Solve the integral
1
I=
2
Z
dx
=
π/4
sec x −
0
!
1
2
2
(cos x + sin x)
dx
1. Solution 1:
Note that
sec2 x
Z
(1 + tan x)
2
Z
1
du
u2
Z
u−2 du =
=
with u = 1 + tan x, du = sec2 x dx
u−1
1
1
+c=− +c=−
+ c.
−1
u
1 + tan x
Our integral is
I
=
=
=
=
Z
π/4
1
2
!
!
Z
1
1 π/4
sec x −
sec2 x −
dx
=
2
2 0
(cos x + sin x)
0
cos2 x 1 +
!
Z π/4
π/4
1
1
sec2 x
1
tan x +
sec2 x −
dx =
2
2 0
2
1 + tan x 0
(1 + tan x)
1
1
π
1
− tan 0 +
tan +
2
4
1 + tan π4
1 + tan 0
1
1
1
1+
− [0 + 1] = .
2
1+1
4
1
2
sin x 2
cos x
dx
2. Solution 2: Rewrite the integral as
Z
I
π/4
=
0
Z
=
0
π/4
1
2
1
(sec x) −
cos x + sin x
1
2
2
(f (θ)) − (g(θ)) dθ
2
2
2 !
Z
π/4
dx =
0
1
2
If r = f (θ) = sec x, then
r = sec x
1
⇒ r=
cos x
⇒ r cos θ = 1
⇒ x=1
If r = g(θ) =
1
, then
cos x + sin x
1
cos x + sin x
⇒ r (cos x + sin x) = 1
r=
⇒ r cos x + r sin x = 1
⇒ x+y =1
1
2
(sec θ) −
1
cos θ + sin θ
2 !
dθ
Note that θ =
π
is the line y = x. This gives the picture
4
The triangle on the right is the area we want.
This is given by the integral
Z 1
Z
(x − (1 − x)) dx =
1/2
1
2
1
1 1
1
−
(2x − 1) dx = x − x 1/2 = [1 − 1] −
= .
4
2
4
1/2
3. Solution 3:
The area of the triangle in Solution 3 is
1
1
bh = (1)
2
2
2
1
1
= .
2
4