Question

I have a question in Trigonometric Integrals.

Answer 1

\[\int\limits_{0}^{\pi/6} \cos^3 (2x) dx \]

Answer 2

I solved by my self and I got this answer \[\sin2x + 1\frac{ 1 }{ 3 } \sin^3 2x + c \]

Answer 3

is this correct? please help

Answer 4

first off it's a definite integral, so you should have a number answer and no +C

Answer 5

aside from that it looks good though :)

Answer 6

I have problem with Trigonometric Integrals. Does it have a table of integral? if you have a table please send to me

Answer 7

Trig integrals are more about strategy
some common ones like integral sin, cos, sec^2, etc. are well-known, but most require some manipulation

Answer 8

a good run-down for tactics in trig integrals can be found here
http://tutorial.math.lamar.edu/Classes/CalcII/IntegralsWithTrig.aspx

Answer 9

here the move I would make is think
"oh, I can get a cos^2 out of this, which I can make into sin
the leftover cos can be used as du in a u-sub"
and get the answer you got
I don't know, but I bet your reasoning was similar: stripping out a cos and converting cos^2 to 1-sin^2

Answer 10

oh you did make one mistake though

Answer 11

your answer is in fact slightly wrong, recheck your u-sub

Answer 12

@ahmed84 are you still with me here?

Answer 13

yep

Answer 14

what is \[\int\cos(2x)dx\]?

Answer 15

could you right the answer for me so I can follow it correctly. If it is possible

Answer 16

you have everything correct, but I am almost certain the problem lies in the fact that you messed up the above integral
so again, what is\[\int\cos(2x)dx\]after that I will run the whole thing for you...

Answer 17

have almost everything correct*

Answer 18

ok, what is\[\int\cos xdx\]?

Answer 19

i am with you

Answer 20

@ahmed84 c'mon brother please try to answer my questions so I can help you

Answer 21

-sinx

Answer 22

no, it is positive, because
d/dx(sin x)=cos x

Answer 23

so it is sinx only?

Answer 24

yes\[\int\cos xdx=\sin x+C\]now what do we have to do to find\[\int\cos(2x)dx\]???

Answer 25

it is \[\frac{ \sin 2x }{ 2 } + c\]

Answer 26

is it correct?

Answer 27

yes, and that was your only mistake in your answer, you forgot that u-sub

Answer 28

so from the top I will do the whole thing...

Answer 29

thank you very much I will follow your answer and learn it step by step.

Answer 30

there was a typo above\[\int_0^{\pi/6}\cos^3(2x)dx=\int_0^{\pi/6}\cos(2x)cos^2(2x)dx\]\[=\int_0^{\pi/6}\cos(2x)(1-\sin^2(2x))dx=\int_0^{\pi/6}\cos(2x)-\sin^2(2x)\cos(2x)dx\]\[\int_0^{\pi/6}\cos(2x)dx-\int_0^{\pi/6}\sin^2(2x)\cos(2x)dx\]\[u=2x\implies du=2dx\implies dx=\frac12du\]\[v=\cos(2x)\implies dv=-2\sin(2x)dx\implies\sin(2x)dx=-\frac12dv\]\[\frac12\int\cos udu-(-\frac12)\int v^2dv\]

Answer 31

\[=\frac12\sin u+\frac16v^3\]sub back in\[u=2x~~~\text{and}~~~v=\sin(2x)\]and evaluate\[\frac12\sin(2x)|_0^{\pi/6}+\frac16\sin^3(2x)|_0^{\pi/6}\]

Answer 32

thanks a lot brother.

Answer 33

you are a great man.

Answer 34

no, just like to be helpful
I like to do this while I exercise, it's a good body-mind work out to start your day :)
welcome!

Answer 35

hello

Answer 36

oh dear, that is a typo
all it will do is change a -sign, but let me see how fast I can redo this...

Answer 37

bro why did you choice u = 2x and v = sin2x, why this is your choice?

Answer 38

thanks my brother.

Answer 39

excellent

Answer 40

don't thank me yet the answer is wrong...

Answer 41

oh I think I just evaluated wrong...

Answer 42

yeah it was just that I screwed up and put sin(pi/3) as 1/2, it's really sqrt(
)/2

Answer 43

*sqrt(3)/2

Answer 44

\[\int_0^{\pi/6}\cos^3(2x)dx=\int_0^{\pi/6}\cos(2x)cos^2(2x)dx\]\[=\int_0^{\pi/6}\cos(2x)(1-\sin^2(2x))dx=\int_0^{\pi/6}\cos(2x)-\sin^2(2x)\cos(2x)dx\]\[=\int_0^{\pi/6}\cos(2x)dx-\int_0^{\pi/6}\sin^2(2x)\cos(2x)dx\]\[u=2x\implies du=2dx\implies dx=\frac12du\]\[v=\sin(2x)\implies dv=2\cos(2x)dx\implies\cos(2x)dx=\frac12dv\]\[\int\cos u(\frac12du)-\int v^2(\frac12dv)\]\[=\frac12\int\cos udu-\frac12\int v^2dv\]\[=\frac12\sin u-\frac16v^3\]sub back in\[u=2x\]\[v=\sin(2x)\]and evaluate\[\frac12\sin(2x)|_0^{\pi/6}-\frac16\sin^3(2x)|_0^{\pi/6}\]\[\frac12[\sin(\frac\pi3)-\cancel{\sin0}^{\huge0}]-\frac16[\sin^3(\frac\pi3)-\cancel{\sin^30}^{\huge0}]\]\[=\frac12(\frac{\sqrt3}2)-(\frac16\cdot\frac{3\sqrt3}{2^3})\]\[=\frac{4\sqrt3}{16}-\frac{3\sqrt3}{16}=\frac{3\sqrt3}{16}\]

Answer 45

whew, at least this time I know it's right, I checked wolfram

Answer 46

Thanks a lot. I really appreciate.