How Do I Simplify These Tricky Integral Calculations?

[rjs123]

1. ∫(√4x² + 9)/x4

using u = atanθ

x = 3/2tanθ

dx = 3/2sec²θ

I plugged x in and am stuck on simplifying



2. ∫x²/√25 - x²)

u = asinθ

x = 5sinθ

dx = 5cosθ

Again, plugged x into original equation cannot simplify



3. ∫2x^3 - 4x - 8/(x² - x)(x² + 4)

got stuck midway through this problem...

2x^3 - 4x - 8 = A(x-1)(x² + 4) + B(x)(x² + 4) + Cx + D(x)(x-1)

let x = 0 A = 2
let x = 1 B = -2

 

[eumyang]

You need to know your trig identities. For example, in the 2nd one:
$$\int \frac{x^2}{\sqrt{25-x^2}} dx$$
If you make the substitutions you get
$$\int \frac{25\sin^2 \theta \cdot 5\cos \theta}{\sqrt{25-25\sin^2 \theta}} d\theta$$

If I factor out the 25 in the denominator:
$$\int \frac{25\sin^2 \theta \cdot 5\cos \theta}{\sqrt{25(1-\sin^2 \theta)}} d\theta$$
... do you recognize anything?

3. ∫2x^3 - 4x - 8/(x² - x)(x² + 4) dx

Don't forget the dx.

2x^3 - 4x - 8 = A(x-1)(x² + 4) + B(x)(x² + 4) + Cx + D(x)(x-1)

let x = 0 A = 2
let x = 1 B = -2

Are you trying to use partial fractions? Expand the right side by multiplying out the polynomials, and after grouping like terms, equate the coefficients. For example, you'll see an x³ term on the right side. Set its coefficient equal to 2 (the coefficient on the left side).

 

[Mark44]

1. ∫(√4x² + 9)/x4

using u = atanθ

x = 3/2tanθ

dx = 3/2sec²θ

I plugged x in and am stuck on simplifying

Show us what you got. Also, you have omitted dx in your original integral, and dθ in your equation for dx. If you don't substitute for the differentials in trig substitutions, you won't be working with the right expressions.

2. ∫x²/√25 - x²)

u = asinθ

x = 5sinθ

dx = 5cosθ

Again, plugged x into original equation cannot simplify



3. ∫2x^3 - 4x - 8/(x² - x)(x² + 4)

got stuck midway through this problem...

You aren't using enough parentheses. As you have written your integrand, it would be interpreted as 2x³ - 4x - (
$$2x^3 - 4x - \frac{8}{(x^2 - x)(x^2 + 4)}$$

From the work you show below, that's not what you intended.

2x^3 - 4x - 8 = A(x-1)(x² + 4) + B(x)(x² + 4) + Cx + D(x)(x-1)

Again, you need more parentheses. The last expression on the right side above should be (Cx + D)(x(x - 1))

let x = 0 A = 2
let x = 1 B = -2

You can use these values to rewrite you equation above.
2x³ - 4x - 8 = 2(x - 1)(x² + 4) - 2x(x² + 4) + (Cx + D)(x(x - 1))

Pick a couple more values of x to solve for C and D.

 

[dextercioby]

I say nothing about your notation, but the first integral is it supposed to be

[tex]I= \int \frac{\sqrt{4x^{2}+9}}{x^k} \, dx [/itex]<br /> <br /> where k=4 ? If k=1, the integration is almost immediate. For k not 1 it gets messy as Mathematica shows (attachment)[/tex]

 

[rjs123]

You need to know your trig identities. For example, in the 2nd one:
$$\int \frac{x^2}{\sqrt{25-x^2}} dx$$
If you make the substitutions you get
$$\int \frac{25\sin^2 \theta \cdot 5\cos \theta}{\sqrt{25-25\sin^2 \theta}} d\theta$$

If I factor out the 25 in the denominator:
$$\int \frac{25\sin^2 \theta \cdot 5\cos \theta}{\sqrt{25(1-\sin^2 \theta)}} d\theta$$
... do you recognize anything?

thanks for the help, i reduced the denominator before plugging in x...led to my mistake

here is where i am stuck at now:

$$\int\frac {25\sin^2\theta}{5cos\theta} d\theta$$

im not sure which way i should go here.

 

[SammyS]

1. ∫(√4x² + 9)/x4

using u = atanθ

x = 3/2tanθ

dx = 3/2sec²θ

I plugged x in and am stuck on simplifying

Assuming you mean:

$$\int \frac{\sqrt{4x^{2}+9}}{x^4} \, dx$$

Using the substitution you suggested:
If x = (3/2)tanθ , then dx = (3/2)sec²θ dθ. Also 4x² = 9tan²θ.

So, your integral becomes:

$$\int\frac{\sqrt{9\tan^2(\theta)+9}}{\displaystyle ({81}/{16})\tan^4(\theta)}\,\left(\frac{3}{2}\right)\sec^2(\theta)\,d\theta$$

$$=\frac{8}{9}\int\frac{|\sec^3(\theta)|}{\tan^4(\theta)}\,d\theta$$

Write the integrand in terms of sin & cos, then you should be fine. Don't forget to change back from θ to x.

 

[SammyS]

thanks for the help, i reduced the denominator before plugging in x...led to my mistake

here is where i am stuck at now:

$$\int\frac {25\sin^2\theta}{5\cos\theta} d\theta$$

im not sure which way i should go here.

Technically, that's:

$$\int\frac {25\sin^2\theta}{5|\cos\theta|} d\theta$$

Use sin²θ = 1‒cos²θ & divide each term by |cosθ|

 

[Mark44]

thanks for the help, i reduced the denominator before plugging in x...led to my mistake

here is where i am stuck at now:

$$\int\frac {25\sin^2\theta}{5cos\theta} d\theta$$

im not sure which way i should go here.

Replace sin^2(theta) by 1 - cos^2(theta), and then divide both terms by cos(theta).

 

[SammyS]

$$\frac{1-\cos^2(\theta)}{\cos(\theta)}=\frac{1}{\cos(\theta)}-\frac{\cos^2(\theta)}{\cos(\theta)}=\ \ ?$$

It's true that $$\int\sec(\theta)\,d\theta$$ is a bit tricky.

 

[dextercioby]

$$\int \sec\theta \, d\theta = \int \frac{d\sin\theta}{1-\sin^2 \theta} =...$$

The last can be done either using the substitution $\sin\theta = \tanh u$ or by partial fractions.