Question

Integral Help (Calc 1)

Answer 1

\[\int\limits_{}^{} \frac{ 8 }{ 4w^2 + 9 }\]

Answer 2

The answer I got is different than the given answer, and I want to know if it's the same.

Answer 3

So I did:
\[\frac{ 8 }{ 9 }\int\limits_{}^{}\frac{ 1 }{ \frac{ 4 }{ 9 }w^2+1 }\]

v = \[\frac{ 4 }{ 9 } w^2+1\]

and I got:
\[\frac{ 8 }{ 9 } \ln \frac{ 4 }{ 9 }w^2+1+C\]

Answer 4

With the 4/9w^2+1 in the brackets of ln. It's showing a trig substitution and I'm wondering if those are the same, or if I've screwed up and can't solve it this way.

Answer 5

that is not the correct (I should say not the ideal, really) u-sub.

if you do that sub, where is your dv?

Answer 6

I used "incorrect" because of the absence of expression for dv

Answer 7

have you learned about trig sub?

Answer 8

Ohhhh. I completely left that out.

Answer 9

Yeah that makes sense!

Answer 10

Don't do that sub

Answer 11

Even with an expression for dv, don't

Answer 12

Did you learn trigonometric substition?

Answer 13

Yes I've learned them, sort of. Not really great with them.

Answer 14

Examples:

\(\large\color{black}{\displaystyle\int\limits_{~}^{~}\frac{8}{4x^2+9}~dx}\)

\(\large\color{black}{\displaystyle\frac{8}{4}\int\limits_{~}^{~}\frac{1}{x^2+\frac{9}{4}}~dx}\)

\(\large\color{blue}{\displaystyle x=\frac{3}{2}\tan(\theta)}\)

Answer 15

see why that substitution in blue is the sub you need?

Answer 16

Not really.

Answer 17

That's ok, I will do an example.

\(\large\color{black}{\displaystyle\int\limits_{~}^{~}\frac{1}{x^2+a^2}{~~}dx}\) (where a is a real number constant)

This is a general form of a trigonometric substituion where the rule is:
\(x=a\tan(\theta)\), and this is what we will do.

\(\large\color{black}{\displaystyle x=a\tan\theta }\)
\(\large\color{black}{\displaystyle dx=a\sec^2\theta{~}d\theta}\)

then substitute for x and for dx accordingly,

\(\large\color{black}{\displaystyle\int\limits_{~}^{~}\frac{1}{\left(a\tan\theta\right)^2+a^2}{~~}\left(a\sec^2\theta{~}d\theta\right)}\)

\(\large\color{black}{\displaystyle\int\limits_{~}^{~}\frac{a\sec^2\theta}{a^2\tan^2\theta+a^2}{~~}d\theta}\)

\(\large\color{black}{\displaystyle\int\limits_{~}^{~}\frac{a\sec^2\theta}{a^2(\tan^2\theta+1)}{~~}d\theta}\)

\(\large\color{black}{\displaystyle\int\limits_{~}^{~}\frac{a\sec^2\theta}{a^2(\sec^2\theta)}{~~}d\theta=\int\limits_{~}^{~}\frac{a}{a^2}{~~}d\theta}\)

and then integrate and solve for theta

Answer 18

In our case, the \(\color{magenta}{a\tan(\theta)}\) peace is \(\color{blue}{\displaystyle \frac{3}{2}\tan(\theta)}\)

Answer 19

(for the example, recall that a is a constant and so is a²)

Answer 20

Thank you! I get it now.

Answer 21

Ok, so lets get back to our problem

\(\large\color{black}{\displaystyle\int\limits_{~}^{~}\frac{8}{4x^2+9}~dx}\)

\(\large\color{black}{\displaystyle\frac{8}{4}\int\limits_{~}^{~}\frac{1}{x^2+\frac{9}{4}}~dx}\)

\(\large\color{blue}{\displaystyle x=\frac{3}{2}\tan(\theta)}\)

Answer 22

what will your \(dx\) be equivalent to?