# Help! i beg of you

1. Jun 26, 2004

### JonF

i for the life of me can't figure out how to integrate this:

$$\int \sin{(x^{1/2})} dx$$

2. Jun 26, 2004

### Muzza

Make the substitution $$u = \sqrt{x}$$ which will turn the integral into something like $$\int \left(\sin{u} \times 2\sqrt{x}\right) du$$. Apply the substitution again and use integration by parts...

Last edited: Jun 26, 2004
3. Jun 26, 2004

We don't have enough x's in there, so use integration by parts to get some. Let u = sin(x^(1/2)) and dv = dx. After that, you'll get an integral that can be solved through a substitution and then a couple of applications of by parts.

4. Jun 26, 2004

### JonF

thanks, got it now

5. Jun 29, 2004

### JonF

Ok a new integral I can’t figure out how to take

$$\int \frac{1}{4y^{2} - 4y - 3} dy$$

Any suggestions?

6. Jun 29, 2004

Partial fractions.

7. Jun 30, 2004

### JonF

Oops I posted the wrong integral, I meant:
$$\int \frac{1}{(4y^{2} - 4y - 3)^{1/2}} dy$$

8. Jun 30, 2004

### Zurtex

Try completing the square and using a substitution of what you get inside the ()^2 bit. It should then be a simple matter of knocking it into standard form.

9. Jun 30, 2004

### TenNen

Try making a change till you get the folowing formula
$$\int \frac{1}{((Y-A)^2+B^2)^{1/2}} dY$$
Sorry I forgot the formula of how to go on, you can check it out in your textbooks.

10. Jun 30, 2004

### Zurtex

This might help (a standard form)

$$\int \frac{1}{(x^2 + a^2)^{\frac{1}{2}}} dx = \ln \left(x + \sqrt{x^2 + a^2} \right)$$

Though it would appear to me you are on the wrong track. Have checked your integral out at http://integrals.wolfram.com (you need to check their syntax on how to input function it's a little temperamental)

11. Jun 30, 2004

### Parth Dave

My suggestion is complete the square on the denominator. Than it should be a matter of using a trig substitution. Looks to me like it will be a secant substitution.