MATH 137
Trigonometric Integrals
Solutions
1. Standard Trig Functions:
(i) ∫ sin x dx = − cos x + C
(ii) ∫ cos x dx = sin x + C
(iii) ∫ tan x dx = − ln cos x + C
(iv) ∫ cot x dx = ln sin x + C
Proof. Let u = cos x . Then
and
du
= − sin x
dx
Proof. Let u = sin x . Then
du
= dx . Then
− sin x
sin x
sin x
du
∫ tan x dx = ∫ cos x dx = ∫ u × − sin x
= −∫
and
du
= cos x
dx
du
= dx . Then
cos x
cos x
cos x
du
∫ cot x dx = ∫ sin x dx = ∫ u × cos x
1
du = − ln u + C
u
=∫
= − ln cos x + C .
1
du = ln u + C
u
= ln sin x + C.
(v) ∫ sec x dx = ln sec x + tan x + C
(vi) ∫ csc x dx = − ln csc x + cot x + C
Proof. We alter the integral as follows:
Proof. We alter the integral as follows:
sec x + tan x
∫ sec x dx = ∫ sec x × sec x + tan x dx
csc x + cot x
∫ csc x dx = ∫ csc x × csc x + cot x dx
=∫
sec 2 x + sec x tan x
dx .
sec x + tan x
Now let u = sec x + tan x so that we
du
have
= sec x tan x + sec 2 x . Then
dx
∫ sec x dx = ∫
sec 2 x + sec x tan x
dx
sec x + tan x
sec 2 x + sec x tan x
du
×
u
sec x tan x + sec 2 x
1
= ∫ du = ln u + C = ln sec x + tan x + C.
u
=∫
=∫
csc2 x + csc x cot x
dx .
csc x + cot x
Now let u = csc x + cot x so that we
du
have
= − csc x cot x − csc2 x . Then
dx
∫ csc x dx = ∫
csc2 x + csc x cot x
dx
csc x + cot x
csc2 x + csc x cot x
du
×
u
− csc x cot x − csc 2 x
1
= − ∫ du = − ln u = − ln csc x + cot x + C .
u
=∫
2. Standard Hyperbolic Trig Functions:
(i) ∫ sinh x dx = cosh x + C
(ii) ∫ cosh x dx = sinh x + C
(iii) ∫ tanh x dx = ln(cosh x) + C
(iv) ∫ coth x dx = ln sinh x + C
Proof. Let u = cosh x . Then
and
du
= sinh x
dx
du
= dx . Then
sinh x
sinh x
sinh x
du
∫ tanh x dx = ∫ cosh x dx = ∫ u × sinh x
=∫
Proof. Let u = sinh x . Then
and
du
= cosh x
dx
du
= dx . Then
cosh x
cosh x
cosh x
du
∫ coth x dx = ∫ sinh x dx = ∫ u × cosh x
1
du = ln u + C
u
=∫
= ln cosh x + C = ln(cosh x ) + C,
1
du = ln u + C
u
= ln sinh x + C .
because cosh x > 0 for all x .
(v) ∫ sech x dx = 2 tan −1 (e x ) + C
(vi) ∫ csch x dx = −2 tanh −1(e x ) + C
Proof. We alter the integral as follows:
Proof. We alter the integral as follows:
1
2
∫ sech x dx = ∫ cosh x dx = ∫ x − x dx
e +e
=∫
1
2
∫ csch x dx = ∫ sinh x dx = ∫ x − x dx
e −e
2
2
dx = ∫ 2x
dx
1
e +1
ex + x
e
ex
=∫
ex
= 2 ∫ 2x
dx
e +1
Now let u = e x so that du = e x dx . Then
ex
sech
x
dx
=
2
dx
∫
∫ 2x
e +1
1
= 2∫ 2
du = 2 tan −1 (u) + C
u +1
= 2 tan −1 (e x ) + C
2
2
dx = ∫ 2x
dx
1
e −1
ex − x
e
ex
ex
ex
= 2 ∫ 2x
dx = −2 ∫
dx .
e −1
1 − e 2x
Now let u = e x so that du = e x dx . Then
∫ csch x dx = −2 ∫
= −2 ∫
ex
dx
1 − e2x
1
−1
2 du = −2 tanh (u) + C
1−u
= −2 tanh −1 (e x ) + C
3. Squares of Standard Trig Functions: We need the following identities:
cos2 x + sin 2 x = 1
1 + tan 2 x = sec 2 x
cot 2 x + 1 = csc 2 x
cos2 x − sin 2 x = cos(2x ) or (1 − sin 2 x ) − sin 2 x = cos(2x ) , which gives
2
1 − 2sin 2 x = cos(2 x) , and then sin x =
1 − cos(2 x)
.
2
Likewise, cos2 x − (1 − cos2 x ) = cos(2x ) , which gives
2
2 cos2 x − 1 = cos(2 x) , and then cos x =
x 1
− sin(2x ) + C
2 4
2
(i) ∫ sin x dx =
2
Proof. Using sin x =
1 + cos(2 x)
.
