1. ## Integration

I hate it. Stuck on 2 integration problems

a). $\int\frac{1}{\sqrt{x}}\sin\sqrt{x}\ .dx\ where\ x = u^2$

b). $\int\frac{2}{e^{2x} + 4}\ where\ u = e^{2x} + 4$

Thanks for any help

2. Originally Posted by Beard
I hate it. Stuck on 2 integration problems

a). $\int\frac{1}{\sqrt{x}}\sin\sqrt{x}\ .dx\ where\ x = u^2$

b). $\int\frac{2}{e^{2x} + 4}\ where\ u = e^{2x} + 4$

Thanks for any help
Your first substitution should work giving

$2\int \sin u\, du$.

For the second try

$u = e^{x}$

3. For the second, I prefer $u = e^{-x}$. No partial fractions decomposition required, just simple sub.

4. Originally Posted by Chop Suey
For the second, I prefer $u = e^{-x}$. No partial fractions decomposition required, just simple sub.
Definitely quicker (as a little easier) but not obvious for most.

5. Thanks for the first part, however I am unable to do the second part of your reply where you said try:

$u = e^x$

because the book has given me a 'set' substitution so it must remain as

$u = e^{2x} + 4$

6. Originally Posted by Beard
Thanks for the first part, however I am unable to do the second part of your reply where you said try:

$u = e^x$

because the book has given me a set 'substitution' so it must remain as

$u = e^{2x} + 4$
In that case then, if $u=e^{2x}+4$, then $\,du=2e^{2x}\,dx$

Thus, the integral becomes $\int\frac{2\,dx}{u}=\int\frac{2e^{2x}\,dx}{e^{2x}u }$

Since $u=e^{2x}+4$, that means that $e^{2x}=u-4$

So the integral becomes $\int\frac{\,du}{u\left(u-4\right)}$

And now, you'll need to apply the technique of partial fractions.

7. Originally Posted by Chris L T521

you'll need to apply the technique of partial fractions.
No, haha, because of $u-(u-4)=4.$

8. Thus, the integral becomes

Confused by the second part of this unless you are saying that you are multiplying by 1 in the form of $\frac{e^{2x}}{e^{2x}}$

9. Originally Posted by Beard
Thus, the integral becomes

Confused by the second part of this unless you are saying that you are multiplying by 1 in the form of $\frac{e^{2x}}{e^{2x}}$
Yes. I multiplied by 1 in the form of $\frac{e^{2x}}{e^{2x}}$ because now we can replace $2e^{2x}\,dx$ with $\,du$ in the numerator. In the denominator, you can now replace $e^{2x}$ with $u-4$.

10. So how do you integrate

$\frac{du}{u^2 - 4u}$

?

11. $\int\frac{1}{u(u-4)}du$

One way is split into partial fractions.

$\frac{1}{4}\int\frac{1}{u-4}du-\frac{1}{4}\int\frac{1}{u}du$

Now, it's easy to see each one is in terms of ln.