# Intergration at SIGHT !?!?!?!?!?

• Mar 9th 2008, 10:08 AM
Stylis10
Intergration at SIGHT !?!?!?!?!?
Firsty i dont know what intergration at sight means, and here is the question:

By Intergrating at sight, find:

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

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

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

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

5. $\int (x + 1). \sqrt{x^2 + 2x + 3}$ $dx$
• Mar 9th 2008, 10:16 AM
Plato
Quote:

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 $u^2 du$.

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

In #3 we see a form of $e^u du$.
• Mar 9th 2008, 10:22 AM
Simplicity
Quote:

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

By Intergrating at sight, find:

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

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

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

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

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

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

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

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

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

5)
$\frac{1}{3}(x(x+2) + 3)^{\frac{3}{2}}$
• Mar 9th 2008, 10:24 AM
Stylis10
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