Some integration examples
Useful identities
sin2 x + cos2 x = 1
tan2 x + 1 = sec2 x
cos 2x = 2 cos2 x − 1 = 1 − 2 sin2 x
sin 2x = 2 sin x cos x
cos(a + b) + cos(a − b) = 2 cos a cos b
sin(a + b) + sin(a − b) = 2 sin a cos b
Z
Z
sin xdx = − cos x + C
cos xdx = sin x + C
Z
Z
sec2 dx = tan x + C
cosh2 x − sinh2 x = 1
tan xdx = ln | sec x| + c
1 − tanh2 x = sech2 x
cosh 2x = 2 cosh2 x − 1 = 2 sinh2 x + 1
Z
Zsinh 2x = 2 sinh x cosh x
cosh xdx = sinh x + C
sinh xdx = cosh x + C
Examples
Z
1. Find
π
4
2 sin2 2xdx.
0
We re-arrange the integrand using the formula cos 2x = 1 − 2 sin2 x. This is because we can
integrate a double angle directly.
π
4
Z
π
4
Z
2
2 sin 2xdx =
0
(1 − cos 4x)dx
0
x−
=
π
.
4
=
Z
2. Find
π
2
sin 4x
4
π
4
0
sin3 xdx.
0
We re-arrange the integrand using the formula sin2 x = 1 − cos2 x.
Z
π
2
sin3 xdx =
Z
sin x(1 − cos2 x)dx
0
Z
=
π
2
(sin x − sin x cos2 x)dx.
0
Now we observe
that − sin x is the derivative of cos x so that the second term is a sight integral
R 0
of the form f (x)f n (x)dx.
Z
Thus
π
2
sin3 xdx =
0
π
2
Z
(sin x − sin x cos2 x)dx
0
"
=
cos3 x
− cos x +
3
1
= − −1 +
3
2
=
.
3
1
#π
2
0
Z
3. Find
π
3
2 tan3 xdx.
0
We re-arrange the integrand using the formula tan2 x = sec2 x − 1.
π
3
Z
2 tan3 xdx =
0
π
3
Z
(2 tan x)(sec2 x − 1)dx
0
π
3
Z
=
(2 tan x sec2 x − 2 tan x)dx.
0
Now we observe
that sec2 x is the derivative of tan x so that the first term is a sight integral
R 0
of the form f (x)f n (x)dx.
π
3
Z
2 tan3 xdx =
π
3
Z
0
(2 sec2 x tan x − 2 tan x)dx
0
=
h
tan2 x − 2 ln | sec x|
iπ
3
0
= 3 − 2 ln 2.
Z
4. Find
π
3
2 sin 2x cos 3xdx.
0
We re-arrange the integrand using the formula sin(a + b) + sin(a − b) = 2 sin a cos b.
Z
π
3
Z
2 sin 2x cos 3xdx =
0
π
3
(sin(5x) + sin(−x))dx
0
Z
=
π
3
(sin(5x) − sin(x))dx
0
π
3
− cos(5x)
=
+ cos x
5
0
1
1 1
−2
= − + + −1=
.
10 2 5
5
Z
5. Find
p
2 x2 − 1dx, for x ≥ 1.
It is often a√good idea √
to try the substitutions x = cos u or x = cosh u for integrals containing
2
the terms 1 − x or x2 − 1 respectively (where there is no factor x).
p
√
√
dx
Since x ≥ 1 let x = cosh u. Then x2 − 1 = cosh2 u − 1 = sinh2 u = sinh u and
=
du
sinh u.
Z
p
2 x2 − 1dx =
Z
Z
=
2 sinh2 udu
(cosh 2u − 1)du using the ‘double angle’ formula
sinh 2u
−u+C
2
= sinh u cosh u − u + C
=
p
= x x2 − 1 − cosh−1 (x) + C.
Note that the integral is only defined if x ≤ −1 or x ≥ 1. In the former case we would let
x = − cosh u and then proceed in the same way.
2