Question

Integration by parts

Evaluate the definite Integral

Answer 1

\[\int\limits_{0}^{\Pi/2}\cos(x)e^3dx\]

Answer 2

What did you try ?

Answer 3

\[\large{\int\limits_{0}^{\Pi/2}\cos(x)e^{3\color{red}x}dx}\]

Is this the problem or the one you wrote earlier ?

Answer 4

is u = cos(x) dv=e^x

du= -sin(x) v=e^x

Answer 5

yes this is the problem

Answer 6

Why not choose dv = e^(3x) ?

Answer 7

=> v = \(\large{\cfrac{e^{3x}}{3}}\)

Answer 8

im sorry the problem is e^x not 3 my mistake

Answer 9

Okay

Answer 10

Then you are correct

Answer 11

So, after this step, what did you try ?

Answer 12

\[\large{\int_{a}^{b} u \ dv = [uv]_{a}^{b} - \int_{a}^{b} v \ du}\]

Answer 13

This is the formula.

Answer 14

I have to do integration by parts twice right?

Answer 15

May be

Answer 16

im pretty sure I have to just not sure of my u and du

Answer 17

Well they are perfectly correct according to me

Answer 18

ok

Answer 19

not sure on how to solve the rest

Answer 20

Okay lets see

Answer 21

We can easily solve the uv part.

Answer 22

The only problem would be the v du part. So, first lets evaluate it.

Answer 23

\[\large{I = \int_{0}^{\cfrac{\pi}{2}} e^x \ (-\sin x) \ dx}\]

This is the v du part

Answer 24

Okay ?

Answer 25

And let:

\[\large{I' = \int_{0}^{\cfrac{\pi}{2}} \cos x \ e^x \ dx}\]

Answer 26

Now, we simplify I.

Answer 27

ok i got that part

Answer 28

\large{I = \int_{0}^{\cfrac{\pi}{2}} e^x \ (-\sin x) \ dx}

\[\large{I = -\int_{0}^{\cfrac{\pi}{2}} e^x \sin x \ dx}\]

Answer 29

*
\[\large{I = \int_{0}^{\cfrac{\pi}{2}} e^x \ (-\sin x) \ dx}\]

Answer 30

Now, we may again apply integration by parts.

Answer 31

Here, u = e^x ; v = sin x

=> du = e^x dx; dv = cos x dx

Answer 32

So,

\[\large{I = -([e^x \sin x]_{0}^{\cfrac{\pi}{2}} - \int_{0}^{\cfrac{\pi}{2}} \sin x \ e^x dx)}\]

Answer 33

\[\large{\implies I = -[e^x \sin x]_{0}^{\cfrac{\pi}{2}} + \int_{0}^{\cfrac{\pi}{2}} \sin x \ e^x \ dx}\]

\[\large{\implies I = -[e^x \sin x]_{0}^{\cfrac{\pi}{2}} + (-I)}\]

\[\large{\implies 2I = -[e^x \sin x]_{0}^{\cfrac{\pi}{2}}}\]

Answer 34

\[\large{\implies I = -\cfrac{[e^x \sin x]_{0}^{\cfrac{\pi}{2}}}{2}}\]

Answer 35

Thus,

\[\large{I' = [e^x \cos x]_{0}^{\cfrac{\pi}{2}} - I}\]

\[\large{\implies I' = [e^x \cos x]_{0}^{\cfrac{\pi}{2}} + -(-\cfrac{[e^x \sin x]_{0}^{\cfrac{\pi}{2}}}{2})}\]

\[\large{\implies I' = [e^x \cos x]_{0}^{\cfrac{\pi}{2}} + \cfrac{[e^x \sin x]_{0}^{\cfrac{\pi}{2}}}{2}}\]

Now you can simplify this and get the answer

Answer 36

did u get 2.47 for the final answer I can't get it??