Trig Integrals

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

A. I= ∫( sin^mx 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 ∫(tan^mx*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

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