Question

compute the indefinite integral.
(sin 2x)^2 (cos2x)^2 dx

please show me in a detail process.

Answer 1

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

Answer 2

\[x^2 (\cos 3x/3)-\int\limits (-\cos3x/3)(2x) \]

something like that right??

Answer 3

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

Answer 4

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

Answer 5

\[\int\limits (\sin2x)^2(\cos2x)^2dx= \int\limits [(\sin2x)(\cos2x)]^2dx\]
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:
\[\int\limits \frac{1}{4}[\frac{1}{2}-\frac{\cos(8x)}{2}]dx \rightarrow \frac {1}{8} \int\limits dx- \frac {1}{8} \int\limits \cos(8x)dx\]
\[Thus:\frac {1}{8} \int\limits\limits dx- \frac {1}{8} \int\limits\limits \cos(8x)dx= \frac{x}{8}-\frac {\sin(8x)}{64}+C\]

Answer 6

thanks!

Answer 7

no problem

Answer 8

can you help me with another problem ?

Answer 9

sure

Answer 10

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

Answer 11

I thought I already did this....?

Answer 12

and what is the error in the approximation.

Answer 13

no lol

Answer 14

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

compute the integral .

Answer 15

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)
\[\int\limits x^2\sin(3x)dx \rightarrow \]
\[f(x)=\sin(3x), F(x)=-\frac {\cos(3x)}{3}, g(x)=x^2, g'(x)=2x\]
\[-\frac{1}{3}x^2\cos(3x)+\frac{2}{3}\int\limits xcos(3x)dx\]
\[f(x)=\cos(3x), F(x)=\frac{\sin(3x)}{3}, g(x)=x, g'(x)=1\]
\[-\frac{1}{3}x^2\cos(3x)+\frac{2}{9}xsin(3x)-\frac{2}{9} \int\limits \sin(3x)dx\]
\[u=3x, du=3dx, \frac{du}{3}=dx\]
\[-\frac{1}{3}x^2\cos(3x)+\frac{2}{9}xsin(3x)-\frac{2}{27} \int\limits\limits \sin(u)du\]
\[-\frac{1}{3}x^2\cos(3x)+\frac{2}{9}xsin(3x)+\frac{2}{27} \cos(u)+C \rightarrow u=3x\]
\[-\frac{1}{3}x^2\cos(3x)+\frac{2}{9}xsin(3x)+\frac{2}{27} \cos(3x)+C\]

Answer 16

omg thanks you are a big help!

Answer 17

your welcome

Answer 18

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??

Answer 19

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.