Q: ∫sin(√x)dx
A: Start with the following u-substitution:
u = √x
So, differentiate to get:
du = 1/(2√x) dx
2du = 1/√x dx
Notice that we cannot just plug u and du into the original problem because the original only shows one “√x”… We would need a “√x” in the denominator as well in order to fit our “du” into the problem….
Hmmm. This requires some magic! Watch like so:
2du = 1/√x dx
√x*(2du) = dx
However, u = √x, so:
u*(2du) = dx
2u*du = dx
Now, follow my color-coded substitution: ∫sin(√x)dx = ∫sin(u)*(2u)du
Clean-house to get:
= 2∫ u sin(u) du
Now, integrate by parts [normally, in integration by parts we use the letters u and v… My problem already involves a u, so I am going to use w and v]
So, integration by parts tells us: ∫w dv = wv – ∫v dw
So, let w = u and dv = sin(u) du
w = u v = -cos(u)
dw = du dv = sin(u)du
Therefore: (don’t forget the 2 multiplier out in front):
2∫ u sin(u) du = 2[-u*cos(u) – ∫ -cos(u) du]
= 2[-u*cos(u) + ∫cos(u) du]
= 2[-u*cos(u) + sin(u) + C]
= -2ucos(u) + 2sin(u) + C
And, put u back… u = √x, so we get:
= -2√xcos(√x) + 2sin(√x) + C
Wow.
🙂 🙂 that makes perfect sense now!!
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