Some integration formulas

[hgaon001]

i'm having trouble with two of my homework problems.

1) Integration of 3/xsqrt(16x^2-7) dx

I know that the first step would be to take the 3 out but then i cant figure out which formula to use, i thought it would be the formula that is intregration 1/u[sqrt(u^2-a^2)] dx but it doesnt work..


2) Intregration 2x/sqrt(9-4x^4) dx

the 2x on the top of the equation has me confused with how to approach it.

 

[galactus]

Fix later. Made a booboo. Thanks for the catch, Tom.

EDIT:

Thanks to Tom catching my mistake, I have decided to tackle this one a different way, then trig sub.

Note that \(\displaystyle sin(sin^{-1}(x))=x\)

By the chain rule, \(\displaystyle sin^{-1}(x)'cos(sin^{-1}(x))=1\)

But \(\displaystyle cos(sin^{-1}(x))=\sqrt{1-x^{2}}\)

Therefore,\(\displaystyle \frac{d}{dx}[sin^{-1}(x)]=\frac{1}{\sqrt{1-x^{2}}}\)

So, from \(\displaystyle \int\frac{2x}{\sqrt{9-4x^{4}}}dx\)

\(\displaystyle =\int\frac{2x}{\sqrt{9-(2x^{2})^{2}}}dx\)

Make the sub \(\displaystyle u=2x^{2}, \;\ \frac{du}{2}=2xdx\)

\(\displaystyle \frac{1}{2}\int\frac{1}{\sqrt{9-u^{2}}}du\)

Now, rewrite by factoring:

\(\displaystyle \frac{1}{\sqrt{9-u^{2}}}=\frac{1}{3}\frac{1}{\sqrt{1-(\frac{u}{3})^{2}}}\)

Let \(\displaystyle w=\frac{u}{3}, \;\ dw=\frac{1}{3}du, \;\ 3dw=du\)

This gives:

\(\displaystyle \frac{1}{2}\int\frac{1}{\sqrt{1-w^{2}}}dw\)

\(\displaystyle =\frac{1}{2}sin^{-1}(w)\)

Resub:

\(\displaystyle \frac{1}{2}sin^{-1}(\frac{u}{3})\)

Resub:

\(\displaystyle \boxed{\frac{1}{2}sin^{-1}(\frac{2x^{2}}{3})}\)

 

[galactus]

1)\(\displaystyle \int\frac{3}{x\sqrt{16x^{2}-7}} dx\)

You can do trig sub with this one as well. Let \(\displaystyle x=\frac{\sqrt{7}}{4}sec(t), \;\ dx=\frac{\sqrt{7}}{4}sec(t)tan(t)dt\) and it whittles down to something super easy. Just be careful with the algebra. That's the booger.

 

[BigGlenntheHeavy]

Following galactus's lead (using trig substitution), we get:

\(\displaystyle \int\frac{3dx}{x\sqrt(16x^{2}-7)} \ = \ \frac{3}{\sqrt7}arcsec(\frac{4x}{\sqrt7} ) + C.\)

But how do we know if our answer is true? An easy check is to take the derivative of your answer.
If your answer is true, you will get the original integral back, viz,.:

\(\displaystyle D_x \ [\frac{3}{\sqrt7}arcsec(\frac{4x}{\sqrt7} ) + C] \ = \ \frac{3}{x\sqrt(16x^{2}-7)}.\)

Hence, for the sake of brevity and not to deny you the joy of exercising your algebra expertise, I have omitted the necessary steps to arrive at the following solution(s).

Addendum: I am assuming that x > 0.

 

[BigGlenntheHeavy]

Following galactus's lead (using trig substitution), we get:

\(\displaystyle \int\frac{3dx}{x\sqrt(16x^{2}-7)} \ = \ \frac{3}{\sqrt7}arcsec(\frac{4x}{\sqrt7} ) + C.\)

But how do we know if our answer is true? An easy check is to take the derivative of your answer.
If your answer is true, you will get the original integral back, viz,.:

\(\displaystyle D_x \ [\frac{3}{\sqrt7}arcsec(\frac{4x}{\sqrt7} ) + C] \ = \ \frac{3}{x\sqrt(16x^{2}-7)}.\)

Hence, for the sake of brevity and not to deny you the joy of exercising your algebra expertise, I have omitted the necessary steps to arrive at the following solution(s).

Addendum: I am assuming that x > 0.

\(\displaystyle Added \ note: \ In \ regards \ to \ the \ above \ problem, \ \int f(x)dx \ = \ F(x) \ implies \ F'(x) \ = \ f(x). \ Why?\)