Question

∫sinxcosx from pi/6 to pi/3

Answer 1

let u=sinx
du=cosx dx
\[\int\limits_{}^{}u du=\frac{u^{1+1}}{1+1}+C=\frac{u^2}{2}+C=\frac{(sinx)^2}{2}+C\]

Answer 2

plug in limits

Answer 3

i got that then i plugged in the values and got 0

Answer 4

actually i got sin2x/2

Answer 5

\[\frac{(\sin(\frac{\pi}{3}))^2}{2}-\frac{(\sin(\frac{\pi}{6}))^2}{2}\]
=\[\frac{(\frac{\sqrt{3}}{2})^2}{2}-\frac{(\frac{1}{2})^2}{2}\]
=\[\frac{1}{2}*\frac{3}{4}-\frac{1}{2}*\frac{1}{4}=\frac{3}{8}-\frac{1}{8}=\frac{2}{8}=\frac{1}{4}\]

Answer 6

can u explain to me how you integrated it

Answer 7

do you see my substitution above?

Answer 8

i let u=sinx
so du=cosxdx

Answer 9

you could also do :
remember sin(2x)=sin(x+x)=sinxcosx+sinxcosx=2sinxcosx
\[\int\limits_{}^{}\frac{1}{2}*\sin(2x)dx\]

Answer 10

well first the question asked to show that sin2x=2sinxcosx

i proved it using Sin(A+B).

Answer 11

yeah i got that answer

Answer 12

but once i put in the values it ends up being 0

Answer 13

let u=2x
du=2 dx
\[\frac{1}{2}\int\limits_{}^{}\frac{1}{2}*2\sin(2x)dx=\frac{1}{2}*\frac{1}{2}\int\limits_{}^{}sinudu=\frac{1}{4}*(-cosu)+C\]
=\[\frac{-1}{4}\cos(2x)+C\]

Answer 14

oh lol i forgot to integrate 1/2 (sin2x)

Answer 15

\[\frac{-1}{4}(\cos(2*\frac{\pi}{3})-\cos(\frac{2*\pi}{6}))=\frac{-1}{4}(\cos(\frac{2\pi}{3})-\cos(\frac{\pi}{3}))\]
\[=\frac{-1}{4}(\frac{-1}{2}-\frac{1}{2})=\frac{-1}{4}(-1)=\frac{1}{4}\]

Answer 16

which way you like better?

Answer 17

brb

Answer 18

i think the second one lol

Answer 19

whenever substitution is invovled i always manage to stuff it up