MHB Calculus II U-Substitution Questions

[ardentmed]

This thread is only addressing question 6.

6a and 6b seemed relatively simple, with 6c seeming to be the most troublesome. I got 1/4 arcsin (x^2 / 2 ) + c for 6a after substituting u for x^2.
As for 6b, I got (x^2)/2 -2ln(x^2 +4) + c after using algebraic division and taking u=x^2 + 4.

Finally, for 6c, I took u as √(x), followed by algebraic division since the nominator should always be smaller than the denominator in order for partial fractions to be usable. This gave me 2√(x) + 2ln(√x -2 ) - 2ln(√x + 2) + c.

 

[Prove It]

This thread is only addressing question 6.

6a and 6b seemed relatively simple, with 6c seeming to be the most troublesome. I got 1/4 arcsin (x^2 / 2 ) + c for 6a after substituting u for x^2.
As for 6b, I got (x^2)/2 -2ln(x^2 +4) + c after using algebraic division and taking u=x^2 + 4.

Finally, for 6c, I took u as √(x), followed by algebraic division since the nominator should always be smaller than the denominator in order for partial fractions to be usable. This gave me 2√(x) + 2ln(√x -2 ) - 2ln(√x + 2) + c.

6.c)

$\displaystyle \begin{align*} \int{ \frac{\sqrt{x}\,\mathrm{d}x}{x - 4}} &= \int{ \frac{\sqrt{x}\,\mathrm{d}x}{ \left( \sqrt{x} \right) ^2 - 4 } } \\ &= \int{ \frac{ \left( \sqrt{x} \right) ^2 \, \mathrm{d}x}{ \sqrt{x} \, \left[ \left( \sqrt{x} \right) ^2 - 4 \right] } } \\ &= 2 \int{ \frac{ \left( \sqrt{x} \right) ^2}{ \left( \sqrt{x} \right) ^2 - 4 } \, \left( \frac{1}{2\,\sqrt{x}} \right) \, \mathrm{d}x } \end{align*}$

Now substitute $\displaystyle \begin{align*} u = \sqrt{x} \implies \mathrm{d}u = \frac{1}{2\,\sqrt{x}} \, \mathrm{d}x \end{align*}$ and the integral becomes

$\displaystyle \begin{align*} \int{ \frac{ \left( \sqrt{x} \right) ^2 }{ \left( \sqrt{x} \right) ^2 - 4} \, \left( \frac{1}{2\,\sqrt{x} } \right) \, \mathrm{d}x } &= \int{ \frac{u^2}{u^2 - 4} \, \mathrm{d}u } \\ &= \int{ \frac{u^2 - 4 + 4}{u^2 - 4} \, \mathrm{d}u } \\ &= \int{ 1 + \frac{4}{u^2 - 4}\, \mathrm{d}u} \end{align*}$

I think you can go from here, you should use partial fractions.

 

[ardentmed]

6.c)

$\displaystyle \begin{align*} \int{ \frac{\sqrt{x}\,\mathrm{d}x}{x - 4}} &= \int{ \frac{\sqrt{x}\,\mathrm{d}x}{ \left( \sqrt{x} \right) ^2 - 4 } } \\ &= \int{ \frac{ \left( \sqrt{x} \right) ^2 \, \mathrm{d}x}{ \sqrt{x} \, \left[ \left( \sqrt{x} \right) ^2 - 4 \right] } } \\ &= 2 \int{ \frac{ \left( \sqrt{x} \right) ^2}{ \left( \sqrt{x} \right) ^2 - 4 } \, \left( \frac{1}{2\,\sqrt{x}} \right) \, \mathrm{d}x } \end{align*}$

Now substitute $\displaystyle \begin{align*} u = \sqrt{x} \implies \mathrm{d}u = \frac{1}{2\,\sqrt{x}} \, \mathrm{d}x \end{align*}$ and the integral becomes

$\displaystyle \begin{align*} \int{ \frac{ \left( \sqrt{x} \right) ^2 }{ \left( \sqrt{x} \right) ^2 - 4} \, \left( \frac{1}{2\,\sqrt{x} } \right) \, \mathrm{d}x } &= \int{ \frac{u^2}{u^2 - 4} \, \mathrm{d}u } \\ &= \int{ \frac{u^2 - 4 + 4}{u^2 - 4} \, \mathrm{d}u } \\ &= \int{ 1 + \frac{4}{u^2 - 4}\, \mathrm{d}u} \end{align*}$

I think you can go from here, you should use partial fractions.

Wow, I completely missed that simple substitution and it was right in front of me the entire time. Thanks a ton!

 

[ardentmed]

Can anyone confirm a and b as well?

Thanks again.

 

[MarkFL]

Can anyone confirm a and b as well?

Thanks again.

Have you tried differentiating your results to see if you get back to the original integrand?

 

[ardentmed]

Have you tried differentiating your results to see if you get back to the original integrand?

Yes, but a came out with an extra 1/4 in front of the expression.

 

[MarkFL]

well, let's look at 6a). we are given to evaluate:

$$I=\int\frac{x}{\sqrt{4-x^4}}\,dx$$

Let:

$$x^2=2\sin(\theta)\,\therefore\,x\,dx=\cos(\theta)\,d\theta$$

and we now have:

$$I=\frac{1}{2}\int\,d\theta=\frac{1}{2}\theta+C$$

Back-substitute for $\theta$:

$$I=\frac{1}{2}\sin^{-1}\left(\frac{x^2}{2}\right)+C$$

Do you see where you may have gone wrong?