Integral of sec^3(x) = (sec(s) + tan(x))/2 + ln(sec(x) + tan(x))/2

Here is another solution that doesn't rely on knowing that integral, submitted by Igor Volkov: First, let I(x,y) stand for "the integral from x to y of ...", and sqr stand for "the square root of...".

To find the answer we must solve I(0,1) I(0,1) I(0,1) I(0,1) sqr((x-a)^{2}+(y-b)^{2}) db da dy dx.

Even the first integration of this is very messy, requiring a big table of integrals, and I think at best a small book of integrals would only take you through the second one. So we need to be more creative to make the problem more managable. First we introduce a change of variables:

- u=x-a
- s=x
- v=y-b
- t=y

Or:

- x=s
- y=t
- a=s-u
- b=t-v

**(1)**: Now we have I(0,1) I(t,t-1) I(0,1) I(s,s-1) sqr(u^{2} + v^{2}) * |J| du ds dv dt

Where |J| = the Jacobian =

| db/du db/ds db/dv du/dt | | 0 0 -1 1 | | | | | | da/du da/ds da/dv da/dt | | -1 1 0 0 | | | = | | = -1 | dy/du dy/ds dy/dv dy/dt | | 0 0 0 1 | | | | | | dx/du dx/ds dx/dv dx/dt | | 0 1 0 0 |

Now lets solve the following inner two integrals of equation (1):

**(2)**: I(0,1) I(s,s-1) sqr(u^{2} + v^{2}) du ds.

The area of integration is bounded by the quadrilateral with vertices (0,0), (1,1), (0,1), and (-1,0), with s being the horizontal axis and u being the vertical.

Changing the order of integral we get:

I(-1,0) I(0,u+1) sqr(u^{2} + v^{2}) ds du +
I(0,1) I(u,1) sqr(u^{2} + v^{2}) ds du =

I(1,0) sqr(u^{2} + v^{2}) * (1+u) du +
I(0,1) sqr(u^{2} + v^{2}) * (1-u) du =

-I(1,0) sqr(p^{2} + v^{2}) * (1-p) dp +
I(0,1) sqr(u^{2} + v^{2}) * (1-u) du
*(letting p=-u in the first integral)* =

I(0,1) sqr(p^{2} + v^{2}) * (1-p) dp +
I(0,1) sqr(u^{2} + v^{2}) * (1-u) du *(Notice the two terms are equal)*=

2* I(0,1) sqr(u^{2} + v^{2}) * (1-u) du.

Now put this into equation (1), and moving the constant terms to the left:

-2* I(0,1) I(t,t-1) I(0,1) sqr(u^{2} + v^{2}) * (1-u) du dv dt =

-2* I(0,1) I(0,1) I(t,t-1) sqr(u^{2} + v^{2}) * (1-u) dv dt du *(changing the order of integration)*=

-2* I(0,1) (1-u) I(0,1) I(t,t-1) sqr(u^{2} + v^{2}) dv dt du =

Note that the inner two integral are of the same form as equation (2), thus:

4* I(0,1) I(0,1) sqr(u^{2} + v^{2}) * (1-u) * (1-v) dv du =

4* I(0,1) I(0,1) sqr(u^{2} + v^{2}) * (1-u) * (1-v) du dv *(changing the order of integration)*

Next we make another substitution:

- u=r*sin(w)
- v=r*cos(w)

This time the Jacobian shall equal:

| du/dr du/dw | | sin(w) r*cos(w) | | | = | | = -r | dv/dr dv/dw | | cos(w) -r*sin(w) |

Thus the integral as a function of w and r is:

-8 * I(0,pi/4) I(0,1/cos(w)) -r * sqr(r*sin^{2}(w) + r*cos^{2}(w)) * (1-r*cos(w)) * (1-r*sin(w)) dr dw.

