Calculus (Integration)

Question

∫ln(x) sin^-1x dx

Answer 1

First integrate sin^-1 x

 

This can be done by integration by parts

 

u = sin^-1 x, du = dx/√(1 - x²), dv = dx, v = x 

 

∫ sin^-1 x dx = x sin^-1 x - ∫ [x/√(1-x²)] dx

 

Do a u-substitution: u = 1 - x², du = -2x dx

 

∫ sin^-1 x dx = x sin^-1 x + (1/2) ∫ u^-1/2 du

 

= x sin^-1 x + √u + C = x sin^-1 x + √(1 - x²) + C

 

 

Now for your integral: ∫ ln x sin^-1 x dx

 

Integrate by parts: u = ln x, du = dx, dv = sin^-1 x, v = x sin^-1 x + √(1 - x²)

 

∫ ln x sin^-1 x dx = [x sin^-1 x + √(1 - x²)]ln x - ∫ [x sin^-1 x + √(1 - x²)]/x dx

 

= [x sin^-1 x + √(1 - x²)]ln x - ∫ sin^-1 x dx - ∫ √(1 - x²) / x dx

 

= [x sin^-1 x + √(1 - x²)](ln x - 1) - ∫ √(1 - x²) / x dx

 

Do a trig-substitution on the rightmost integral: x = sin θ, dx = cos θ dθ

 

= [x sin^-1 x + √(1 - x²)](ln x - 1) - ∫ [√(1 - sin² θ) / sin θ] cos θ dθ

 

= [x sin^-1 x + √(1 - x²)](ln x - 1) - ∫ [(1 - sin² θ) / sin θ] dθ

 

= [x sin^-1 x + √(1 - x²)](ln x - 1) + ∫ sin θ dθ - ∫ csc θ dθ

 

= [x sin^-1 x + √(1 - x²)](ln x - 1) - cos θ - ∫ csc θ dθ

 

= [x sin^-1 x + √(1 - x²)](ln x - 1) - √(1 - x²) + ∫ -(csc² θ + csc θ cot θ)/(csc θ + cot θ) dθ

 

Do a u-substitution: u = csc θ + cot θ, du = -(csc² θ + csc θ cot θ) dθ

 

= [x sin^-1 x + √(1 - x²)](ln x - 1) - √(1 - x²) + ∫ 1/u du

 

= [x sin^-1 x + √(1 - x²)](ln x - 1) - √(1 - x²) + ln |u| + C

= [x sin^-1 x + √(1 - x²)](ln x - 1) - √(1 - x²) + ln |csc θ + cot θ| + C

 

= [x sin^-1 x + √(1 - x²)](ln x - 1) - √(1 - x²) + ln |[1 + √(1 - x²)]/x| + C

 

= (ln x - 1) x sin^-1 x + (ln x - 2)√(1 - x²) + ln [1 + √(1 - x²)] - ln |x| + C

Answer 2

We have here an integral of a product.  This is where we use integration by parts method.  But I warn you, this method can be a a real, real pain.  You will see why.  The formula for this method is

 

∫udv = uv - ∫vdu

 

We let u represent the factor you can derive.

We let dv represent the factor to integrate.

 

 

so:

 

u = sin^-1x                              dv = ln(x)dx

du = 1 / √(1 - x²) dx                   v = ?

 

 

We must use integration by parts to integrate ln(x)dx, before moving forward.

 

∫ln(x)dx

 

u = ln(x)                    dv = dx

du = x^-1dx                    v = x

 

 

Plug these values into the formula.

 

 

∫ln(x)dx = xln(x) - ∫x x^-1 dx

             = xln(x) - ∫dx

             = xln(x) - x

 

 

Going back to the original integral,

 

u = sin^-1x                              dv = ln(x)dx
du = 1 / √(1 - x²)dx                 v = xln(x) - x

 

 

Plug in these values into the formula

 

∫ln(x) sin^-1x dx = sin^-1x(xlnx - x) - ∫[(xlnx - x) / √(1 - x²)]dx

 

 

Next, reapply this method.  Find the integral of the rational part  [(xlnx - x) / √(1 - x²)]dx.

 

u = (1 - x²)^-1/2                                  dv = xln(x) - x  dx = x(lnx - 1)dx

du = -2x * (-1/2)(1 - x²)^1/2 dx             v = ?

du = x(1 - x²)^1/2 dx

 

 

You can find the integral of x(lnx - 1)dx using integration by parts.   Repeat the steps that were shown to you earlier until you obtain something you can factor or simplify.  Didn't I tell you this method can be a real, real pain? 

 

Good luck!

Answer 3

Zuhaib,

 

Yikes!  So, if you're doing an integral like this, I'm assuming that you're not half bad at calculus!  :)  So, I'm going to skip some steps in here, but happy to offer some additional hints if needed.

 

Step 1a:  Well, when you've got a mix of functions, you think integration by parts?  Conventional wisdom says that you make u be the ln part, and then dv is arcsin.  (Many of my students have learned "LIPET" as an acronym to figure out which function to try as u, so you'd try the ln first.  If there's no ln, then the next best bet is an inverse trig function, and so on.)

 

Step 1b:  To figure out what v is, you have to do the anti-derivative of arcsin.  The trick for that one is to use integration by parts again, with u is arcsin, and dv is just dx. 

 

Step 1c:  Within thaaaat one, you end up with the integral of (x / (√ (1 - x²)).  That's a u-substitution?

 

So, allll of that, if you can sift through it, gives you uv minus an integral, and that integral has a fraction that you can split into pieces that are subtracted.

 

Step 2a:  the first piece, if I did mine right, is just arcsin.  I say "just" because based on Step 1b, we actually know how to do that one again.

 

Step 2b:  the second piece is an integral with a square root on top.  Trig substitution, maybe?

 

Step 3:  the trig substitution, if I did it right, leads to an integral of cosecant.  That's an integral you'll find in some integral tables, or maybe you memorized it...or there's a trick to doing it.

 

That's a lot to take in, so I'll pause.  Any of that help?