1. Intergration at SIGHT !?!?!?!?!?

Firsty i dont know what intergration at sight means, and here is the question:

By Intergrating at sight, find:

1.$\displaystyle \int x^3. ({x^4}+4)^2$ $\displaystyle dx$

2.$\displaystyle \int \sin\theta \sqrt{1-\cos\theta}$ $\displaystyle d\theta$

3.$\displaystyle \int x.e^{x^2 + 1}$ $\displaystyle dx$

4.$\displaystyle \int e^x. (1-e^x)^3$ $\displaystyle dx$

5.$\displaystyle \int (x + 1). \sqrt{x^2 + 2x + 3}$ $\displaystyle dx$

2. Originally Posted by Stylis10
Firsty i dont know what intergration at sight means,
It means that looking at the problem we see basic forms.
In #1 we see a form of $\displaystyle u^2 du$.

In #2 we see a form of $\displaystyle \sqrt{u} du$.

In #3 we see a form of $\displaystyle e^u du$.

3. Originally Posted by Stylis10
Firsty i dont know what intergration at sight means, and here is the question:

By Intergrating at sight, find:

1.$\displaystyle \int x^3. ({x^4}+4)^2$ $\displaystyle dx$

2.$\displaystyle \int \sin\theta \sqrt{1-\cos\theta}$ $\displaystyle d\theta$

3.$\displaystyle \int x.e^{x^2 + 1}$ $\displaystyle dx$

4.$\displaystyle \int e^x. (1-e^x)^3$ $\displaystyle dx$

5.$\displaystyle \int (x + 1). \sqrt{x^2 + 2x + 3}$ $\displaystyle dx$
EDIT: With practice, you will be able to spot some integration by sight. Mainly, the chain rule is the one that can be differentiated and integrated by sight easily.

1)
$\displaystyle \frac{1}{12}(x^4 + 4)^3$

2)
$\displaystyle \frac{2}{3} (1 - cos \theta)^\frac{3}{2}$

3)
$\displaystyle \frac{1}{2}e^{x^2 +1}$

4)
$\displaystyle -\frac{1}{4}(1 - e^x)^4$

5)
$\displaystyle \frac{1}{3}(x(x+2) + 3)^{\frac{3}{2}}$

4. im still clueless, reply to plato

thank you AIR, i was trying to use the intergration product rules once ofund out u were supposed to do it in your head, but the chain rule makes more sense