Note that we halved the area of integration, going from 0 to pi/4 as opposed to pi/2, and multiplying by 8 instead of the 4. This is permissible because of the symetry of the two regions.

= 8 * I(0,pi/4) I(0,1/cos(w)) r^{2} * (1-r*cos(w)) * (1-r*sin(w)) dr dw.

= 8 * I(0,pi/4) I(0,1/cos(w)) r^{2} - r^{3}*(cos(w)+sin(w)) +
r^{4}*(cos(w)+sin(w)) dr dw.

= 8 * I(0,pi/4) r^{3}/3 - r^{4}*(cos(w)+sin(w))/4 +
r^{5}*(cos(w)+sin(w))/5 ( from 0 to 1/cos(w) ) dw.

= 8 * I(0,pi/4) (3*cos^{3}(w))^{-1} - (cos(w)+sin(w))/(4*cos^{4})
+ (cos(w)+sin(w))/(5*cos^{5}(w)) dw.

= 8 * I(0,pi/4) (3*cos^{3}(w))^{-1} -
(4*cos^{3}(w))^{-1} -
sin(w)*(4*cos^{4}(w))^{-1} +
sin(w)*(5*cos^{4}(w))^{-1} dw.

= 8 * I(0,pi/4) (12*cos^{3}(w))^{-1} -
sin(w)*(20*cos^{4}(w))^{-1} dw.

**(3)** = 2/15 * I(0,pi/4) 5/cos^{3}(w) -
3*sin(w)/cos^{4}(w) dw.

Next we must dust off our table of integrals to help us with the integral of cos^{-3}(w).

I dx/cos^{n}(x) = (1/(n-1)) * sin(x)/cos^{n-1}(x) + ((n-2)/(n-1)) * I dx/cos^{n-2}(x).

In our case n=3 so,

I dx/cos^{3}(x) = (1/2) * sin(x)/cos^{2}(x) + (1/2) * I sec(x) dx.

Now lets look up the integral of sec(x):

I sec(x) dx = (1/2) * ln((1+sin(x))/(1-sin(x))).

Now we are ready to integrate cos^{-3}(w):

I(0,pi/4) cos^{-3}(w) dw = sin(w)/(2*cos^{2}(w)) (from 0 to pi/4) + (1/2)*I(0,pi/4) sec(x) dx.

= (1/2)*(2^{-1/2}/2^{-1} - 0) + (1/4)*ln((2+sqr(2))/(2-sqr(2))).

= (1/2)*sqr(2) + (1/4)*ln(((2+sqr(2))/2+sqr(2))/((2-sqr(2))*(2+sqr(2))).

= 1/sqr(2) + (1/4)*ln((4+4*sqr(2)+2)/(4-2)).

= 1/sqr(2) + (1/4)*ln(3+2*sqr(2)).

= 1/sqr(2) + (1/2)*ln(3+2*sqr(2))^{1/2}.

= 1/sqr(2) + (1/2)*ln(1+sqr(2)).

Now lets integrate sin(w)/cos^{4}(w).

let m=cos(w), thus dm=-sin(w) dw.

I(0,pi/4) sin(w)/cos^{4}(w).

= I(1,1/sqr(2)) -1*m^{-4} dm.

= 1/(3*m^{3}) from 0 to 1/sqr(2).

= (1/3) * (2*sqr(2) - 1).

Not it is time to plus these integrals into equation (3):

(2/15) * [ 5*(1/sqr(2) + (1/2)*ln(1+sqr(2))) + 3*((1/3) * (2*sqr(2) - 1)) ]

= (2/15) * [ 5*sqr(2)/2 + (5/2)*ln(1+sqr(2)) - 4*sqr(2)/2 + 1 ]

= (2/15) * [ sqr(2)/2 + (5/2)*ln(1+sqr(2)) + 1 ]

= (sqr(2) + 2 + 5*ln(1+sqr(2))) / 15.

Congradulations and thanks to

Michael Shackleford, A.S.A.