I am needing help figuring out which rules of integration to use.

Question

*integral* of (x^2)(sin(3x))

*integral* of (x^2) / (sqrt ( 9-x^2))

Answer 1

The first integral can be done using integration by parts. let u= x² and dv = sin3xdx so that du =2xdx and

v=-cos3x/3. The integral becomes -x²cos3x/3 +2/3 ∫xcos3xdx. Use integration by parts again and let u=x and dv=cos3xdx du=dx and v=sin3x/3 so the second integral ∫xcos3xdx = xsin3x/3 -1/3∫sin3xdx = xsin3x/3 +1/9

cos3x. Putting all the pieces together the original integral ∫x²sin3xdx = -x²cos3x/3 +2xsin3x/9 +2cos3x/27.

The second integral can be integrated using a trig substitution. Let x=3sinΘ so dx =3cosΘdΘ. The integral now becomes ∫9sin²Θ3cosΘdΘ/ 3cosΘ. This gives ∫9sin²ΘdΘ. Integrating this using sin²Θ =(1-cos2Θ)/2 which becomes 9(Θ/2 -sin2Θ/4). So since x=3sinΘ then Θ=sin^-1(x/3) and sin2Θ =2sinΘcosΘ then 2sinΘcosΘ can be written as 2x√(9-x²)/9 where the sinΘ=x/3 and the cosΘ= √(9-x²)/3 using a triangular relationship for Θ and the vertical leg =x and the horizontal leg = √(9-x²). Piecing this together gives 9sin^-1(x/3)/2 -x√(9-x²)/2 +C.

Answer 2

The second integral can be done as follows:

∫ dx x² / [9 - x²]^1/2

Perform Integration by Parts

u = x² dv = dx / [9 - x²]^1/2

du = 2x dx v = ∫ dx / [9 - x²]^1/2 = sin^-1(x/3) since in general ∫ dx / [a² - x²]^1/2 = sin^-1(x/|a|)

∫ u dv = uv - ∫ v du

So,

∫ dx x² / [9 - x²]^1/2 = x² sin^-1(x/3) - 2 ∫x sin^-1(x/3) dx (****)

We can attack this integral by Integration by Parts.

∫x sin^-1(x/3) dx = ?

u = x v = ∫ sin^-1(x/3) dx

du = dx v = x sin^-1(x/3) + 3[ 1 - (x/3)²]^1/2 where in general

∫ sin^-1(ax) dx = xsin^-1(ax) + (1/a)[ 1 - a²x²]^1/2

So

∫ u dv = uv - ∫ v du

∫x sin^-1(x/3) dx = x{x sin^-1(x/3) + 3[ 1 - (x/3)²]^1/2} - ∫x sin^-1(x/3) dx - 3∫ [ 1 - (x/3)²]^1/2 dx

We can combine the first integral on the right with the integral on the left to get

2 ∫x sin^-1(x/3) dx = x{x sin^-1(x/3) + 3[ 1 - (x/3)²]^1/2} - 3∫ [ 1 - (x/3)²]^1/2 dx

Let's clean up the integral on the right-hand side.

2 ∫x sin^-1(x/3) dx = x{x sin^-1(x/3) + 3[ 9/9 - (x²/9]^1/2} - 3∫ [ 9/9 - x²/9]^1/2 dx

2 ∫x sin^-1(x/3) dx = x{x sin^-1(x/3) + (3/3)[ 9 - x²]^1/2} - (3/3)∫ [ 9 - x²]^1/2 dx

2 ∫x sin^-1(x/3) dx = x² sin^-1(x/3) + x [ 9 - x²]^1/2 - ∫ [ 9 - x²]^1/2 dx

In general, ∫ [ a² - x²]^1/2 dx = (1/2){x [ a² - x²]^1/2 + a² sin^-1(x/|a|)}

So,

2 ∫x sin^-1(x/3) dx = x² sin^-1(x/3) + x [ 9 - x²]^1/2 - (1/2) {x [ 9 - x²]^1/2 + 9 sin^-1(x/3)}

2 ∫x sin^-1(x/3) dx = x² sin^-1(x/3) + x [ 9 - x²]^1/2 - (1/2) x [ 9 - x²]^1/2 + (9/2) sin^-1(x/3)

∫x sin^-1(x/3) dx = (1/2) x² sin^-1(x/3) + (1/2) x [ 9 - x²]^1/2 - (1/4) x [ 9 - x²]^1/2 + (9/4) sin^-1(x/3)

∫x sin^-1(x/3) dx = (1/2) x² sin^-1(x/3) + (1/4) x [ 9 - x²]^1/2 + (9/4) sin^-1(x/3)

Go back to Eq. (****) above.

