**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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