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I have checked and rechecked my problem the whole way through and I can't figure out what is wrong with it, I would really really like to know what I did wrong.

 

\(\int_{-\pi/3}^{0}(\sqrt{\cos x})(\sin^{3}x) dx\)

 

Next I just rearranged things:

 

\(-1 \int_{-\pi /3}^{0} (\cos x)^{1/2}(\sin^{2}x)(-\sin x) dx\)

 

which equals:

 

\(-1 \int_{-\pi /3}^{0} (\cos x)^{1/2}(1-\cos^{2}x)(-\sin x) dx\)

 

then I defined variables to use for substitution:

 

u = cos x         and          du = -sin x dx

 

then substitute them in:

 

\(-1 \int_{\cos -\pi /3}^{\cos 0} (u)^{1/2}(1-u^{2})du\)

 

distribute and find the endpoints:

 

\(-\int_{1/2}^{1}(u^{1/2}-u)du\)

 

then power rule:

 

\(-[\frac{2}{3}u^{3/2}-\frac{1}{2}u^{2}]\)and that should have a straight line at the end that means from one half to one, i just dont know how to do it on here.

Next just distribute that -1:

 

\([-\frac{2}{3}u^{3/2}+\frac{1}{2}u^{2}]\)once again there should be a line at the end that means from one half to one.

 

then use the fundamental theorem of calculus to find the definite integral:

 

\([-\frac{2}{3}(1)^{3/2}+\frac{1}{2}(1)^{2}]-[-\frac{2}{3}(1/2)^{3/2}+\frac{1}{2}(1/2)^{2}]\)

 

then after some simplification I end up with:

 

\(\frac{-2}{3} +\frac{1}{2}+\frac{2}{3}(\frac{1}{2\sqrt{2}})-\frac{1}{8}\)

 

then I get:

 

\(-\frac{7}{24}+\frac{\sqrt{2}}{6}\)

 

That is my answer, but the correct answer should be:

 

\(-\frac{8}{21}+\frac{25\sqrt{2}}{168}\)

 Feb 7, 2017

Best Answer 

 #2
avatar+25597 
+10

I have checked and rechecked my problem the whole way through and I can't figure out what is wrong with it, I would really really like to know what I did wrong.

\(\int_{-\pi/3}^{0}(\sqrt{\cos x})(\sin^{3}x) dx\)

 

Until \(-1 \int_{\cos -\pi /3}^{\cos 0} (u)^{1/2}(1-u^{2})du\) it is correct.

 

distribute and find the endpoints:

\(\begin{array}{|rcll|} \hline && -\int \limits_{\frac12}^{1} u^{\frac12}(1-u^{2})\ du \\ &=& -\int \limits_{\frac12}^{1}( u^{\frac12}-u^{\frac12}\cdot u^{2})\ du \quad & | \quad u^{\frac12}*u^2 \text { is not }u !!!\\ &=& -\int \limits_{\frac12}^{1}( u^{\frac12}-u^{\frac12+2})\ du \\ &=& -\int \limits_{\frac12}^{1}( u^{\frac12}-u^{\frac12+\frac42})\ du \\ &=& -\int \limits_{\frac12}^{1}( u^{\frac12}-u^{\color{red}{\frac52}})\ du \\ &=& - \left[\frac23 u^{\frac32}- \frac27u^{\frac72}) \right]_{\frac12}^{1} \\ &=& \left[-\frac23 u^{\frac32}+ \frac27u^{\frac72}) \right]_{\frac12}^{1} \\ &=& -\frac23\cdot (1) + \frac27 \cdot(1)-[-\frac23\cdot (\frac12)^{\frac32}+\frac27\cdot (\frac12)^{\frac72} ] \\ &=& -\frac23 + \frac27 +\frac{1}{3\sqrt{2}}-\frac{1}{28\sqrt{2}} \quad & | \quad \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2} \\ &=& -\frac23 + \frac27 +\frac{\sqrt{2}}{6}-\frac{\sqrt{2}}{56} \\ &=& \frac{-14+6}{21} + \frac{56-6}{6\cdot 56}\cdot \sqrt{2} \\ &=& \frac{-8}{21} + \frac{50}{336}\cdot \sqrt{2} \\ &=& -\frac{8}{21} + \frac{25}{168}\cdot \sqrt{2} \\ \hline \end{array}\)

 

 

laugh

 Feb 7, 2017
 #2
avatar+25597 
+10
Best Answer

I have checked and rechecked my problem the whole way through and I can't figure out what is wrong with it, I would really really like to know what I did wrong.

\(\int_{-\pi/3}^{0}(\sqrt{\cos x})(\sin^{3}x) dx\)

 

Until \(-1 \int_{\cos -\pi /3}^{\cos 0} (u)^{1/2}(1-u^{2})du\) it is correct.

 

distribute and find the endpoints:

\(\begin{array}{|rcll|} \hline && -\int \limits_{\frac12}^{1} u^{\frac12}(1-u^{2})\ du \\ &=& -\int \limits_{\frac12}^{1}( u^{\frac12}-u^{\frac12}\cdot u^{2})\ du \quad & | \quad u^{\frac12}*u^2 \text { is not }u !!!\\ &=& -\int \limits_{\frac12}^{1}( u^{\frac12}-u^{\frac12+2})\ du \\ &=& -\int \limits_{\frac12}^{1}( u^{\frac12}-u^{\frac12+\frac42})\ du \\ &=& -\int \limits_{\frac12}^{1}( u^{\frac12}-u^{\color{red}{\frac52}})\ du \\ &=& - \left[\frac23 u^{\frac32}- \frac27u^{\frac72}) \right]_{\frac12}^{1} \\ &=& \left[-\frac23 u^{\frac32}+ \frac27u^{\frac72}) \right]_{\frac12}^{1} \\ &=& -\frac23\cdot (1) + \frac27 \cdot(1)-[-\frac23\cdot (\frac12)^{\frac32}+\frac27\cdot (\frac12)^{\frac72} ] \\ &=& -\frac23 + \frac27 +\frac{1}{3\sqrt{2}}-\frac{1}{28\sqrt{2}} \quad & | \quad \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2} \\ &=& -\frac23 + \frac27 +\frac{\sqrt{2}}{6}-\frac{\sqrt{2}}{56} \\ &=& \frac{-14+6}{21} + \frac{56-6}{6\cdot 56}\cdot \sqrt{2} \\ &=& \frac{-8}{21} + \frac{50}{336}\cdot \sqrt{2} \\ &=& -\frac{8}{21} + \frac{25}{168}\cdot \sqrt{2} \\ \hline \end{array}\)

 

 

laugh

heureka Feb 7, 2017
 #3
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Ooohhh okay okay thank you heureka!!!!!! smiley

 Feb 7, 2017

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