1. ## Integrate this equation

I am unable to initiate the first few steps.....
solve

$\displaystyle \int sin4x - cos6x$

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

$\displaystyle \int sin4x - cos6x$
You're expected to know the following standard forms:

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

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

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

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

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

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

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

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

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

Therefore
$\displaystyle I=I_1+I_2 = {\color{red} - } \frac{cos4x}{4} - \frac{sin6x}{6} +c$
Note the corrections in red.