Calculus II Integration by parts

Question

Evaluate the integral: ∫x³lnxdx

Evaluate the integral: ∫e^xcosxdx

Answer 1

remember: D E T A I L

D stands for the differential dv...

E exponential

T trig

A algebraic/rational

i inverse trig

L log

we got log against algebraic, and the

algebraic is first, before the log in DETAIL,

so dv = x^3

IBP says:

U = ln x dv = x^3

dU = dx/x v = (1/4)x^4

U V - integral [ V du]

lnx* (1/4) x^4 - integral [ (1/4)x^4 [dx/x]]

(1/4) x^4 lnx - (1/4) integral [ x^3 dx ]

(1/4) x^4 ln x - (1/4) (1/4) x^4

[(1/4) x^4] ( ln x - 1/4) <--- this is the anti-derivative

check by differentiation:

(1/4)x^4 /x + (ln x - 1/4)x^3

(1/4)x^3 + 1n x *x^3 - (1/4)x^3

x^3 * ln x <--- which is the original function

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DETAIL...

exponential comes before trig

(The arguments are just 1x, so the chain

rule will not add any factors. The argument

x shall be omitted)

IBP #1:

U = cos dv = exp

dU = -sin V = exp

the integral becomes:

U V - integral [ V du ] =

cos * exp + integral [ exp * sin]

IBP #2:

U = sin dv = exp

dU = cos v = exp

cos*exp + [ sin * exp - integral [ cos * exp ]]

notice the last term being subtracted is the same

integral that we are trying to find....

so integral [ cos * exp] = cos*exp + [ sin * exp - integral [ cos * exp ]]

= cos * exp + sin * exp - integral [ cos * exp]

2 * integral [ cos * exp] = cos * exp + sin * exp

= exp ( cos + sin)

The answer is then (1/2) exp* ( cos + sin) <-- this is the anti-derivative

check by differentiation

(1/2) exp* ( cos - sin) + (1/2) exp*( cos + sin)

=exp * cos, which is the original function

the anti-derivative is correct

Answer 2

2) Evalute ∫e^xcosxdx

This one is fun and a bit more tricky!

Let's let:

u = cosx and

dv = e^x so

du = -sinxdx and

v = e^x .

We integrate like before, so:

∫e^xcosxdx = e^xcosx + ∫e^xsinxdx .

Now, the integral on the right is also a product, so we need to do integration by parts again. This time, let :

u = sinx so

du = cosxdx and

dv = e^x so

v= e^x.

So we can say that

∫e^xsinxdx = e^xsinx - ∫e^xcosxdx

and we can plug this back into our first integration by parts. We then get

∫e^xcosxdx = e^xcosx + ∫e^xsinxdx

∫e^xcosxdx = e^xcosx + e^xsinx - ∫e^xcosxdx

and we notice that the integral on the far right side of the equation is the same exact integral we started with! This means we can add it to the left side of the equation, giving us

2∫e^xcosxdx = e^xcosx+ e^xsinx

2∫e^xcosxdx = e^x(cosx +sinx)

∫e^xcosxdx = e^x(cosx +sinx) / 2 .

Hope this was helpful!