Question

Evaluate the integral. (Use C for the constant of integration.)

Answer 1

\[\int\limits 71x \cos^2x dx\]

Answer 2

Apply Cosine Half-Angle Formula,\[\large\rm \frac{71}{2}\int x(x+\cos2x)dx\]Distribute, then integration by parts on the second term.

Answer 3

\[\frac{ 71 }{ 2 }\int\limits x(1+\cos 2x) dx=\frac{ 71 }{ 2 }\int\limits x dx+\frac{ 71 }{ 2 }\int\limits x \cos 2x~dx+c\]

Answer 4

Please check my work, i made an error somewhere.
\[71 \int\limits xcos^2x dx\]
\[71 \int\limits x \left[ \frac{ 1 }{ 2 } (1+\cos2x)\right]\]
\[71 \int\limits x \left[ \frac{ 1 }{ 2 } +\frac{ \cos2x }{ 2 } \right]\]
\[71 \int\limits \left[ \frac{ x }{ 2 } + \frac{ x \cos2x }{ 2 }\right] \]
\[\frac{ 71 }{ 2 } \int\limits x+x \cos2x\]
Is that correct so far?

Answer 5

Yes, looks correct so far.
What I wrote was nonsense, sorry about that :)
\(\large\rm cos^2x\ne \frac12(x+cos2x)\)

Answer 6

As my answer I got\[\frac{ 71x^2 }{ 4 }+\frac{ 71 }{ 4 }xsin2x-\frac{ 71 }{ 8}\cos2x\]

Could someone please verify? I only have 1 more attempt at the answer.

Answer 7

Oh yea + the constant as well

Answer 8

the last one should be +71/8
bec \[\int\limits_{ }^{} \sin(2x) dx = -\frac{1}{2} \cos(2x)\]

Answer 9

Oh right, so..
\[\frac{ 71x^2 }{ 4 }+\frac{ 71 }{ 4 }xsin2x+\frac{ 71 }{ 8 }\cos2x\]

Answer 10

+Chocolates
and yes that's what i got

Answer 11

yes :)

Answer 12

Okay thank you guys!