Question

integrate 1-2sin^2x

Answer 1

Using the identity sin^2(x)=(1/2)(1-cos(2x)) you have:
\[\int\limits 1-2(1/2)(1-\cos(2x))dx=\int\limits dx-\int\limits dx+\int\limits \cos(2x)dx=\frac{1}{2}\sin(2x)+C\]
Then you know that sin(2x)=2sin(x)cos(x)
So you could write it as:
sin(x)cos(x)+C

Answer 2

okay thanks malevolence

Answer 3

No problem :P

Answer 4

but wait

Answer 5

What?

Answer 6

oh nevermind...

Answer 7

Sure? I don't mind explaining if you're lost :P

Answer 8

i was wondering why you said sin2x but then you change the answer...okay

Answer 9

2sin^x+2cos^2x=1
so \[\int\limits_{}^{}2cos^2x=INTEGRAL2(1/2+1/2cos2x)=\]
\[\int\limits_{}^{}1+cos2x=x+1/2sin2x+c\]

Answer 10

sorry this is wrong the first part equal 2 not 1 so that means
2cos^2x+2sin^x=2 so 1-2sin^2x=2cos^2x-1


so the answer should have a -dx which would make that x go away so the final answer should be 1/2sin2x +c

Answer 11

Either way. You have (1/2)sin(2x)=(2/2)sin(x)cos(x)=sin(x)cos(x)...?

Answer 12

yeah i guess so i just got creatice you can do 1-2sin^2x=1-(1/2-1/2cos2x)=1-1/2+1/2cos2x=x/2+1/4sin2x+c

i hated these integrals too much algebra thats my final attempted

just use the idenities
sin^2x=1/2-1/2cos2x
cos^2x=1/2+1/2cos2x

Answer 13

lol thank bnut for the other version

Answer 14

But bnut, you have a 2 outside the sin^2(x) so you have 1-1-cos(2x) so you're only left with the integral of -cos(2x).