Integral of (cosx+sinx)^2dx

Sep 2017
5
0
Mexico
I did this but not sure if it is correct:
(cosx+sinx)^2dx=
(cos^2x+2cosxsinx+sin^2x)=
∫(2cosxsinx)=
2∫cosxsinx=
2sinx-cosx+C <--------End
 
Last edited:

skeeter

MHF Helper
Jun 2008
16,217
6,765
North Texas
$(\cos{x}+\sin{x})^2 = \cos^2{x}+2\sin{x}\cos{x}+\sin^2{x} = 1+2\sin{x}\cos{x} = 1 + \sin(2x)$



$\displaystyle \int 1 + 2\sin{x}\cos{x} \, dx = x + \sin^2{x} + C$

or ...

$\displaystyle \int 1 + 2\sin{x}\cos{x} \, dx = x - \cos^2{x} + C$

or ...

$\displaystyle \int 1 + \sin(2x) \, dx = x - \dfrac{\cos(2x)}{2} + C$


Now a question for you ...

Why are all three antiderivative forms above valid?
 
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Sep 2017
5
0
Mexico
I think i got the answer:
∫sin(2x)=
(using sustitution where u=2x) ∫sin(u)du/2=
1/2∫sin(u)du=
1/2-cos(u)=
1/2-cos(2x)+C <---End
Thread is close then!
 
Sep 2017
5
0
Mexico
Failed in 2cosxsinx because it is sin2x, thanks man!
 

skeeter

MHF Helper
Jun 2008
16,217
6,765
North Texas
I think i got the answer:
∫sin(2x)=
(using sustitution where u=2x) ∫sin(u)du/2=
1/2∫sin(u)du=
1/2-cos(u)=
1/2-cos(2x)+C <---End
Thread is close then!
No ...

$\displaystyle \int \sin(2x) \, dx = -\dfrac{1}{2}\cos(2x) + C$

not $\dfrac{1}{2}-\cos(2x) + C$
 
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