by simple inspection, and knowing the rules of derivatives for sine and cosine ....
no need for substitution.
∫sin(x)cos^2(x)dx
= (-1/3) cos^3 x + c
Hi, could you help me with this one please?
Evaluate
∫sin(x)cos^2(x)dx by using the substitution u=sinx
Ive done this by using u=cosx no problem but can't get this one :(
Thank you so much in advance!
4 answers
Awkward choice, but let
u = sinx
du = cosx dx
∫sinx cos^2x dx
= ∫sinx cosx cosx dx
= ∫u√(1-u^2) du
Now let
v = 1-u^2
dv = -2u du
and your integral now becomes
∫-1/2 √u du
= (-1/2)(2/3)u^(3/2)
= -1/3 u^(3/2)
= -1/3 sin^3 x + C
Now you just have to show that the two answers are equivalent (differ only by using different values of C)
u = sinx
du = cosx dx
∫sinx cos^2x dx
= ∫sinx cosx cosx dx
= ∫u√(1-u^2) du
Now let
v = 1-u^2
dv = -2u du
and your integral now becomes
∫-1/2 √u du
= (-1/2)(2/3)u^(3/2)
= -1/3 u^(3/2)
= -1/3 sin^3 x + C
Now you just have to show that the two answers are equivalent (differ only by using different values of C)
Thank you so much! ^^
My typo made it wrong. I hope you caught it. Let's pick up at
Now let
v = 1-u^2
dv = -2u du
and your integral now becomes
∫-1/2 √v dv
= (-1/2)(2/3)v^(3/2)
= -1/3 v^(3/2)
-1/3 (1-u^2)^(3/2)
= -1/3 cos^3 x + C
That's better!
Now let
v = 1-u^2
dv = -2u du
and your integral now becomes
∫-1/2 √v dv
= (-1/2)(2/3)v^(3/2)
= -1/3 v^(3/2)
-1/3 (1-u^2)^(3/2)
= -1/3 cos^3 x + C
That's better!