Trig Integrals

There are two parts of the techniques of calculating the integral of a given trigonometry.

A. I= ∫( sinmx cosn dx) (m and n are positive numbers)

A1: n is odd (n=1,3,5,7.... m doesn't matter)

Consider I= ∫(sin2x cos3x dx)

= ∫( sin2x cos2x*cosx dx)

use sin2x+cos2 =1

then cos2x=1-sin2x

I= ∫(sin2x(1-sin2x)cosx dx)

use u=sin(x), so that du=cos(x)

I= ∫(u2 (1-u2)du)

=∫((u2-u4)du)

=c+u3/3-1/5 u5

=c+sin3x-1/5 sin5

What to do? Isolate cos(x), use cos2x =1-sin2x

Denote: sin(x)=u

A2:m is odd, n doesn't matter consider

I= ∫ (sin5x*cos3x dx)

=∫(sin4x * cos3xsinx dx)

=- ∫(sin4x*cos3x* cosx' dx)

u=cosx so that sin2x=1-u2

=-∫((1-u*u)2 u3 du

=-∫(1+u4-2u2)*u3du)

=-∫((u3+u7-2u5) du)

=c-u4/4-u8/8+2u6/6

what to do? Isolate sin(x) use sin2x=1-cos2x

substitution u(x)=cos(x)

get Integral of some polynomial u

A3: Both m,n are even numbers consider

I = ∫( sin2x cos2x dx)

since cos 2x=1-2sin2x=2cos2x-1

cos2x= (cos2x+1)/2, sin2x=(1-cos2x)/2

I= 1/4 * ∫(cos2x+1)(1-cos2x) dx

=1/4 * ∫((1-cos22*2x)dx)

=1/4 *x-1/4 ∫((1+cos4x)/2)dx

=1/4 *x-1/2*x-1/8 ∫(cos4x dx)

=-1/4*x-1/8sin4x *1/4

=-1/4*x-1/32* sin4x

B ∫(tanmx*secnxdx) (m,n are positive integers)

B1:n is even

I= ∫(tanx*sec2x dx)

keep sec2x alone, sec2x=(tanx)'

I=∫ (tanx*(tanx)' dx)

let u=tanx

I= ∫(u dx)

=c+1/2 * u2

=1/2 tanx2+c

what to do: Isolate a factor sec2x and denote u=tanx

B2: n is odd

I= ∫(tan2x secx*dx)

Isolate tanxsecx

I= ∫(tanx (tanx secx)dx)

=∫(tanx(secx)' dx) u=secx

I= ∫(tanx*du)

=ln |secx|+c