Indefinite integrals are written in the form of int f(x) dx = F(x) +C
where: f(x) as the integrand
F(x) as the anti-derivative function
C as the arbitrary constant known as constant of integration
For the given problem int sin(-7x)cos(6x) dx or intcos(6x)sin(-7x) dx has a integrand in a form of trigonometric function. To evaluate this, we apply the identity:
cos(A)sin(B) =[sin(A+B) -sin(A-B)]/2
The integral becomes:
intcos(6x)sin(-7x) dx = int[sin(6x+(-7x)) -sin(6x-(-7x))]/2dx
= int[sin(6x-7x) -sin(6x+7x)]/2dx
Apply the basic properties of integration: int c*f(x) dx= c int f(x) dx .
int[sin(6x-7x) -sin(6x+7x)]/2dx= 1/2int[sin(6x-7x) -sin(6x+7x)]dx
Apply the basic integration property: int (u+v) dx = int (u) dx + int (v) dx .
1/2 *[int (sin(6x-7x))dx - int sin(6x+7x)dx]
Then apply u-substitution to be able to apply integration formula for cosine function: int sin(u) du=-cos(u) +C .
For the integral: intsin(6x-7x)dx , we let u = 6x-7x=-x then du= - dx or (-1)du =dx .
intsin(6x-7x)dx=intsin(-x) dx
=intsin(u) *(-1)du
=(-1) int sin(u)du
=(-1)(-cos(u) )+C
=cos(u) +C
Plug-in u =-x on cos(u) +C , we get:
intsin(6x-7x)dx= cos(-x) +C
For the integral: intsin(6x+7x)dx , we let u = 6x+7x=13x then du= 13 dx or (du)/13 =dx .
intsin(6x+7x)dx=intsin(13x) dx
=intsin(u) *(du)/13
= 1/13 int sin(u)du
= 1/13( -cos(u))+C or -1/13cos(u) +C
Plug-in u =13x on -1/13 cos(u) +C , we get:
intsin(6x+7x)dx= -1/13 cos(13x) +C
Combing the results, we get the indefinite integral as:
intsin(-7x)cos(6x) dx= 1/2*[ cos(-x) -(-1/13 cos(13x))] +C
or 1/2 cos(-x) +1/26 cos(13x) +C
Since cosine is an even function, cos(-x) = cos(x) , so we get:
intsin(-7x) cos(6x)dx=1/2 cos(x) +1/26 cos(13x) +C
No comments:
Post a Comment