Question

integral of sin^3 (2x)/ 1+ cos 2x

Answer 1

Is this your question?
\[
\int\frac{\sin^3(2x)}{1+\cos(2x)}dx
\]

Answer 2

yes

Answer 3

Hint:
\[
\sin^3(2x)=(\sin(2x))^3\\
\cos(2x)=1-2\sin^2(x)
\]

Answer 4

\[\int\limits(\sin ^{2} 2x)(\sin 2x)=\frac{ \int\limits(1+\cos 2x)(1-\cos2x)(\sin2x }{ 1+\cos2x }\]

could this be possible?

Answer 5

just wondering an alternative method :
s= sin 2x, c = cos 2x

s^3 = s*s^2 = s (1-c^2) = s(1+c)(1-c)

so we get
integral of s(1-c) = integral of s -sc = sin 2x -(1/2) sin 4x

Answer 6

aha! pooh, you tried the same thing :)
and ofcourse thats possible !

Answer 7

then is it
=-1/2(cos 2x) -1/4(cos 2x) + c

Answer 8

I got \(\frac{1}{4}\cos^4(x)+c\)...

Answer 9

-1/2(cos 2x) is correct first term, but how did u get
-1/4(cos 2x)
?

Answer 10

im quite confused...can you let u as cos 2x and du= 2 sin 2x?

[sin^2 (2x)][sin 2x] / 1+cos 2x dx

= [1+cos 2x][1-cos 2x](sin 2x) / 1+cos 2x dx

=[1-cos 2x][sin 2x] dx

= (sin 2x) - (cos 2x sin 2x) dx

=-1/2(cos 2x) -1/4(cos 2x) + c

Answer 11

du= -2sin 2x

Answer 12

i just asked how did u calculate integral (cos 2x sin 2x) dx as 1/4(cos 2x) + c


what i would do is
(cos 2x sin 2x) dx = 1/2 (2 cos 2x sin 2x) = 1/2 sin 4x dx
integrating we get

1/2 (cos 4x)/4 +c
or 1/8 cos 4x +c for the 2nd term

Answer 13

if you let u as cos 2x and du= 2 sin 2x

then
(cos 2x sin 2x) dx becomes 1/2 (u) dx
integrating
u^2/4 +c

(cos ^2 2x)/4 +c

did you missed the square ?

Answer 14

and both are correct and equivalent ways to integrate :)

Answer 15

Your way is more clever than mine's.

Answer 16

To be honest, I have absolutely no idea what poohtooth is trying to do. Why there is \(\int(\sin^2(2x)\sin(2x))dx\)?

Answer 17

\(\sin^3 2x = \sin 2x \times \sin ^2 2x\)

Answer 18

ah i missed something... i let u=sin2x and du= 2cos2x that's why i got \[\frac{ 1 }{ 2 }\frac{ \cos2x }{ 2 }=\frac{ 1 }{ 4 }\cos 2x\] am i correct?

Answer 19

@ thomas it is to cancel your denominator..

Answer 20

you still get integral u/2 du
right ??

integrating it gives you

u^2/4 +c ... you doesn't seem to get the square term

Answer 21

ah ok...so you mean\[(\frac{ \sin 2(2x) }{ 2})\]?

Answer 22

I am not sure whether or not you know this but I will just point this out.
\[
\int\frac{f(x)}{g(x)}dx{\neq\frac{\int f(x)\,dx}{\int g(x)\,dx}}\neq\frac{1}{g(x)}\int f(x)\,dx
\]

Answer 23

Which is the integral, a,b or neither?
\[
\begin{align*}
&\int\frac{\sin^3(2x)}{1+\cos(2x)}dx &\text{a}\\
&\frac{\int\sin^3(2x)\,dx}{1+\cos(2x)}&\text{b}
\end{align*}
\]

Answer 24

a

Answer 25

Have you learned integration by substitution?

Answer 26

yes of course...