∫ dx x² / [9 - x²]^1/2 = x² sin^-1(x/3) - 2 ∫x sin^-1(x/3) dx

Now, substitute our expression for ∫x sin^-1(x/3) dx.

∫ dx x² / [9 - x²]^1/2 = x² sin^-1(x/3) - 2{(1/2) x² sin^-1(x/3) + (1/4) x [ 9 - x²]^1/2 + (9/4) sin^-1(x/3)}

∫ dx x² / [9 - x²]^1/2 = x² sin^-1(x/3) - x² sin^-1(x/3) - (1/2) x [ 9 - x²]^1/2 + (9/2) sin^-1(x/3) + C where C is an arbitrary constant, which I have been neglecting.

The first two terms on the right-hand side cancel.

∫ dx x² / [9 - x²]^1/2 = - (1/2) x [ 9 - x²]^1/2 + (9/2) sin^-1(x/3) + C

Answer 3

(1) ∫ x² sin(3x)dx

(2) ∫ ^{x^2} / _√(9-x^2) dx

The tricky thing about integration techniques is that it is often a guessing game. You observe the problem, making a guess about which technique to use, and see if it works. The hardest part is not getting frustrated when one technique doesn't seem to work.

Looking at (1), it appears that Integration by Parts will work:

∫udv = uv - ∫vdu

This is a good candidate because if u = x², then du = 2x dx. Using integration by parts as second time with u = x will make du = dx and so the x² disappears. Let's give it a try: [WARNING: There may be typos.]

∫ x² sin(3x)dx; u = x², du = 2x dx, dv = sin(3x)dx, v = -¹/₃ cos(3x)

∫ x² sin(3x)dx = ( x² ) ( -¹/₃ cos(3x) ) - ∫ ( -¹/₃ cos(3x) ) (2x dx)

= -¹/₃ x² cos(3x) + ²/₃ ∫ xcos(3x)dx (use Integration by Parts again)

∫ xcos(3x)dx; u = x, du = dx, dv = cos(3x)dx, v = ¹/₃ sin(3x)

∫ xcos(3x)dx = ¹/₃ x sin(3x) - ¹/₃ ∫ sin(3x)dx

= ¹/₃ x sin(3x) + ¹/₉ cos(3x) + C

We must now put this all together:

∫ x² sin(3x)dx = -¹/₃ x² cos(3x) + ²/₃ ∫ xcos(3x)dx

= -¹/₃ x² cos(3x) + ²/₃ [ ¹/₃ x sin(3x) + ¹/₉ cos(3x) + C ]

= -¹/₃ x² cos(3x) + ²/₉ x sin(3x) + ²/₂₇ cos(3x) + C

For problem (2), I think trigonometric substitution will work. I notice this because of √(9 - x²). If we let

x = 3sin θ, dx = 3cos θ dθ

then √(9 - x²) = √(9 - 9cos² θ) = 3√(1 - cos² θ) = 3sin θ.

It follows that

∫ ^{x^2} / _√(9-x^2) dx = ∫ ^{(9 sin^2 θ)} / _{(3 sin θ)} (3cos θ dθ)

= 9 ∫ sin θ cos θ dθ

You may be tempted to use integration by parts now. That may work. But let's just try substitution:

∫ sin θ cos θ dθ; u = sin θ, du = cos θ dθ

∫ sin θ cos θ dθ = ∫ u du

= ¹/₂ u² + C

= ¹/₂ sin² θ + C

Therefore, putting this all together,

∫ ^{x^2} / _√(9-x^2) dx = 9 ∫ sin θ cos θ dθ

= 9 [ ¹/₂ sin² θ + C ]

= ⁹/₂ sin² θ + C

= ⁹/₂ (^x/₃)² + C (since x = 3sin θ)

= ¹/₂ x² + C

Answer 4

The 2nd one is easy: just set x = 3 sin θ and then you will need to remember that sin²θ = (1 - cos 2θ)/2.

The you will have some which is relatively easy to integrate..

I am not sure about the 1st one, but I think you could try two approaches, one of which may get the answer for you. First you could try integration by parts...and apply it twice I think. Or you can you use the identity

sin 3x = 3 sin x - 4 sin²x. I will work on this one a bit more and get back to you if I figure it out.