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# Evaluate the integral shown below using the given method

∫sin(x)cos(x) dx

1. substitution where u=sin(x)
2. substitution where u=cos(x)
3 .integration by parts
4. using the identity sin(2x)=2sin(x)cos(x)

1) u = sin x
du = cos x dx

Substitute

Int( u du) = (u^2)/2 + c = .5(sin x)^2+ c

2) u = cos x

du = - sin x dx

Int(-u du) = (-u^2)/2 + c = -.5(cos x)^2 + c

3) u = sin x

du = cos x dx

dv = cos x dx

v = sin x

Int( u*dv) = uv - Int(v*du)

Int(sin x * cos x) = (sin x)^2 - Int(sin x * cos x)

2*Int(sin x * cos x) = (sin x)^2

Int(sin x * cos x) = .5(sin x)^2 + c

4) Int(sin x * cos x) = Int(.5sin 2x) = -.25cos 2x + c

I only see one comment Surendra.
To say , the following expressions give us the same result

0.5(sinx)^2
-0.5(cosx)^2.
-0.25(cos2x)

For that constant of integrations should be different in all the three cases.

Thanks so much for your help