 1\, +\, \cos(x),\, du\, =\, -\sin(x)\, dx\)
\(\displaystyle \mbox{C. }\, u\, =\, \cos(x),\, du\, =\, \sin(x)\, dx\)
\(\displaystyle \mbox{D. }\, u\, =\, \cos(x),\, du\, =\, -\sin(x)\, dx\)
\(\displaystyle \mbox{E. }\, u\, =\, \sin(x),\, du\, =\, \cos(x)\, dx\)
The first thing I did was solve the left function getting ln(3/2) and then I thought the next step would be to solve f(u)du by subbing in the options being
1 + cosx, cosx, and sinx to find the integral but I can't get right side to = left side. Anybody know how to go about this?
\(\displaystyle \displaystyle \mbox{If }\, \int_{\pi/3}^{\pi/2}\, \dfrac{\sin(x)}{1\, +\, \cos(x)}\, dx\, =\, \int_{3/2}^1\, f(u)\, du,\)
. . .\(\displaystyle \mbox{then which of the following statements is true?}\)
\(\displaystyle \mbox{A. }\, u\, =\, 1\, +\, \cos(x),\, du\, =\, \sin(x)\, dx\)
\(\displaystyle \mbox{B. }\, u\, =\, 1\, +\, \cos(x),\, du\, =\, -\sin(x)\, dx\)
\(\displaystyle \mbox{C. }\, u\, =\, \cos(x),\, du\, =\, \sin(x)\, dx\)
\(\displaystyle \mbox{D. }\, u\, =\, \cos(x),\, du\, =\, -\sin(x)\, dx\)
\(\displaystyle \mbox{E. }\, u\, =\, \sin(x),\, du\, =\, \cos(x)\, dx\)
The first thing I did was solve the left function getting ln(3/2) and then I thought the next step would be to solve f(u)du by subbing in the options being
1 + cosx, cosx, and sinx to find the integral but I can't get right side to = left side. Anybody know how to go about this?
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