Trig Integrals

From UBC Wiki
Jump to navigation Jump to search
MathHelp.png This article is part of the MathHelp Tutoring Wiki


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