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[lisasayzhi]

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[lisasayzhi]

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[Unco]

Re: By Parts

1.
\(\displaystyle \mbox{ \fbox{\int xsin(2x) dx \\ \\

u = x du = dx \\ \\

dv = sin(2x) dx}}\)

\(\displaystyle \mbox{ \Rightarrow \fbox{s = 2x \Rightarrow \frac{1}{2} ds = dx \\ \\

\frac{1}{2} \int sin(s) ds = \int -\frac{1}{2}cos(s) = -\frac{1}{2}cos(2x)=v}}\)

\(\displaystyle \mbox{ \fbox{\int xsin(2x) dx = -\frac{1}{2}xcos(2x) - \int -\frac{1}{2}cos(2x) dx \\ \\

= -\frac{1}{2}xcos(2x) + \frac{1}{2} \int cos(2x) dx}}\)

\(\displaystyle \mbox{ \Rightarrow \fbox{{\int cos(2x) dx \\ \\

u = 2x \Rightarrow \frac{1}{2}du = dx \\ \\

\frac{1}{2} \int cos(u) du = \frac{1}{2} sin(u) = \frac{1}{2} sin(2x) + C}}\)

\(\displaystyle \mbox{ \fbox{\int xsin(2x) dx = -\frac{1}{2}xcos(2x) + \frac{1}{2}(\frac{1}{2} sin(2x)) \\ \\

= -\frac{1}{2}xcos(2x) + \frac{1}{4}(sin(2x)) + C}}\)

Correct.

2. Correct but can be simplified.

3. You've given the antiderivative of (3x + 5), in a weird way, which I'm too sleepy to decipher.

4. Nope; you might write tan(x)/cos(x) as sin(x)/cos^2(x) and u-sub; be careful with your signs.

5. Yes.

You should always differentiate to check your integral.