Search for notes by fellow students, in your own course and all over the country.
Browse our notes for titles which look like what you need, you can preview any of the notes via a sample of the contents. After you're happy these are the notes you're after simply pop them into your shopping cart.
Title: Derivative and integral of elementary trigonometric function
Description: The document has derivative and integral of elementary trigonometric functions. The skill to find derivative and integration have been welded as one interactive concept
Description: The document has derivative and integral of elementary trigonometric functions. The skill to find derivative and integration have been welded as one interactive concept
Document Preview
Extracts from the notes are below, to see the PDF you'll receive please use the links above
y = sin �
Differentiating with respect to �
dy
d�
=
d
d�
d
d�
(sin �) = cos �
(sin �) = cos �
d(sin �) = cos �d� --------- (1)
Since dC = 0, for constant C
⟹
d(sin �) = d(sin �) + 0 = d(sin �) + dC = d(sin � + C) ----- (2)
From equation (1) and equation (2)
cos �d� = d(sin � + C)
Taking integrals (antiderivatives) on both sides
∫cos � d�
= ∫d(sin � + C)
∫cos �d� = sin � + C
...
y=
sin �
cos �
Using the formula
d u
d� v
du
=
dv
v d� - u d�
v2
Putting
u = sin �, v = cos �
d
sin �
d
=
d� cos �
d
cos � � (sin �) - sin � d� (cos �)
cos 2 �
As
d
(sin �) = cos �,
(cos �) = - sin �
d�
d�
Therefore
d
d
sin �
d� cos �
d
sin �
d� cos �
=
cos � cos � - sin �(-sin �)
=
cos 2 �
cos 2 � + sin 2 �
2
cos �
=
1
2
cos �
= sec 2 �
Since
sin �
= tan �
cos �
d
sin �
=
d� cos �
⟹
d
d�
d
d�
(tan �) = sec 2 �
(tan �) = sec 2 �
d(tan �) = sec 2 �d� ------- (5)
d(tan �) = d(tan �) + 0 = d(tan �) + d(C) = d(tan � + C) -------- (6)
From equation (5) & equation (6)
sec 2 �d� = d(tan � + C)
Taking integrals on both sides
∫sec 2 �d� = ∫d(tan � + C) = tan � + C
∫sec 2 �d� = tan � + C
...
y=
1
sin �
Using the formula
d u
d� v
du
=
dv
v d� - u d�
v2
Here
u = 1, v = sin �
d
1
d� sin �
sin �
=
d
d
(1) - (1) d� (sin �)
d�
sin 2 �
Here
d
d
(1) = 0,
(sin �) = cos �
d�
d�
d
1
d� sin �
d
=
(sin �)(0) - (1)(cos �)
sin 2 �
1
= -
d� sin �
cos �
1
sin �
sin �
=
0 - cos �
sin 2 �
= -
cos �
sin 2 �
= - cot � cosec �
Putting
1
sin �
d
1
d� sin �
d
d�
=
d
d�
= cosec �
(cosec �) = - cot � cosec �
(cosec �) = - cot � cosec �
d(cosec �) = - cot � cosec � d� ----------- (9)
For some constant K
d(cosec �) = d(cosec �) + 0 = d(cosec �) + d(K) = d(cosec � + K) ---------- (10)
From equation (9) and equation (10)
-cot � cosec �d� = d(cosec � + K)
Taking integrals on both sides
∫ - cot � cosec �d� = ∫d(cosec � + K)
-∫cot � cosec �d� = cosec � + K
∫cot � cosec �d� = -cosec � - K
Putting C = -K
∫cot � cosec � d� = -cosec � + C
...
∫tan �d� = ∫ sin � d� = ∫ sin �d�
cos �
Since
cos �
d(cos �) = - sin �d�
-d(cos �) = sin � d�
Therefore
∫tan �d� = ∫ sin �d� = ∫ -d(cos �) = - ∫ d(cos �)
cos �
cos �
cos �
Using the formula
∫ dx = ln x
x
∫tan �d� = -∫ d(cos �) =
cos �
- ln cos � + C
---------------------------------
∫cot �d� = ∫ cos � d� = ∫ d(sin �)
sin �
sin
= ln sin � + C
Title: Derivative and integral of elementary trigonometric function
Description: The document has derivative and integral of elementary trigonometric functions. The skill to find derivative and integration have been welded as one interactive concept
Description: The document has derivative and integral of elementary trigonometric functions. The skill to find derivative and integration have been welded as one interactive concept