Question

Evaluate the integral sin x {f′(cos x) − 2} dx from 0 to pi/2
when f(0)=1 and f(1)=5

Answer 1

\[\int_{0}^{\frac{\pi}{2}} Sin(x)[f\prime(Cos(x))-2] \;\;dx =f(Cos(x))+2Cos[x] \|_{0}^{\frac{\pi}{2}} \]

So,
\[f(Cos(\frac{\pi}{2}))+2Cos[\frac{\pi}{2}]-(f(Cos(0))+2Cos[0])\]

then

\[f(0)+2(0)-(f(1)+2)\]

Eventually you end up with this
\[f(0)-f(1)-2= 1-5-2=-6\]

Answer 2

Thanks for the help, but it's not the right answer. You did help get me on the right track. :)

Answer 3

is the answer 2? thats what im getting using a u=cos x substitution.

Answer 4

Answer is 2, sorry I forgot the negative

Answer 5

yes. Can you show me how to do it using u sub?

Answer 6

posting in a sec, scanner is scanning.

Answer 7

when you do the u sub, make sure to remember to switch the limits of integration. thats what always gets me >.< lol

Answer 8

THANKS! :)

Answer 9

\[f((g))'=f'(g(x))g'(x)\]

so \[\int f((g))'dx= \int f'(g(x))g'(x) dx\]

\[g(x)= cos(x)\]
so
\[g(x)= -sin(x)\]

I forgot the negative but everything else is fine. ^^

Answer 10

Correction \[g(x)'=-sin(x)dx\]

Answer 11

∫ π 2 0 Sin(x)[f′(Cos(x))−2]dx=f(Cos(x))+2Cos[x]∥ π 2 0

So,
f(Cos(π 2 ))+2Cos[π 2 ]−(f(Cos(0))+2Cos[0])

then

f(0)+2(0)−(f(1)+2)

Eventually you end up with this
f(0)−f(1)−2=1−5−2=−6

Answer 12

luckey can u help mE??

Answer 13

yeah., i'll try to
now m leaving, have some work u mail ur prblm at lokeshsonee@gmail.com