FP2 - Integrating Inverse Trigonometric Functions

Integrals of the form $\frac{1}{(a^2 - x^2)^{1/2}}$ require the substituion $x = a cos u$ and you can quote the result that integrating $\frac{1}{(a^2 - x^2)^{1/2}}$ gives $arcsin(\frac{x}{a})$

Zeidj if you have the time please would you give the full solutions to a couple of questions.. that would benefit me the most

Please could you do 4 e) and f) and 5 c) ?

Thanks, however the book suggests to use the substituion x=cu, is your method any more difficult that using that?

I read the example where they integrated $\frac{1}{a^2 + b^2x^2}$ using the subsitution bx = au.

I used this for question 4a) and get the correct answer. It also works for $\frac{1}{\sqrt{a^2 - b^2x^2}}$

Basically using that substition turns your integral into $\frac{number}{number} \frac{1}{1 + u^2}$. Personally, I found this method much easier. I am absolutley terrible with trigonometry (I sat my C3 and C4 exams a year ago and have never looked book since I got an A on both). I just don't remember them! I do sort-of understand your working but I look at the trig and shut down

Thanks for the help, your example did show me the method of to use the substituion and after reading I was able to answer it, in my own way atleast

I always use sin, but I guess either one works.

$\displaystyle \int \frac{1}{2 + 3x^2}$

I know that I want it to be "1 plus something squared", so I take out whatever is necessary to make the constant term on the bottom = 1.
In this case, I will take out a factor of 2 from the demoninator, leaving me with:

$\displaystyle \frac{1}{2} \int \frac{1}{1 + \frac{3x^2}{2}}$

Now I have "1 plus something squared", I can use my trig identities.

I know that $\sin^2 \theta + \cos^2\theta \equiv 1$, so dividing both sides by $\cos \theta$ gives:

$1 + \tan^2\theta \equiv \sec^2\theta$

The left hand side looks a lot like my denominator, except instead of $\frac{3x^2}{2}$, it has $\tan^2\theta$

So, I'll make the substitution $\tan u = \frac{\sqrt{3}x}{\sqrt2}$

Now, I need to change my $dx$ into $du$, so I need to find what $dx$ and $du$ are in terms of each other.

I can do this by differentiating the substitution I made, $\tan u = \frac{\sqrt{3}x}{\sqrt2}$

Differentiating both sides [implicitly] gives:

$\sec^2 u \frac{du}{dx} = \frac{\sqrt{3}}{\sqrt{2}}$

Rearranging to make dx the subject:

$\frac{\sqrt{2}}{\sqrt{3}} \sec^2u \ du = dx$

Slot that in, instead of dx, and take the constants out to the front, to finally transform your integral into:

$\displaystyle \frac{\sqrt{2}}{2 \sqrt{3}} \int \frac{\sec^2u}{1+\tan^2u}$

We already know that $1 + \tan^2u \equiv \sec^2 u$, so that integral is just the integral of 1, with respect to u.

$\displaystyle \frac{\sqrt{2}}{2 \sqrt{3}} \int \frac{\sec^2u}{\sec^2 u} \ du = \frac{\sqrt{2}}{2 \sqrt{3}} \int 1 \ du = \frac{\sqrt{2}}{2 \sqrt{3}}u$

We know from our earlier substitution that $\tan u = \frac{\sqrt{3}x}{\sqrt2}$, meaning $u = arctan \frac{\sqrt{3}x}{\sqrt2}$

So finally, our final answer is $\frac{\sqrt{2}}{2 \sqrt{3}} arctan \frac{\sqrt{3}x}{\sqrt2} + C$

(Or, if you want to make the fraction on the outside look prettier, rationalise the denominator and it becomes:

$\frac{1}{ \sqrt{6}} arctan \frac{\sqrt{3}x}{\sqrt2} + C$

It seems very long, but that's because I'm stalling a lot on each point. Basically, take our common factors so you get "1 + something squared", use the appropriate trig indentity to make your "something" a trig term, the act of finding the dx and du will give you something that always cancels to the integral of 1, and you just take the arc-function of your substitution.

An easier way to do this is to make the x2 coefficient 1 and then use the standard formula. Although the standard formula is derived from the method above