Question

please need help with these two integral problems

Answer 1

\[\int\limits_{0}^{\pi/2} \sin^2x*\cos^2x\]

Answer 2

and \[\int\limits_{?}^{?} \sin8x * \cos5x\]

Answer 3

@Concentrationalizing @perl @zepdrix @freckles

Answer 4

how do i do them?

Answer 5

in the first one just multiply and divide by 4
make it sin^2 2x
then convert it using 1-cos 2x = 2sin^2 x
it will become a basic integral

Answer 6

why by 4?

Answer 7

and for the second one
2Sin A cos B = Sin (A+B) + Sin (A-B)

Answer 8

\[\sin(x+y)=\sin(x)\cos(y)+\sin(y) \cos(x) \\ \sin(x-y)=\sin(x)\cos(y)-\sin(y)\cos(x) \\ \text{ add these identities } \sin(x+y)+\sin(x-y)=2 \sin(x) \cos(y) +0 \\ \text{ now divide by } 2 \\ \frac{\sin(x+y)+\sin(x-y)}{2}=\sin(x) \cos(y) \\ \frac{1}{2} \sin(x+y)+\frac{1}{2} \sin(x-y)=\sin(x)\cos(y)\]
for the second one (Which I'm late for )

Answer 9

coz sin 2x = 2 sin x cos x
and sin ^2 2x = 4sin^2 x cos^2 x
this will replace those two terms

Answer 10

okay so i got sin(13)+sin(3) for the second one

Answer 11

so now i divide the whole thing by 2?

Answer 12

check this exp again sin (13) + sin (3)
isn't there something missing?

Answer 13

oh the 1/2's right

Answer 14

its 'x' dear

Answer 15

sin (13x) + sin (3x)

Answer 16

x's where do they go?

Answer 17

got it now?

Answer 18

so thats the final answer?

Answer 19

no, now you have to integrate them wrt x

Answer 20

how?

Answer 21

oh you mean turn them to cosines

Answer 22

yeah

Answer 23

okay so it would be 13cos(13x) + 3cos(3x)

Answer 24

right?

Answer 25

you should see your notes again
those 12 and 3 will be in the denom

Answer 26

okay fixed it so for the first one its a trig identity right?

Answer 27

yeah

Answer 28

so when its sin^2x*cos^2x you always change it to 4sin^2x*cos^2x?

Answer 29

yep

Answer 30

okay

Answer 31

okay so i got