Question

Integrate (sinx + cosx)^2?

Answer 1

can you expand \[(\sin x + \cos x)^2\]

Answer 2

\[(sinx+cosx)^2=\sin^2x+2sinxcosx+\cos^2x\]\[\int\limits{(sinx+cosx)^2}dx=\int\limits{(\sin^2x+\cos^2x)dx}+\int\limits{2sinxcosxdx}\]you know that\[\sin^2x+\cos^2x=1\]and\[2sinxcosx=\sin2x\]so the integral becomes\[\int\limits{(sinx+cosx)^2dx}=\int\limits{dx}+\int\limits{\sin(2x)dx}\]can u take it from here?

Answer 3

Yes i think so thanks for the great explanation !

Answer 4

:)

Answer 5

I just didnt realize you had to expand :s but i get it now

Answer 6

could someone explain to me how sin2x becomes sin^2x?

Answer 7

@lalaly is so awesome isnt she? she's like SOOOO GREAT!!!

Answer 8

no no no... sin(2x)=2sinxcosx

Answer 9

lol ewgeeee thanks

Answer 10

the answer in the text book says the answer is x + (sinx)^2 and i was confused how they got that

Answer 11

x is the \(\int dx\) you know that right?

Answer 12

yes i get that one but its the sin2x i dont get

Answer 13

hehe that's all i got too :P but i do remember this =_=

Answer 14

http://www.wolframalpha.com/input/?i=integral+%28sinx%2Bcosx%29%5E2

Answer 15

should be \[-\frac{\cos (2x)}{2}\] right?

Answer 16

yeah

Answer 17

:S it says x + (sinx)^2

Answer 18

so i geuss typo or something?

Answer 19

well wolfram approves my answer so i guess the answer u have is wrong

Answer 20

okay thanks for all the help

Answer 21

ur welcome

Answer 22

well \[\int 2\sin x \cos x dx\] let u = \(\sin x\)

\[\int 2udu = u^2 = \sin^2 x\]

Answer 23

that's how your book did it

Answer 24

okay i see thanks !

Answer 25

lol nice@lgbasallote

Answer 26

<tips hat>