# Int sin[sqrt[t]]

1. Sep 16, 2005

### SplinterIon

I'm drawing a blank as to how to approach this one. I've been looking over all trig identities and substitutions I could possibly make - but to no avail .

$$\int_{1}^{x^2} \sin{(\sqrt{t})} \ dt$$

2. Sep 16, 2005

### MalleusScientiarum

Try making a u-substitution for the square root of t.

3. Sep 17, 2005

### Orion1

indefinite integral...

The closest identity that I could determine is:

identity:
$$\int u \sin u \; du = \sin u - u \cos u + C$$

$$\int \sin \sqrt{t} \; dt = 2 \left( \sin \sqrt{t} - \sqrt{t} \cos \sqrt{t} \right) + C$$

4. Sep 18, 2005

### HallsofIvy

Staff Emeritus
Good try but it looks to me like that gives you
$$\int u^{\frac{1}{2}}sin u du$$ which doesn't look any more hopeful.

I'd be willing to bet that this doesn't have an elementary anti-derivative.

5. Sep 19, 2005

### dextercioby

Don't bet too much, Halls.

$$\sqrt{t}=u$$

implies $t=u^{2} \ \mbox{and} \ dt= 2 u du$

and the antiderivative becomes

$$\int 2 u \sin u \ du$$ which can be easily tackled with the part integration method.

Daniel.

6. Sep 20, 2005

### HallsofIvy

Staff Emeritus
One of these days, I really have to learn algebra!