Question

Evaluate the definite integral

∫_0^pi/2 3sinxcosx/(1+3sin^2x)^(1/2) dx

Answer 1

is it from 0 till pi/2 ?

Answer 2

yes sorry i'm trying to clear it up now

Answer 3

and the bottom is sin^2(x) ?

Answer 4

\[\int\limits_{0}^{\pi/2}{\frac{3sinxcosx}{\sqrt{1+3\sin^2x}}}\]

Answer 5

yes i hope that clears it up

Answer 6

I would try with substitution , u=sinx
du=cosxdx
than it will be
3u/(1+3u^2)^1/2

Answer 7

u=1+3sin^2x no?

Answer 8

maybe that would work too

Answer 9

but I cannot see that through, but with u=sinx, I think I have it

Answer 10

3u/(1+3u^2)^1/2 until this I guess it is clear
now lets use another sub for 1+3u^2=t
dt=6udu
so it will be
1/(2*(t)^1/2)

Answer 11

okay

Answer 12

is this clear?

Answer 13

not thae last part

Answer 14

the last line

Answer 15

the antiderivative of u^-1/2

Answer 16

3u/(1+3u^2)^1/2
we have this and use
t=1+3u^2
so dt=6udu
so from the nominator 3udu there will be 1/2dt

Answer 17

I still have not done any integration

Answer 18

im following you i see what u did in order to get it do be the same as the numerator u divide by 1/2

Answer 19

yep and integrationg now we get
t^1/2+C

Answer 20

that is (1+3u^2)^1/2
that is (1+3sin^2x)^1/2

Answer 21

yes it is correct :-) Just checked it on wolfgramalpha

Answer 22

and now you just have to put pi/2 and 0 into this and subtract

Answer 23

hmm okay i see haha lol

Answer 24

sinpi/2=1
sin0=0
2-1=1

Answer 25

okay thank you!

Answer 26

you are welcome, I like integrals. Lot of work but they are interesting

Answer 27

yeah they confuse me though but i have to practice...maybe you can help me with a question i posted earlier...please...

Answer 28

OK, lets see

Answer 29

which one, you posted quite a few :-)

Answer 30

lol i think i may have deleted it let me repost