Thread: trig substitution

1. trig substitution

hey there

I can't get this problem I've been trying it for a while now and it is really frustrating me...some help would be greatly appreciated

integral of 5 / sqrt (48-16x-64x^2) dx

I tried completing the square by getting 49-64(x+1/8)^2...is this right? I've seen so many ways to complete the square I don't know which I should use for integrals...

and then I went ahead and tried to solve it by trig substitution but I just can't get it :S I used u= 7sint

thanking you in advance

2. Integral

Hello zaddie21
Originally Posted by zaddie21
hey there

I can't get this problem I've been trying it for a while now and it is really frustrating me...some help would be greatly appreciated

integral of 5 / sqrt (48-16x-64x^2) dx

I tried completing the square by getting 49-64(x+1/8)^2...is this right? I've seen so many ways to complete the square I don't know which I should use for integrals...

and then I went ahead and tried to solve it by trig substitution but I just can't get it :S I used u= 7sint

thanking you in advance
Have you noticed that you can take out a common factor of 16? It makes the numbers easier to handle.

$48-16x-64x^2 = 16(3-x-4x^2)$

... and there's really only one way to complete the square, and that is:

$3 - x- 4x^2 = -4(x^2 + \tfrac{1}{4}x - \tfrac{3}{4})$

$= -4[(x+\tfrac{1}{8})^2- \tfrac{3}{4}-\tfrac{1}{64}]$

$= -4[(x+\tfrac{1}{8})^2-\tfrac{49}{64}]$

$= 4[\tfrac{49}{64}-(x+\tfrac{1}{8})^2]$

So $48-16x-64x^2 = 16\cdot 4[\tfrac{49}{64}-(x+\tfrac{1}{8})^2]$

$=64[(\tfrac{7}{8})^2 -(x+\tfrac{1}{8})^2]$

Put this into your integral, and you'll have something in the form

$\int\frac{kdx}{\sqrt{a^2 - (x+b)^2}}$

If you now substitute $(x+b) = a\sin\theta$, that should do it.

Grandad