http://www.math.ucdavis.edu/~kouba/CalcTwoDIRECTORY/trigintdirectory/TrigInt.htmlStrategy: The strategy is not obvious. Multiply and divide by (sec x + tan x); use Substitution. 1.sec x dx = sec xsec x + tan x

sec x + tan xdx

set u = sec x + tan x then we find du = (sec x tan x + sec2 x) dx substitute du = (sec x tan x + sec2 x) dx, u = sec x + tan x sec xsec x + tan x

sec x + tan xdx =(sec2 x + sec x tan x) dx

sec x + tan x

=du

u

solve integral = ln |u| + C substitute back u=sec x + tan x = ln |sec x + tan x| + C 2. tan x dx =sin x

COs xdx

set u = COs x. then we find du = - sin x dx substitute du=-sin x, u=COs x sin x

COs xdx = -(-1) sin x dx

COs x

= -du

u

Solve the integral = - ln |u| + C substitute back u=COs x = - ln |COs x| + C Q.E.D. 2. Alternate Form of Resulttan x dx = - ln |COs x| + C = ln | (COs x)-1 | + C = ln |sec x| + C Therefore: tan x dx = - ln |COs x| + C = ln |sec x| + C

3.cot x dx =cos x

sin xdx

set u = sin x. then we find du = cos x dx substitute du=cos x, u=sin x cos x

sin xdx =du

u

solve integral = ln |u| + C substitute back u=sin x = ln |sin x| + C

4.csc x dx = csc xcsc x + cot x

csc x + cot xdx

set u = csc x + cot x then we find du = (- csc x cot x - csc2 x) dx substitute du = (- csc x cot x - csc2 x) dx, u = csc x + cot x csc xcsc x + cot x

csc x + cot xdx = -(- csc2 x - csc x cot x) dx

csc x + cot x

= -du

u

solve integral = - ln |u| + C substitute back u=csc x + cot x = - ln |csc x + cot x| + C

Integral trigonometri1. Strategy: Use definition of sinh. sinh x =ex - e-x

2

sinh x dx =ex - e-x

2dx

= (1/2) ex dx - (1/2) e-x dx solve left equation = (1/2) ex - (1/2) e-x dx set u = - x then we find du = - dx substitute du= - dx, u= - x

= (1/2) ex + (1/2) - e-x dx = (1/2) ex + (1/2) eu du solve the right integral = (1/2) ex + (1/2) eu + C substitute back u= - x = (1/2) ex + (1/2) e-x + C =ex + e-x

2+ C

which by definition = cosh x + C

2. cosh x =ex + e-x

2

cosh x dx =ex + e-x

2dx

= (1/2) ex dx + (1/2) e-x dx solve left equation = (1/2) ex + (1/2) e-x dx set u = - x then we find du = - dx substitute du= - dx, u= - x = (1/2) ex - (1/2) - e-x dx = (1/2) ex - (1/2) eu du solve the right integral = (1/2) ex - (1/2) eu + C substitute back u= - x = (1/2) ex - (1/2) e-x + C =ex - e-x

2+ C

which by definition = sinh x + C

3.tanh x =sinh x

cosh x=(ex - e-x) / 2

(ex + e-x) / 2

tanh x dx =ex - e-x

ex + e-xdx

set u = ex + e-x then we find du = (ex - e-x) dx substitute du= (ex - e-x) dx, u = ex + e-x =du

u

solve = ln |u| + C substitute back u = ex + e-x = ln |ex + e-x| + C since ex and e-x are always positive = ln (ex + e-x) + C

since (ex + e-x)/2 = cosh(x)

= ln (2 cosh x) + C = ln 2 + ln (cosh x) + C

ln 2 is merely a constant that can be combined with C

= ln (cosh x) + C

4. coth x =cosh x

sinh x=(ex + e-x) / 2

(ex - e-x) / 2

coth x dx =ex + e-x

ex - e-xdx

set u = ex - e-x then we find du = (ex + e-x) dx substitute du= (ex + e-x) dx, u = ex - e-x =du

u

solve = ln |u| + C substitute back u = ex - e-x = ln |ex - e-x| + C since (ex - e-x)/2 = sinh(x) = ln |2 sinh x| + C = ln 2 + ln |sinh x| + C ln 2 is merely a constant that can be combined with C = ln |sinh x| + C http://math2.org/math/integrals/tableof.htm integral berpangkatLet u = cos^(n-1) x, dv = cos x dx => v = sin x, du = (n-1) cos^(n-2)x (-sin x) dx

Let I = cos^n x dx = u dv = uv - v du= cos^(n-1) x sin x - -(n-1) cos^(n-2) x sin^2 x= cos^(n-1) x sin x + (n-1) cos^(n-2) x (1 - cos^2 x) dx= cos^(n-1) x sin x + (n-1) [ cos^(n-2) x dx - I]So I + (n-1) I = cos^(n-1) x sin x + (n-1) cos^(n-2) x dx=> I = 1/n cos^(n-1) x sin x + ((n-1)/n) cos^(n-2) x dx tan^n(x) dx = tan^(n-2)(x) * tan^2(x) dx. . . . . . . . . = tan^(n-2)(x) * (sec^2(x) - 1) dx. . . . . . . . . = tan^(n-2)(x) * sec^2(x) dx - tan^(n-2)(x) dx

Now use substitution:u = tan(x)du = sec^2(x) dx

. . . . . . . . . = u^(n-2) du - tan^(n-2)(x) dx. . . . . . . . . = u^(n-1) / (n-1) - tan^(n-2)(x) dx. . . . . . . . . = tan^(n-1)(x) / (n-1) - tan^(n-2)(x) dx

2. I = sec^n(x) dx= sec^(n-2)(x) sec^2(x) dx= sec^(n-2)(x) dtan(x)= sec^(n-2)(x) tan(x) - tan(x) (n-2) sec^(n-2) tan(x) dx= sec^(n-2)(x) tan(x) - (n-2) sec^(n-2) tan^2(x) dx= sec^(n-2)(x) tan(x) - (n-2) sec^n(x) + (n-2)sec^(n-2)(x) dx, since tan^2(x) = sec^2(x) - 1= sec^(n-2)(x) tan(x) - (n-2) I + (n-2)sec^(n-2)(x) dxSolve for I, I = sec^(n-2)(x)tan(x))/(n-1)) + ((n-2)/(n-1)) sec^(n-2)(x) dx

3. Use integration by parts via sec^n(x) = sec^(n-2)(x) * sec^2(x).

Let u = sec^(n-2)(x), dv = sec^2(x) dxdu = (n-2) sec^(n-3)(x) * sec x tan x dx, v = tan x.

So, sec^n(x) dx= sec^(n-2)(x) tan x - (n-2) sec^(n-2)(x) tan^2(x) dx= sec^(n-2)(x) tan x - (n-2) sec^(n-2)(x) (sec^2(x) - 1) dx= sec^(n-2)(x) tan x - (n-2) sec^n(x) dx - (n-2) sec^(n-2)(x) dx.

Now, solve for sec^n(x) dx: sec^n(x) dx = sec^(n-2)(x) tan x - (n-2) sec^n(x) dx - (n-2) sec^(n-2)(x) dx==> sec^n(x) dx + (n-2) sec^n(x) dx = sec^(n-2)(x) tan x - (n-2) sec^(n-2)(x) dx==> (n-1) sec^n(x) dx = sec^(n-2)(x) tan x - (n-2) sec^(n-2)(x) dx==> sec^n(x) dx = [sec^(n-2)(x) tan x]/(n-1) - [(n-2)/(n-1)] sec^(n-2)(x) dx.