substitution rule?

**infraRed** · January 21st 2013, 08:13 PM

Hi. I'm somewhat frustrated because of a homework question. I don't know the mathhelpforum graphical formatting code, so I'm using '[' to represent that integral S-like symbol:

[ (dx / sqrt(7 - x^2)

I guess that can be translated into...

[ (7 - x^2)^(-1/2) dx

So, my natural inclination is to let u = 7 - x^2. But I do not see 2x elsewhere in the integral. So do I insert it? This doesn't look right:

[ (7 - x^2)^(-1/2) dx = [ u^(-1/2)2xdu = 2 [ u^(-1/2)xdu

Or is it...??? What am I supposed to do with that extra x?

**topsquark** · January 21st 2013, 08:52 PM

Originally Posted by infraRed

Hi. I'm somewhat frustrated because of a homework question. I don't know the mathhelpforum graphical formatting code, so I'm using '[' to represent that integral S-like symbol:

[ (dx / sqrt(7 - x^2)

I guess that can be translated into...

[ (7 - x^2)^(-1/2) dx

So, my natural inclination is to let u = 7 - x^2. But I do not see 2x elsewhere in the integral. So do I insert it? This doesn't look right:

[ (7 - x^2)^(-1/2) dx = [ u^(-1/2)2xdu = 2 [ u^(-1/2)xdu

Or is it...??? What am I supposed to do with that extra x?

What you need to do is replace the x with the function of u.

$u = \sqrt{7 - x^2} \implies x = \sqrt{7 - u}$

This does nothing nice for your integral.

Try this substitution: $x = \sqrt{7}~sin(u)$ .

-Dan

**ibdutt** · January 21st 2013, 08:52 PM

**Prove It** · January 21st 2013, 09:04 PM

Originally Posted by infraRed

Hi. I'm somewhat frustrated because of a homework question. I don't know the mathhelpforum graphical formatting code, so I'm using '[' to represent that integral S-like symbol:

[ (dx / sqrt(7 - x^2)

I guess that can be translated into...

[ (7 - x^2)^(-1/2) dx

So, my natural inclination is to let u = 7 - x^2. But I do not see 2x elsewhere in the integral. So do I insert it? This doesn't look right:

[ (7 - x^2)^(-1/2) dx = [ u^(-1/2)2xdu = 2 [ u^(-1/2)xdu

Or is it...??? What am I supposed to do with that extra x?

I suggest you read up about trigonometric and hyperbolic substitution, as that is what is appropriate in this case, not a u-substitution.

$\displaystyle \begin{align*} \int{\frac{dx}{\sqrt{7 - x^2}}} \end{align*}$

Let $\displaystyle \begin{align*} x = \sqrt{7}\sin{\theta} \implies dx = \sqrt{7}\cos{\theta}\,d\theta \end{align*}$ and the integral becomes

$\displaystyle \begin{align*} \int{\frac{dx}{\sqrt{7 - x^2}}} &= \int{\frac{\sqrt{7}\cos{\theta}\,d\theta}{\sqrt{7 - \left( \sqrt{7}\sin{\theta} \right)^2}}} \\ &= \int{\frac{\sqrt{7}\cos{\theta}\,d\theta}{\sqrt{7 - 7\sin^2{\theta}}}} \\ &= \int{ \frac{ \sqrt{7}\cos{\theta}\,d\theta }{ \sqrt{7 \left( 1 - \sin^2{\theta} \right) } } } \\ &= \int{\frac{\sqrt{7}\cos{\theta}\,d\theta}{\sqrt{7 \cos^2{\theta}}}} \\ &= \int{\frac{\sqrt{7}\cos{\theta}\,d\theta}{\sqrt{7} \cos{\theta}}} \\ &= \int{1\,d\theta} \\ &= \theta + C \\ &= \arcsin{\left( \frac{\sqrt{7} \, x}{7} \right)} + C \end{align*}$

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