i 4got how to do this prob....i know we do intgrate by parts but how do we start it?

Question

can u refresh me up....
....pi/2
I= (x^2 +5) sinx dx
....0
can u show me the steps ..i already know the answer (pi+3) but i need the work

Answer 1

First break the integral into 2 pieces:
We get
∫ (0,..π/2) x² sin x dx + ∫ (0,..π/2) 5 sin x dx
The second one is easy:
It is -5 cos x(0,..π/2) = 5.
Now let's go to work on the first one, ignoring the limits for now. Let's call this integral I
Let u = x² dv = sin x dx
du = 2x dx v = -cos x
Then I = -x² cos x + 2 ∫ x cos x dx
Now integrate by parts once more:
Let u = x dv = cos x dx
du = dx v = sin x
So 2 ∫ x cos x dx = 2x sin x - 2 ∫ sin x dx
= 2x sin x + 2cos x
Thus I = -x² cos x + 2x sin x + 2cos x
Now let's evaluate I from 0 to π/2.
The first term is 0 at both ends.
The second term gives π and the last term gives
-2.
So I get 3 + π, as desired.

Answer 2

∫(x^2 + 5)sinx dx

=>∫x^2 sinx dx + 5∫sinx dx

you have to integrate ∫x^2 sinx dx by parts

you know that d(uv) = vdu + udv

integrating both sides

∫d(uv) = ∫vdu + ∫udv

but integral of differential of uv = uv itself

so uv = ∫vdu + ∫udv

so ∫udv = uv - ∫vdu

in the given integral ∫x^2 sinx dx

let u = x^2 , then du = 2x dx

let dv = sinx , then v = -cosx

so ∫udv = ∫x^2sinx

uv = -x^2 cosx and

∫vdu = ∫-cosx 2x dx

so ∫x^2 sinx = -x^2 cosx - ∫2x (-cosx)dx

∫x^2 sinx = - x^2cosx + 2∫x cosx dx

again integrating ∫x cosx dx by parts by letting

u1 = x : du1 = dx

dv1 = cosx : v1 = sinx

∫x^2 sinx = -x^2 cosx + 2[u1v1 - ∫v1du1

∫x^2sinx = -x^2cosx + 2xsinx - 2∫sinx dx

∫x^2sinx = -x^2cosx + 2xsinx + 2cosx

so ∫(x^2+5)sinx dx = -x^2cosx + 2xsinx + 2cosx +5∫sinx dx

∫(x^2+5)sinx dx = -x^2cosx + 2xsinx + 2cosx - 5cosx

∫(x^2+5)sinx dx = -x^2cosx + 2xsinx - 3cosx

now evaluate definite integral between 0 to pi/2

[-(pi/2)^2 cos(pi/2) + 0] + 2[(pi/2)sin(pi/2) + 0] - 3(cos(pi/2- cos(0)

[0] + 2[(pi/2)*(1) - 3(0 - 1]

=>pi +3

Answer 3

I assume you are trying to integrate (x^2 + 5)sinx dx from 0 to pi/2?

Well first find the integral.

u = x^2 + 5 dv = sinx dx
du = 2x dx v = -cosx

So we do

uv - integral(vdu)

-(x^2 + 5)cosx + int(2x cosx)dx

Integrate by parts yet again

u = 2x dv = cosx dx
du = 2 dx v = sinx

So we get

-(x^2 + 5)cosx + 2x sinx - 2*int(sinx dx)

-(x^2 + 5)cosx + 2xsinx + 2cosx


Last but not least we want to integrate from o to pi/2

At pi/2 we get

-((pi/2)^2 + 5)cos(pi/2) + 2*pi/2*sin(pi/2) + 2*cos(pi/2)

cos(pi/2) = 0 and sin(pi/2) = 1

So we get just pi at x = pi/2

Now at x=0 we get

-(0^2 + 5)cos(0) + 2*0*sin(0) + 2*cos(0)

cos(0) = 1 and sin(0) = 0

-5 + 2 = -3

So subtracting the two numbers we get

pi - (-3) = pi+3

Your answer is right and thats how you get the right answer


Remember

int(udv) = uv - int(vdu)

Answer 4

I = Integral of (x^2 +5) sinx dx
Integration by parts
let u = x^2 +5 ---> du = 2xdx
let dv =sinx ---> v = -cosx
I = -cosx (x^2+5) - Integral of (-2xcosx )dx
let u =2x ---> du = 2dx
let dv = cosx ---> v = sinx
I = -cosx (x^2+5) +2xsinx +2cosx
Now evaluate this at 0 and pi/2 to get
I = pi -{-5 +2} = pi +3

Answer 5

ok so you have the intergral from 0-pi/2 of (x^2+5)sinx
I stands for intergral
= I (x^2sinx + 5sinx)dx
= I x^2sinx dx + I 5sinx dx
so you have to do integration by parts twice to the first part....
u=x^2 dv=sinx
du=2x V=-cosx
this gives you
=-x^2cosx - I (-2xcosx)dx + -5cosx
then integration by parts again
u=-2x dv=cosx
du=-2 V= sinx
= -x^2cosx -(-2xsinx - I (-2sinx)dx) - 5 cosx
= -x^2cosx +2xsinx + 2 cosx - 5cosx
= -x^2cosx +2xsinx - 3cosx
then just evaluate the integral from 0 to pi/2....i checked it and you get the right answer

Answer 6

u = (x^2 + 5) du = 2x dx
dv = sinx dx v = - cosx

int (u dv)
= uv - Int( v du)
= - cosx (x^2 + 5) - Int( - 2x cosx dx)
= - cosx (x^2 + 5) + 2 Int( x cosx dx)

Handling the second integral:
Int( x cosx dx)
u = x du = dx
dv = cosx dx v = sinx

= x sin x - Int (sin x dx)
= x sin x + cos x

And sub into the previous iteration :
= - cosx (x^2 + 5) + 2( x sin x + cos x)
= - x^2 cosx + 2x sinx - 3cos x


Now, apply the bounds of integration and away you go.