math problem

sir sickolot II SG

how do you integrate (X^2 +1)^0.5 dx

the question is.. prove that the above integration is equal to..

(1/2)X( X^2 + 1)^0.5 + 0.5ln (X + (X^2 +1)^0.5) + C

Do you do integration by parts?

Dammit.. hate university maths. I think it's simple actually... but not for arithmetically challenged individuals like me =(

Y_Shun SG

edit error! Laughing

speakupforyourrights SG

1+1=2

tanjun SG

deleted cos careless in first part

eagle SG

does your question allow you to differentiate the bottom to get the top instead?

Or die die need to do integration?

angelfairy SG

substitution, then simplify and get integrate sec^3x, then by parts and resub back

hisoka SG

differentiate the other side can or not?

tanjun SG

deleted again after discovering another careless mistake.

angelfairy SG

Originally posted by tanjun:

Thanks for the hint.
Let x = tan u
dx = sec^2 u du

Sub above eqn into original eqn.

Integrate (x^2 + 1)^1/2 = Integrate (tan^2 u + 1)^1/2 (sec^2 u)du
=Integrate (sec^2 u)^1/2 (sec^2 u)du
= Integrate (sec^3 u) du

By parts
Spilt sec^3 u into sec u and sec^2 u. U can choose your u to be either one.

Formula: Integrate u(dv/dx) dx = uv - Integrate v(du/dx) dx

I make U to be sec u.
dU/dx = sec u tan u

= Integrate (sec u)(sec^2 u) du = sec^3 u - integrate sec^2 u(sec u tan u) du
= sec^3 u - 1/3 sec^3 u + c
= 2/3 sec^3 u + c

Drawing a right-angled triangle,
sec u = (x + 1)^2
= 2/3(x + 1)^6 + c

Am i correct?

secutanu, not sec^3u, and tan^2usecu, not sec^2u(secutanu) if u using U = secu

eagle SG

Usually proving questions have a pre-question

I wonder if there is any other question parts before this because there's a very much simpler method if there is,

ie before this question, you were asked to prove

integrate (X^2 +1)^-0.5 dx = ln (X + (X^2 +1)^0.5)

tanjun SG

= Integrate (sec u)(sec^2 u) du = sec^3 u - integrate sec^2 u(sec u tan u) du
= sec^3 u - 1/3 sec^3 u + c
= 2/3 sec^3 u + c

U = sec u
dU/dx = sec u tan u

V = sec^2 u
dV/dx = 2sec^2 u tan u

UV - integrate V(dU/dx) = integrate U(dV/dx)
sec^3 u - integrate sec^2 u(sec u tan u)

as stated above.

However, ans is still vastly different from the one given above with a ln function.

Differentiate the above ans.
1/4(1 + x^2)^-1/2 + (2x + 2(x^2 + 1)^1/2 )^-1

Combining these two fractions will make things very complicated.

eagle SG

Try doing by parts right at the very start without substitution first Wink

at least it seems so to me
Calculations are much more simplified Very Happy

tanjun SG

(1 + x^2)^0.5

Very hard to do by parts leh.

Cos it is not a product of two simpler terms. It functions like a prime number with factors 1 and the number itself.

^tamago^ SG

since the qn gives u the two sides and prove, differentiate the result.

eagle SG

let dv/dx be 1
u be (1 + x^2)^0.5

u will be able to reduce the question to integrating only (1 + x^2)^-0.5, which when u do substituition of x=tan u, you only need to integrate sec u. This gives you ln (sec u + tan u), which when converted to x, will give you the answer

edit: careless typo

eagle SG

actually I tried both sides. It seems to me integration as slightly less tedious in this question

But I'm still wondering if TS has posted the full question.

unclebutcher SG

oh God this just reminds me that my math is gg

tanjun SG

Differentiate the above ans.
1/4(1 + x^2)^-1/2 + (2x + 2(x^2 + 1)^1/2 )^-1

as stated in above reply.

However, the main problem I think is the different denominator existed in both terms. Even if I combine, it is of no use. One denominator contains 1 + x^2 while the other has a 2x added on to 2(x^2 + 1). Addition of terms is hard to cancel out.

I am doubting the accuracy of the ans.

eagle SG

the answer is correct. I managed to get it

sir sickolot II SG

thanks all and eagle. it's the full question.

the solution i'm very sure does not involve trigo, although it would defintely be easier to solve using trigo.

reason being my maths course does not cover trigo integration

i think integration by parts is my best bet.... or by dy/dx the RHS......

haiz.