2
2
(ii) ∫ cos x dx =
x 1
+ sin(2x ) + C
2 4
1 − cos(2 x)
1 + cos(2 x)
2
, we have Proof. Using cos x =
, we have
2
2
1 1

2
∫ sin x dx = ∫  2 − 2 cos(2x ) dx
1
1 1
x − × sin(2 x) + C
2
2 2
x 1
= − sin(2 x) + C .
2 4
=
1 1

2
∫ cos x dx = ∫  2 + 2 cos(2x ) dx
1
1 1
x + × sin(2 x) + C
2
2 2
x 1
= + sin(2 x) + C .
2 4
=
(iii) ∫ sec 2 x dx = tan x + C
(iv) ∫ csc 2 x dx = − cot x + C
(v) ∫ tan 2 x dx = tan x − x + C
(vi) ∫ cot 2 x dx = − cot x − x + C
2
2
Proof. Using 1 + tan x = sec x , we have
2
2
∫ tan x dx = ∫ (sec x − 1) dx
= tan x − x + C.
2
2
Proof. Using cot x = csc x − 1 , we have
2
2
∫ cot x dx = ∫ (csc x − 1) dx
= − cot x − x + C.
4. Squares of Standard Hyperbolic Trig Functions: We need the following identities:
cosh 2 x − sinh 2 x = 1
1 − tanh 2 x = sech 2 x
coth 2 x − 1 = csch 2 x
cosh 2 x + sinh2 x = cosh(2x ) or (1 + sinh 2 x ) + sinh 2 x = cosh(2 x) , which gives
cosh(2 x) − 1
.
2
2
1 + 2 sinh2 x = cosh(2x ) , and then sinh x =
Likewise, cosh 2 x + (cosh 2 x − 1) = cosh(2x ) , which gives
2
2 cosh 2 x − 1 = cosh(2x ) , and then cosh x =
2
(i) ∫ sinh x dx =
Proof.
With
1
x
sinh(2 x) − + C
4
2
sinh2 x =
have
cosh(2 x) − 1
,
2
1
2
∫ sinh x dx = ∫  2 cosh(2 x ) −
cosh(2 x) + 1
.
2
1
x
sinh(2 x) + + C
4
2
2
(ii) ∫ cosh x dx =
2
we Proof. Uisng cosh x =
1
 dx
2
1 1
1
= × sinh(2 x) − x + C
2 2
2
1
x
= sinh(2 x) − + C .
4
2
cosh(2 x) + 1
,
2
1
1
2
∫ cosh x dx = ∫  2 cosh(2x ) + 2  dx
1 1
1
× sinh(2x ) + x + C
2 2
2
1
x
= sinh(2 x) + + C.
4
2
=
(iii) ∫ sech 2 x dx = tanh x + C
(iv) ∫ csch 2 x dx = − coth x + C .
(v) ∫ tanh 2 x dx = x − tanh x + C
(vi) ∫ coth 2 x dx = − coth x + x + C
2
2
2
2
Proof. Using 1 − tanh x = sech x , we have Proof. Using coth x = csch x + 1 , we have
2
2
∫ tanh x dx = ∫ (1 − sech x) dx
= x − tanh x + C.
2
2
∫ coth x dx = ∫ (csch x + 1) dx
= − coth x + x + C .
5. Cubes of Standard Trig Functions:
cos3 x
(i) ∫ sin x dx = − cos x +
+C
3
sin3 x
(ii) ∫ cos x dx = sin x −
+C
3
Proof.
Proof.
3
3
3
2
∫ sin x dx = ∫ sin x × sin x dx
3
2
∫ cos x dx = ∫ cos x × cos x dx
= ∫ sin x × (1 − cos2 x ) dx
= ∫ cos x × (1 − sin 2 x ) dx
= ∫ sin x dx − ∫ sin x × (cos 2 x ) dx
= ∫ cos x dx − ∫ cos x × (sin2 x ) dx
= − cos x +
cos3 x
+C
3
(The last integral is by int. by sub. using
u = cos x .)
(iii) ∫ tan 3 x dx =
tan 2 x
+ ln cos x + C
2
= sin x −
(The last integral is by int. by sub. using
u = sin x .)
∫ sec3 x dx
(iv)
=
Proof.
3
2
∫ tan x dx = ∫ tan x × tan x dx
= ∫ tan x × (sec2 x − 1) dx
= ∫ tan x sec 2 x dx − ∫ tan x dx
2
=
tan x
+ ln cos x + C
2
(The second to last integral is by int. by
sub. using u = tan x .)
sin3 x
+C
3
1
1
sec x tan x + ln sec x + tan x + C
2
2
Proof. By int. by parts:
u = sec x
u′ = sec x tan x
v ′ = sec 2 x dx
v = tan x
∫ sec3 x dx = u v − ∫ u′v
= sec x tan x − ∫ sec x tan 2 x dx
= sec x tan x − ∫ sec x (sec 2 x − 1) dx
= sec x tan x − ∫ sec3 x dx + ∫ sec x dx
So,
2 ∫ sec x dx = sec x tan x + ∫ sec x dx
3
and
1
1
3
∫ sec x dx = 2 sec x tan x + 2 ∫ sec x dx
=
1
1
sec x tan x + ln sec x + tan x + C
2
2