Eric had a good answer, but I'll give my approaches since it's not about whether they know what to do, but whether they can figure it out without someone telling them what to do.
I prefer to start from the fundamentals to avoid any mistakes. So first off, the trig substitution trio:
(a^2 + x^2)(a2+x2) reminds me of 1 + tan^2theta = sec^2theta1+tan2θ=sec2θ, and I would use x = asecthetax=asecθ.
(a^2 - x^2)(a2−x2) reminds me of 1 - cos^2theta = sin^2theta1−cos2θ=sin2θ, and I would use x = asinthetax=asinθ.
(x^2 - a^2)(x2−a2) reminds me of sec^2theta - 1 = tan^2thetasec2θ−1=tan2θ, and I would use x = atanthetax=atanθ.
Now let's look at the problems.
a)
With this problem, I see the sqrt(1-9x^2)√1−9x2 and I think about trig substitution. If you see something like sqrt(pmx^2pma^2)√±x2±a2, consider trig substitution. If you don't see a square root, don't go straight to trig substitution. Think about if you could have done something else, like u-substitution or partial fraction decomposition (hopefully not that).
In this case, we can modify the terms to give a coefficient of 1 on x^2x2. Then we get:
int 1/(sqrt(9)sqrt(1/9-x^2))dx = 1/3int 1/(sqrt(1/9-x^2))dx∫1√9√19−x2dx=13∫1√19−x2dx
So, a = 1/3a=13. At this point I usually consider what trig identity this reminds me of. When I see a^2 - x^2a2−x2, if I replace xx with a trig term, I'm drawn to sin^2theta + cos^2theta = 1sin2θ+cos2θ=1, because cos^2theta = 1-sin^2thetacos2θ=1−sin2θ. So, let's see what this reduces to.
Let x = asintheta = 1/3sinthetax=asinθ=13sinθ. Then that means dx = 1/3costhetadx=13cosθddthetaθ. Now we can substitute.
1/3int 1/(sqrt(1/9-x^2))dx = 1/3int 1/(sqrt(1/9-1/9sin^2theta))1/3costheta13∫1√19−x2dx=13∫1√19−19sin2θ13cosθ
= 1/3int costheta/(cancel(3)cancel(1/3)sqrt(1-sin^2theta))dtheta
= 1/3int cancel(costheta/(sqrt(cos^2theta)))dtheta = 1/3intdtheta
= 1/3theta + C
Now, we can work backwards and look at how we said x = 1/3sintheta. Therefore, theta = arcsin(3x). So, we get:
1/3arcsin(3x) + C
So that one was pretty straightforward, but I don't think it's a very comprehensive practice problem. You didn't even get to draw a right triangle. Ah well.
b)
At first I thought about multiplying it out, but then I realized, it was written as x-2 for a reason.
int 1/(9 + (x-2)^2)dx = 1/(a^2 + (x-2)^2)
a = 3
x - 2 = ?
I look at this, and it reminds me of 1 + tan^2theta = sec^2theta.
So naturally, we get:
x - 2 = 3tantheta
dx = 3sec^2thetadtheta
Plug it in, and divide out common factors:
int 1/(9 + 9tan^2theta)3sec^2thetadtheta
= int sec^2theta/(3 + 3tan^2theta)dtheta
Then, WOAH, what's this? Another substitution we can do? Yeah... that's right. We can do a u-substitution on this. But let's not, because we can use identities.
= int cancel(sec^2theta)/(3(cancel(sec^2theta)))dtheta
= 1/3theta + C
If you recall, x-2 = 3tantheta, so theta = arctan((x-2)/3). So, we get:
1/3arctan((x-2)/3) + C
c)
int (x-3)/(x^2+1)dx
Already this looks interesting. What might trip people up is the x-3, because it is not a common factor of x^2 + 1. However, you can separate the fraction.
int (x-3)/(x^2+1)dx = int x/(x^2+1)dx - int 3/(x^2+1)dx
You may notice that there is no need for trig substitution here. There isn't. You now have a 1/(1+x^2) integral in the second one, which if you recall, is arctanx + C. The first one gives you x^2 and xdx, which resembles a u-substitution problem.
Let:
u = x^2 + 1
du = 2xdx
xdx = 1/2du
int x/(x^2+1)dx - int 3/(x^2+1)dx = 1/2int 1/udu - 3int 1/(x^2+1)dx
That's about it. Just integrate and get your ln|u| and arctanx:
= 1/2ln|x^2+1| - 3arctanx + C
Eric, note that you wrote:
x = tantheta
which is fine, because a = 1. But you also wrote:
2dx = sec^2thetadtheta
But dx = sec^2thetadtheta, so 2dx = 2sec^2thetadtheta.
So, the answer would not contain 3/2arctanx, but 3arctanx. :) (Also, int1/udu = ln|u| + C, rather than lnu + C, but that's a minor point since x^2 + 1 > 0 anyways)
