Thank you to Derek Wan and Andrei Lazanu for this solution which is composed from Derek's written work and Andrei's diagrams.
y=f(x)
Consider the graph of an arbitrary function y=f(x) defined over the interval a < x < b. We can see from the diagram that the rectangle with base b and height f(b) consists of the three smaller parts: the white part with area bf(b)-af(a) and the parts shaded yellow and green. The area of the part shaded green is òabf(x)dx). So
The yellow area+ ó
õ 		b

a 
 f(x)dx = bf(b) - af(a).

In the diagram mark the lines y=a and y=b in addition to y=f(a) and y=f(b) and reflect the graph of y=f(x) across the line y=x to obtain the inverse function y=f-1(x). Extending the lines y=a and y=b helps us to find x=f(a) and x=f(b) the bounds for integration. This gives us the yellow area and thus the formula:
ó
õ 		f(b)

f(a) 
 f-1(t)dt + ó
õ 		b

a 
 f(x)dx = bf(b) - af(a).

We can find ò14Öt dt by direct integration:
ó
õ 		4

1 
 Öt dt = é
ê
ë 		2
3
 t3/2 ù
ú
û 		4

1 
 = 14
3
 .
Or we can use the formula with f(x)=x2, f(a)=a2=1 so a=1 and f(b)=b2=4 so b=2
ó
õ 		4

1 
 Öt dt
= - ó
õ 		2

1 
 x2 dx + (2×4 - 1× 1)
= - é
ê
ë 		1
3
 x3 ù
ú
û 		2

1 
 +7
= - 7
3
 + 7
= 14
3
and this is in agreement with the previous result.

In a similar fashion we find ò01sin-1tdt with f(x)=sinx, f(a)=sina = 0 so a=0, and f(b)=sinb = 1 so b = p/2 :
ó
õ 		1

0 
 sin-1tdt + ó
õ 		p/2

0 
 sinx dx = p
2
 .
Hence
ó
õ 		1

0 
 sin-1tdt = p
2
 - [-cosx]0p/2 = p
2
 -1 .
Verification by direct integration:
ó
õ 		1

0 
 sin-1tdt
= [tsin-1t+   ____
Ö1-t2
 
 ]01
= (1×sin-11+   ____
Ö1-12
 
 )-(0× sin-10+   ____
Ö1-02
 
 )
=( p
2
 +0)- (0+1)= p
2
 -1
which is agreement with the previous result.

Editors note: this formula provides a method for carrying out integration where the function cannot be integrated directly but the inverse function can be integrated.