integration

**wutever** · June 1st 2008, 02:48 AM

1) integrate sin3x cos7x

i assumed u=3x and another time u=7x
then i came up with the ans:
-1/3 cos3x * 1/7 sin7x

2) integrate sqrrt (1-4x^2) dx
using the integration substitution to get the form of sqrt(a^2-x^2)
let a^2 = 1/4 x=1/2sinx
sqrt(1/4-1/4sin^2x)
but then i couldnt continue...

**wingless** · June 1st 2008, 03:53 AM

1) $\int \sin 3x \cos 7x~dx$
No way! You can't split an integral in two and integrate and then multiply them.. You have to integrate $\int \sin 3x \cos 7x~dx$ as a whole.

The best way is first using the trig identity $\cos a \sin b = \frac{1}{2}\cdot [\sin(a+b)-\sin(a-b)]$ , then integrating.

Remember: $\int f(x)g(x)~dx$ is NOT $\left( \int f(x)~dx \right ) \cdot \left( \int g(x)~dx\right )$

But $\int f(x)+g(x)~dx$ is equal to $\int f(x)~dx+\int g(x)~dx$

-----------------------

2) $\int \sqrt{1-4x^2}~dx$

Can you see that this is equal to $\int \frac{1-4x^2}{\sqrt{1-4x^2}}~dx$

Now you can use u-substitution and then integrate it by parts.

**mr fantastic** · June 1st 2008, 04:01 AM

Originally Posted by wutever

1) integrate sin3x cos7x

i assumed u=3x and another time u=7x
then i came up with the ans:
-1/3 cos3x * 1/7 sin7x

Mr F says: Uh uh. You need to manipulate the compound angle formulae to get the following identity:

${\color{red} \sin A \cos B = \frac{\sin (A + B) + \sin (A - B)}{2}}$ .

So ${\color{red} \sin (3x) \cos (7x) = \frac{\sin (10x) + \sin (-4x)}{2} = \frac{\sin (10x) - \sin (4x)}{2}}$ .

2) integrate sqrrt (1-4x^2) dx
using the integration substitution to get the form of sqrt(a^2-x^2)
let a^2 = 1/4 x=1/2sinx
sqrt(1/4-1/4sin^2x)
but then i couldnt continue...

Mr F says: Perhaps if I write ${\color{red} \frac{1}{4} - \frac{1}{4} \sin^2 x = \frac{1}{4} (1 - \sin^2 x)}$ you'll see how the Pythagorean Identity makes life simple ......

..

**wutever** · June 1st 2008, 04:17 AM

ok i get the first one but i cant fully understand the second one !!!!

**mr fantastic** · June 1st 2008, 04:21 AM

Originally Posted by wutever

ok i get the first one but i cant fully understand the second one !!!!

$1 - \sin^2 x = \cos^2 x .......$

**wutever** · June 1st 2008, 04:28 AM

okk i know that but i have a weired ans :
1/4 sin inverse (2x) + 1/2 x sqrt (1-4x^2)

so how ??
and why $ \int \sqrt{1-4x^2}~dx $

is equal to

$ \int \frac{1-4x^2}{\sqrt{1-4x^2}}~dx $

**mr fantastic** · June 1st 2008, 04:34 AM

Originally Posted by wutever

okk i know that but i have a weired ans :
1/4 sin inverse (2x) + 1/2 x sqrt (1-4x^2)

Mr F says: *Ahem* .... + C

so how ??

Mr F says: The answer you got is correct. If you differentiated it, you would see that you get \sqrt{1 - 4x^2} .......

and why $ \int \sqrt{1-4x^2}~dx $

is equal to

$ \int \frac{1-4x^2}{\sqrt{1-4x^2}}~dx $

Mr F says: Basic algebra. Multiply ${\color{red}\sqrt{1 - 4x^2}}$ by ${\color{red}1 = \frac{\sqrt{1 - 4x^2}}{\sqrt{1 - 4x^2}}}$

..

**smifffy** · June 1st 2008, 06:22 AM

For the second integral you want to use the substitution:

2u = sin(x)

You then get: Sqrt (1 - (sin(x))^2)

Which = sqrt ( (cos(x))^2 )

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