Question

Evaluate the integral:

Answer 1

0

Answer 2

\[\LARGE \int_{0}^{\pi/2}~sin^2x~cos^2x~dx\]

Answer 3

try without the limits first

Answer 4

I'll do the indefinite you do the limits :D

Answer 5

copy cat

Answer 6

copy bat*

Answer 7

Actually no, luigi show me what you've done first...

Answer 8

Can you do it by parts?

Answer 9

you can do anything you want

Answer 10

cos^2x = (1-sin^2x)

Answer 11

\[\int\limits \sin^2x(1-\sin^2x)dx => \int\limits (\sin^2x - \sin^4x)dx \]

Answer 12

This is the identity:
\[\frac{ 1 }{ 4 }\sin^{2}(2x)\]

Answer 13

http://mathworld.wolfram.com/TrigonometricPowerFormulas.html

Answer 14

Lets confuse luigi more

Answer 15

I changed it to \[\LARGE \int (\frac{1+cos2x}{2})(\frac{1-cos2x}{2})~dx\] but got the wrong answer..

Answer 16

laughing out loud

Answer 17

that ultimately reduces to (1/4) sin^2(2x)

Answer 18

\[\int\limits \sin^2xcos^2x dx = \int\limits \sin^2x(1-\sin^2x)dx\]

Because cos^2x = 1-sin^2x, then you can expand it and get,

\[\int\limits (\sin^2x-\sin^4x) dx = \int\limits \sin^2xdx - \int\limits \sin^4x dx\]

You can use the reduction formula here, do you know it?

Answer 19

\[I=\int\limits_0^{\pi/2} \sin^2 x \cos^2 x\ dx\]\[J=\int\limits_0^{\pi/2} \sin^2 x \ dx\]\[J-I=\int\limits_0^{\pi/2} \sin^2 x(1-\cos^2 x) \ dx=\int\limits_0^{\pi/2} \sin^4 x \ dx\] ok now you can solve this more complicated simplification. GL!

Answer 20

\[\frac{ 1 }{ 4 }\sin^{2}(2x)=\frac{ 1 }{ 8 }(1-\cos(4 x)) \]

Answer 21

neat trick, kai

Answer 22

\[I=\int\limits_0^{\pi/2} \sin^2 x-\sin^4 x \ dx\] Although honestly is that any better?

Answer 23

Don't bully me

Answer 24

laughing out loud
he's batman

Answer 25

I'm robbin'

Answer 26

you can use power reduction from the get go

Answer 27

\[\frac{ 1 }{ 4 }\sin^3xcosx+\int\limits \sin^2xdx-\frac{ 3 }{ 4 } \int\limits \left( \frac{ 1 }{ 2 }-\frac{ 1 }{ 2 }\cos(2x) \right)dx\]

eek

Answer 28

ok the real way to do it is like this:

Answer 29

\[\int\limits_{0}^{\Pi/2}\frac{ 1 }{ 4 }\sin^{2}(2x)=\int\limits_{0}^{\Pi/2}\frac{ 1 }{ 8 }(1-\cos(4x))\]

Answer 30

\[\int\limits_0^{\pi/2} \sin^2 x \cos^2 x \ dx=\int\limits_0^{\pi/2} (\sin x \cos x )^2\ dx=\int\limits_0^{\pi/2} (\frac{1}{2}\sin (2x) )^2 \ dx\]

Answer 31

\[\frac{1}{4}\int\limits_0^{\pi/2} \sin ^2 (2x) \ dx =\frac{1}{4}\int\limits_0^{\pi/2} \frac{1-\cos (4x)}{2} \ dx=\frac{1}{8}\int\limits_0^{\pi/2} 1-\cos (4x) \ dx\]

Answer 32

\[\int\limits \sin^2xcos^2x dx = \frac{ 1 }{ 32 }(4x-\sin(4x))+C \] Final answer of indefinite integral, fyi.

Answer 33

So pi/16 with limits

Answer 34

Actually there's more cleverness I suggest you do while solving this:

4x=u to get rid of how frequent it is.

\[\frac{1}{2}\int\limits_0^{2\pi} 1-\cos (u)\ du\]

Notice how that limit of integration changed? That automatically allows us to drop the cosine part! Why? Well every 2pi it repeats right? Just look at this those areas cancel out every 2pi so don't even bother integrating it since you know it's garbage.


\[\frac{1}{2}\int\limits_0^{2\pi}du=\pi\]

That's right right?