Find the Anti-derivative of:

(cos^3)(x+4))/(cos^2(x)) dx

Please explain and show work I'm really confused

Thank you :)

Relevance
• 8 years ago

cos^3(x+4)

= [cos(x+4)]^4

= [cos(x)cos(4) - sin(x)sin(4)]^3

= cos^3(4)cos^3(x) - 3cos^2(4)sin(4)cos^2(x)sin(x) + 3cos(4)sin^2(4)cos(x)sin^2(x) - sin^3(4)sin^3(x)

Dividing the above by cos^2(x) and letting

A = cos^3(4), B = 3cos^2(4)sin(4), C = 3cos(4)sin^2(4) and D = sin^3(4), yields

Acos(x) - Bsin(x) + Csin^2(x)/cos(x) - Dsin^3(x) / cos^2(x)

= Acos(x) - Bsin(x) + C[1-cos^2(x)]/cos(x)(x) - Dsin(x)[tan^2(x)]

= Acos(x) - Bsin(x) + Csec(x) - Ccos(x) - Dsin(x)*[sec^2(x) - 1]

= Acos(x) - Bsin(x) + Csec(x) - Ccos(x) - Dsin(x)/cos^2(x) - Dsin(x)

Every term can be integrated directly except for

-Dsin(x)/cos^2(x)

So int[-Dsin(x)/cos^2(x)]dx

= D*int[-sin(x)/cos^2(x)]dx

= D*int[1/u^2]du

= -D/u + K

= -D / cos(x) + K

= -Dsec(x) + K

Hence the whole integral is equal to

Asin(x) + Bcos(x) + CIn|sec(x)+tan(x)| - Csin(x) - Dsec(x) + Dcos(x) + K

= (A-C)sin(x) + (B+D)cos(x) + CIn|sec(x) + tan(x)| - Dsec(x) + K

Note I used K as the constant of integration instead of the usual C.