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anonymous

  • 5 years ago

compute the indefinite integral. (sin 2x)^2 (cos2x)^2 dx please show me in a detail process.

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  1. nowhereman
    • 5 years ago
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    Substitute 2x, write cos in terms of sin and then use formula for powers of sin.

  2. anonymous
    • 5 years ago
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    \[x^2 (\cos 3x/3)-\int\limits (-\cos3x/3)(2x) \] something like that right??

  3. anonymous
    • 5 years ago
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    wait sorry i was working on a different problem, can you please show me the steps to the previous problem...(sin 2x)^2 (cos2x)^2 dx

  4. anonymous
    • 5 years ago
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    \[\int\limits (\sin2x)^{2} (\cos2x) ^{2} dx \]

  5. anonymous
    • 5 years ago
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    \[\int\limits (\sin2x)^2(\cos2x)^2dx= \int\limits [(\sin2x)(\cos2x)]^2dx\] now use this trig identity: \[\sin(2u)=2\sin(u)\cos(u)\] in our case u=2x , so... \[\sin(4x)=2\sin(2x)\cos(2x)\rightarrow \frac{1}{2}\sin(4x)=\sin(2x)(\cos(2x)\] so... \[\int\limits\limits [(\sin2x)(\cos2x)]^2dx= \int\limits [\frac{1}{2}\sin(4x)]^2dx \] \[\rightarrow \int\limits \frac{1}{4}(\sin(4x))^2dx\] We have to use another trig identity: \[\cos(2u)=1-2(\sin(u))^2\rightarrow 2(\sin(u))^2=1-\cos(2u)\] \[\rightarrow (\sin(u))^2=\frac {1-\cos(2u)}{2}\] in our case u=4x, so... \[(\sin(4x))^2=\frac {1-\cos(8x)}{2}\] Now our integral becomes: \[\int\limits \frac{1}{4}[\frac{1}{2}-\frac{\cos(8x)}{2}]dx \rightarrow \frac {1}{8} \int\limits dx- \frac {1}{8} \int\limits \cos(8x)dx\] \[Thus:\frac {1}{8} \int\limits\limits dx- \frac {1}{8} \int\limits\limits \cos(8x)dx= \frac{x}{8}-\frac {\sin(8x)}{64}+C\]

  6. anonymous
    • 5 years ago
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    thanks!

  7. anonymous
    • 5 years ago
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    no problem

  8. anonymous
    • 5 years ago
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    can you help me with another problem ?

  9. anonymous
    • 5 years ago
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    sure

  10. anonymous
    • 5 years ago
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    \[\int\limits_{0}^{8} (x+e^x)^2 dx \] use the midpoint rule with n =4 to appoximate this integral.

  11. anonymous
    • 5 years ago
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    I thought I already did this....?

  12. anonymous
    • 5 years ago
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    and what is the error in the approximation.

  13. anonymous
    • 5 years ago
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    no lol

  14. anonymous
    • 5 years ago
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    wait actually nvm, i got this question. but what about this one: \[\int\limits x^2 \sin(3x)dx \] compute the integral .

  15. anonymous
    • 5 years ago
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    Use integration by part twice and the use u substitution: \[\int\limits\limits\limits f(x)g(x)dx= F(x)g(x)-\int\limits\limits\limits F(x)g'(x)dx\] Where F(x) \in the antiderivative of f(x) \[\int\limits x^2\sin(3x)dx \rightarrow \] \[f(x)=\sin(3x), F(x)=-\frac {\cos(3x)}{3}, g(x)=x^2, g'(x)=2x\] \[-\frac{1}{3}x^2\cos(3x)+\frac{2}{3}\int\limits xcos(3x)dx\] \[f(x)=\cos(3x), F(x)=\frac{\sin(3x)}{3}, g(x)=x, g'(x)=1\] \[-\frac{1}{3}x^2\cos(3x)+\frac{2}{9}xsin(3x)-\frac{2}{9} \int\limits \sin(3x)dx\] \[u=3x, du=3dx, \frac{du}{3}=dx\] \[-\frac{1}{3}x^2\cos(3x)+\frac{2}{9}xsin(3x)-\frac{2}{27} \int\limits\limits \sin(u)du\] \[-\frac{1}{3}x^2\cos(3x)+\frac{2}{9}xsin(3x)+\frac{2}{27} \cos(u)+C \rightarrow u=3x\] \[-\frac{1}{3}x^2\cos(3x)+\frac{2}{9}xsin(3x)+\frac{2}{27} \cos(3x)+C\]

  16. anonymous
    • 5 years ago
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    omg thanks you are a big help!

  17. anonymous
    • 5 years ago
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    your welcome

  18. anonymous
    • 5 years ago
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    one last question.. consider the integral \[\int\limits_{0}^{1} e ^{-x} dx \] suppose that we want to approximate this integral using the trapezoid rule. if we want our approximation to have an error less than 10^-3, how large should we take n??

  19. nowhereman
    • 5 years ago
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    the error term for the trapezoidrule for function f over interval [a,b] is \[-\frac{(b-a)^3}{12}f''(ζ) = -\frac{1}{12n^3}e^{-ζ} \] so the absolute error is at most \[\frac{1}{12n^2}\] if you sum it over all the n intervals. So you want \[10^{-3} \geq \frac{1}{12n^2} ⇔ n \geq \sqrt{\frac{10^3}{12}}\approx 9.1287\] So n must be greater or equal to 10.

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