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Evaluate the integral shown below using the given method

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1 Answer

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

Comments

Hello Muhammad,
 
Kindly look for my comments and respond please.
 
 
I only see one comment Surendra.
What I said Muhammad,
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.