Question

can you help me with this one
∫0Π/2cos(x)exdx


i know i integrate twice

so u= e^x dv= cos(x)dx
du= e^x v= sin(x)

u= e^x dv= sin(x) dx
du= e^x v = -cos(x)

∫0Π/2cos(x)exdx=[(ex)(sin(x))−(ex)(−cos(x))]

Answer 1

\[\int\limits_{0}^{\Pi/2} \cos(x) e^x dx\]

Answer 2

this is how the problem is written

Answer 3

Have you done integration by parts?

Answer 4

I go the u = e^x dv= cos(x)dx
du= e^x dx v= sin(x)

then u =e^x du=e^x dv=sin(x) v= -cos(x)

Answer 5

so...

\[\int\limits_{0}^{\Pi/2} \cos(x) e^x dx= [e^x \sin(x)-e^x-(\cos(x))]\]

Answer 6

of integral 0 to \[\Pi/2\] + integral 0 to \[\Pi/2\] e^x(-cos(x)) dx

Answer 7

@mathmate

Answer 8

Let I=integral, and when you integrate by parts the first time, you get
\(I=cos(x)e^x+\int sin(x)e^xdx\)
when you integrate the second time,
\(I=cos(x)e^x+sin(x)e^x-\int cos(x)e^xdx\) or
\(I=cos(x)e^x+sin(x)e^x-I\)
\( 2I=cos(x)e^x+sin(x)e^x\)
hence
\(\huge I=\frac{e^x(cos(x)+sin(x))}{2}\)

I will leave it to you to evaluate the definite integral.

Answer 9

is it 2.418040 ?

Answer 10

@mathmate

Answer 11

I get a smaller number. Sure the limits are 0 to \(\pi/2\)?
Want to check you work? Don't forget that e^0=1.