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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

Substitute 2x, write cos in terms of sin and then use formula for powers of sin.

2. anonymous

$x^2 (\cos 3x/3)-\int\limits (-\cos3x/3)(2x)$ something like that right??

3. anonymous

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

$\int\limits (\sin2x)^{2} (\cos2x) ^{2} dx$

5. anonymous

$\int\limits (\sin2x)^2(\cos2x)^2dx= \int\limits [(\sin2x)(\cos2x)]^2dx$ 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

thanks!

7. anonymous

no problem

8. anonymous

can you help me with another problem ?

9. anonymous

sure

10. anonymous

$\int\limits_{0}^{8} (x+e^x)^2 dx$ use the midpoint rule with n =4 to appoximate this integral.

11. anonymous

I thought I already did this....?

12. anonymous

and what is the error in the approximation.

13. anonymous

no lol

14. anonymous

wait actually nvm, i got this question. but what about this one: $\int\limits x^2 \sin(3x)dx$ compute the integral .

15. anonymous

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

omg thanks you are a big help!

17. anonymous

your welcome

18. anonymous

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

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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