# Integrate this equation

• June 6th 2009, 02:20 AM
zorro
Integrate this equation
I am unable to initiate the first few steps.....
solve

$\int sin4x - cos6x$
• June 6th 2009, 02:23 AM
mr fantastic
Quote:

Originally Posted by zorro
I am unable to initiate the first few steps.....
solve

$\int sin4x - cos6x$

You're expected to know the following standard forms:

$\int \sin (kx) \, dx = - \frac{1}{k} \cos (kx) + C$

$\int \cos (kx) \, dx = \frac{1}{k} \sin (kx) + C$
• June 6th 2009, 11:40 PM
zorro
Quote:

Originally Posted by mr fantastic
You're expected to know the following standard forms:

$\int \sin (kx) \, dx = - \frac{1}{k} \cos (kx) + C$

$\int \cos (kx) \, dx = \frac{1}{k} \sin (kx) + C$

I am using a different technique please let me know if it is right or no...
$I = I_1 + I_2$
$I_1 = \int sin(4x)dx$
$4x=u_1$
$4=\frac{du_1}{dx}$
$Therefore \ I_1=\frac{1}{4} \int 4 sin u_1 du_1$
$I_1= \frac{1}{4} cos u_1 +c_1$

$I_2 = \int cos(6x)dx$
$\frac{1}{6} \int 6 cos(6x)dx$
$6x=u_2$
$6=\frac{du_2}{dx}$
$I_2=\frac{1}{6} \int cosu_2 du_2$
$I_2=\frac{-sin6x}{6}+c_2$

Therefore
$
I=I_1+I_2
=\frac{cos4x}{4} - \frac{sin6x}{6} +c
$
• June 7th 2009, 03:05 AM
mr fantastic
Quote:

Originally Posted by zorro
I am using a different technique please let me know if it is right or no...
$I = I_1 + I_2$
$I_1 = \int sin(4x)dx$
$4x=u_1$
$4=\frac{du_1}{dx}$
$Therefore \ I_1=\frac{1}{4} \int 4 sin u_1 du_1$
$I_1= {\color{red} - } \frac{1}{4} cos u_1 +c_1$

$I_2 = \int cos(6x)dx$
$\frac{1}{6} \int 6 cos(6x)dx$
$6x=u_2$
$6=\frac{du_2}{dx}$
$I_2=\frac{1}{6} \int cosu_2 du_2$
$I_2=\frac{-sin6x}{6}+c_2$

Therefore
$
I=I_1+I_2
= {\color{red} - } \frac{cos4x}{4} - \frac{sin6x}{6} +c
$

Note the corrections in red.