integrate sinxcosx please

how does sin (x) cos (x) = (1/2) sin 2x

**0**

# integration chain rule please help

Started by Stewartd7, May 18 2009 01:29 PM

4 replies to this topic

### #1

Posted 18 May 2009 - 01:29 PM

### #2

Posted 18 May 2009 - 02:07 PM

because the expansion of sin(2x) is 2sin(x)cos(x)

sin(a+b)=sin(a)cos(b)+cos(a)sin(b), and if a=x,b=x

sin(x+x)=sin(x)cos(x)+cos(x)sin(x)

sin(2x)=2sin(x)cos(x)

sin(x)cos(x)=(1/2)sin(2x)

so the integral is (remember divide by the derivative of the inside )

-(1/4)cos(2x) +c

sin(a+b)=sin(a)cos(b)+cos(a)sin(b), and if a=x,b=x

sin(x+x)=sin(x)cos(x)+cos(x)sin(x)

sin(2x)=2sin(x)cos(x)

sin(x)cos(x)=(1/2)sin(2x)

so the integral is (remember divide by the derivative of the inside )

-(1/4)cos(2x) +c

**=-=-=Marcus=-=-=**### #3

Posted 18 May 2009 - 02:11 PM

QUOTE (Marcus @ May 18 2009, 03:07 PM) <{POST_SNAPBACK}>

because the expansion of sin(2x) is 2sin(x)cos(x)

sin(a+b)=sin(a)cos(b)+cos(a)sin(b), and if a=x,b=x

sin(x+x)=sin(x)cos(x)+cos(x)sin(x)

sin(2x)=2sin(x)cos(x)

sin(x)cos(x)=(1/2)sin(2x)

so the integral is

-(1/4)cos(2x) +c

sin(a+b)=sin(a)cos(b)+cos(a)sin(b), and if a=x,b=x

sin(x+x)=sin(x)cos(x)+cos(x)sin(x)

sin(2x)=2sin(x)cos(x)

sin(x)cos(x)=(1/2)sin(2x)

so the integral is

-(1/4)cos(2x) +c

where did the half come from???

and how do you do cos^2 (x) then? ta

### #4

Posted 18 May 2009 - 02:19 PM

the half comes from dividing both sides by 2, as sin(2x) =

and for cos²(x) you use a similar principle only using the cos(2x) expansion

cos(a+b)=cos(a)cos(b)-sin(a)sin(b)

cos(2x)=cos²x-sin², but we know cos²x+sin²x=1 so sin²=1-cos²x and subbing this in

cos(2x)=2cos²x-1

so 2cos²x=cos(2x)+1

cos²=(1/2)(cos(2x)) + (1/2)

which if we integrate, (1/2)(cos(2x)) becomes (1/4)sin(2x) and 1/2 becomes x/2

so final answer

(1/2)((1/2)sin(2x)+x)+c

symbolic answer

**2**cos(x)sin(x) and we only want one cos(x)sin(x)and for cos²(x) you use a similar principle only using the cos(2x) expansion

cos(a+b)=cos(a)cos(b)-sin(a)sin(b)

cos(2x)=cos²x-sin², but we know cos²x+sin²x=1 so sin²=1-cos²x and subbing this in

cos(2x)=2cos²x-1

so 2cos²x=cos(2x)+1

cos²=(1/2)(cos(2x)) + (1/2)

which if we integrate, (1/2)(cos(2x)) becomes (1/4)sin(2x) and 1/2 becomes x/2

so final answer

(1/2)((1/2)sin(2x)+x)+c

symbolic answer

**=-=-=Marcus=-=-=**### #5

Posted 18 May 2009 - 03:06 PM

QUOTE (Marcus @ May 18 2009, 03:19 PM) <{POST_SNAPBACK}>

the half comes from dividing both sides by 2, as sin(2x) =

and for cos²(x) you use a similar principle only using the cos(2x) expansion

cos(a+b)=cos(a)cos(b)-sin(a)sin(b)

cos(2x)=cos²x-sin², but we know cos²x+sin²x=1 so sin²=1-cos²x and subbing this in

cos(2x)=2cos²x-1

so 2cos²x=cos(2x)+1

cos²=(1/2)(cos(2x)) + (1/2)

which if we integrate, (1/2)(cos(2x)) becomes (1/4)sin(2x) and 1/2 becomes x/2

so final answer

(1/2)((1/2)sin(2x)+x)+c

symbolic answer

**2**cos(x)sin(x) and we only want one cos(x)sin(x)and for cos²(x) you use a similar principle only using the cos(2x) expansion

cos(a+b)=cos(a)cos(b)-sin(a)sin(b)

cos(2x)=cos²x-sin², but we know cos²x+sin²x=1 so sin²=1-cos²x and subbing this in

cos(2x)=2cos²x-1

so 2cos²x=cos(2x)+1

cos²=(1/2)(cos(2x)) + (1/2)

which if we integrate, (1/2)(cos(2x)) becomes (1/4)sin(2x) and 1/2 becomes x/2

so final answer

(1/2)((1/2)sin(2x)+x)+c

symbolic answer

i got 1/4sin(2x+1) +c n is that wrong???

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