integrating sinxcosx (1 Viewer)

[starkskyy]

using subsitution the answer is wrong, but from what i know, it should work

integral (sinx.cosx)
=integral (u) let u=sinx, du=cosx dx
=1/2u^2 +c
=1/2sin^2x +c
=1/2[1/2(1-cos2x)] +c
=1/4-1/4cos2x +c

but then there's also,

integral (sinx.cosx)
=1/2 integral (sin2x)
=1/2.-1/2cos2x +c
=-1/4cos2x +c

they both seem correct.... i have a feeling the constants (c and 1/4) can be considered as one, but im not so sure about this. which one is correct, or which is the correct method to use.

 

[undalay]

using subsitution the answer is wrong, but from what i know, it should work

integral (sinx.cosx)
=integral (u) let u=sinx, du=cosx dx
=1/2u^2 +c
=1/2sin^2x +c
=1/2[1/2(1-cos2x)] +c
=1/4-1/4cos2x +c

but then there's also,

integral (sinx.cosx)
=1/2 integral (sin2x)
=1/2.-1/2cos2x +c
=-1/4cos2x +c

they both seem correct.... i have a feeling the constants (c and 1/4) can be considered as one, but im not so sure about this. which one is correct, or which is the correct method to use.

yes both are correct, they only differ by a different constant C

 

[clintmyster]

using subsitution the answer is wrong, but from what i know, it should work

integral (sinx.cosx)
=integral (u) let u=sinx, du=cosx dx
=1/2u^2 +c
=1/2sin^2x +c
=1/2[1/2(1-cos2x)] +c
=1/4-1/4cos2x +c

but then there's also,

integral (sinx.cosx)
=1/2 integral (sin2x)
=1/2.-1/2cos2x +c
=-1/4cos2x +c

they both seem correct.... i have a feeling the constants (c and 1/4) can be considered as one, but im not so sure about this. which one is correct, or which is the correct method to use.

this happened to me in a few integration questions when i get a constant value and a +C out the back. What you do is get rid of the constant and just write +C and therefore it will be the same answer. Essentially 1/4 + C is still a C so by convention you write +C

 

[Trebla]

Both are correct. Whether you write "+ c" or "1/4 + c" doesn't really matter, either is acceptable.

 

[starkskyy]

wouldn't the extra constant have effect on doing definite integrals?

 

[Uncle]

using subsitution the answer is wrong, but from what i know, it should work

integral (sinx.cosx)
=integral (u) let u=sinx, du=cosx dx
=1/2u^2 +c
=1/2sin^2x +c
=1/2[1/2(1-cos2x)] +c
=1/4-1/4cos2x +c

but then there's also,

integral (sinx.cosx)
=1/2 integral (sin2x)
=1/2.-1/2cos2x +c
=-1/4cos2x +c

they both seem correct.... i have a feeling the constants (c and 1/4) can be considered as one, but im not so sure about this. which one is correct, or which is the correct method to use.

Use your second method.
Substitution is for losers.

 

[study-freak]

wouldn't the extra constant have effect on doing definite integrals?

No. It's for the same reason as why c does not affect the value of definite integrals.
Constants get eliminated since [c]=+c-c=0.

 

[jchoi]

but the thing is, you can just write a new constant.

integral (sinx.cosx)
=integral (u) let u=sinx, du=cosx dx
=1/2u^2 +c
=1/2sin^2x +c
=1/2[1/2(1-cos2x)] +c
=1/4-1/4cos2x +c => -1/4cos2x + k (where k is a constant).

like that. You can use any other pronumerals as long as you state that it is a constant, and is not used in the question.