We follow the terminology of [1] [2] [3] . The Trigonometry had resulted from the continuous interaction between mathematics and astronomy, and this remained the special relationship between trigonometry and astronomy until the third century AD, when it began to disconnect Nasir al-Din al-Tusi (1201-1274) AD. In the middle of the seventeenth century when calculus has been developed by Issac Newton and that is done by inventing new form and relationship between mathematics and physical phenomena, moreover. Newton’s work proved many functions as a series of infinity. with respect x and then Newton obtained series since x and similar series of cosine of x as well as tangent x with the invention of calculus, and re-considered the trigonometric functions where still play an important role in both pure and applied mathematics analysis [4] [5] .

Theorem 1.1 [6] : For all k ∈ N . then:

1) sin 2 k − 1 ( x ) = ( − 1 ) k − 1 2 2 k − 2 ∑ n = 1 k ( − 1 ) n − 1 ( 2 k − 1 n − 1 ) sin ( 2 k − 2 n + 1 ) x .

2) sin 2 k ( x ) = 1 2 2 k ( 2 k k ) + ( − 1 ) k 2 2 k − 1 ∑ n = 0 k − 1 ( − 1 ) n ( 2 k n ) cos ( 2 k − 2 n ) x .

3) cos 2 k − 1 ( x ) = 1 2 2 k − 2 ∑ n = 1 k ( 2 k − 1 n − 1 ) cos ( 2 k − 2 n + 1 ) x .

4) cos 2 k ( x ) = 1 2 2 k ( 2 k k ) + 1 2 2 k − 1 ∑ n = 0 k − 1 ( 2 k n ) cos ( 2 k − 2 n ) x .#

Theorem 1.2: For all k ∈ N . then:

tan 2 k ( x ) = ( − 1 ) k + sec 2 ( x ) ∑ n = 1 k ( − 1 ) n + 1 tan 2 k − 2 n ( x ) .span class="bracketMark">(1.2.1)

tan 2 k + 1 ( x ) = ( − 1 ) k tan ( x ) + sec 2 ( x ) ∑ n = 1 k ( − 1 ) n + 1 tan 2 k − 2 n + 1 ( x ) .span class="bracketMark">(1.2.2)

cot 2 k ( x ) = ( − 1 ) k + csc 2 ( x ) ∑ n = 1 k ( − 1 ) n + 1 cot 2 k − 2 n ( x ) .span class="bracketMark">(1.2.3)

cot 2 k + 1 ( x ) = ( − 1 ) k cot ( x ) + csc 2 ( x ) ∑ n = 1 k ( − 1 ) n + 1 cot 2 k − 2 n + 1 ( x ) .span class="bracketMark">(1.2.4)

Proof:

1) We use mathematical induction on k, k ≥ 1 .

For k = 1 . we find by direct:

tan 2 ( x ) = − 1 + sec 2 ( x ) .

For k = 2 . then

tan 4 ( x ) = tan 2 ( x ) ( − 1 + sec 2 ( x ) ) = 1 − sec 2 ( x ) + tan 2 ( x ) sec 2 ( x ) = 1 + sec 2 ( x ) ( tan 2 ( x ) − 1 ) = ( − 1 ) 2 + sec 2 ( x ) ∑ n = 1 2 ( − 1 ) n + 1 tan 4 − 2 n (x)

We assume that (1.2.1) is true for k = m . m ≥ 1 .nd prove that is true for k = m + 1 .

tan 2 ( m + 1 ) ( x ) = tan 2 m ( x ) ⋅ tan 2 ( x ) = − tan 2 m ( x ) + tan 2 m ( x ) sec 2 (x)

For hypotheses induction, we get:

tan 2 ( m + 1 ) ( x ) = − [ ( − 1 ) m + sec 2 ( x ) ∑ n = 1 m ( − 1 ) n + 1 tan 2 m − 2 n ( x ) ] + sec 2 ( x ) tan 2 m ( x ) = ( − 1 ) m + 1 + sec 2 ( x ) ∑ n = 1 m + 1 ( − 1 ) n + 1 tan 2 ( m + 1 ) − 2 n (x)

Therefore, the relation is true for all k, k ≥ 1 .

2) Also use mathematical induction on k, k ≥ 1 .

For k = 1 . we find that:

tan 3 ( x ) = tan ( x ) ( − 1 + sec 2 ( x ) ) = − tan ( x ) + tan ( x ) sec 2 (x)

For k = 2 . then

tan 5 ( x ) = tan 3 ( x ) ( − 1 + sec 2 ( x ) ) = − tan 3 ( x ) + tan 3 ( x ) sec 2 ( x ) = tan ( x ) − sec 2 ( x ) tan ( x ) + sec 2 ( x ) tan 3 ( x ) = tan ( x ) + sec 2 ( x ) ( tan 3 ( x ) − tan ( x ) ) = ( − 1 ) 2 tan ( x ) + sec 2 ( x ) ∑ n = 1 2 ( − 1 ) n + 1 tan 5 − 2 n (x)

We assume that (1.2.2) is true for k = m . m ≥ 1 .nd prove that is true for k = m + 1 .

tan 2 ( m + 1 ) + 1 ( x ) = tan 2 m + 1 ( x ) ⋅ tan 2 ( x ) = − tan 2 m + 1 ( x ) + tan 2 m + 1 ( x ) sec 2 (x)

For hypotheses induction, we get:

tan 2 ( m + 1 ) + 1 ( x ) = − [ ( − 1 ) m tan ( x ) + sec 2 ( x ) ∑ n = 1 m ( − 1 ) n + 1 tan 2 m − 2 n + 1 ( x ) ] + sec 2 ( x ) tan 2 m + 1 ( x ) = ( − 1 ) m + 1 tan ( x ) + sec 2 ( x ) ∑ n = 1 m + 1 ( − 1 ) n + 1 tan 2 ( m + 1 ) + 1 − 2 n (x)

Therefore, the relation is true for all k, k ≥ 1 .

3) We can prove it with the same method.

4) We can prove it with the same method. #

Theorem 1.3: For all k ∈ N . then:

sec 2 k ( x ) = sec 2 ( x ) ∑ n = 0 k − 1 ( k − 1 n ) tan 2 n ( x ) . (1.3.1)

csc 2 k ( x ) = csc 2 ( x ) ∑ n = 0 k − 1 ( k − 1 n ) cot 2 n ( x ) . (1.3.2)

Proof:

1) sec 2 k ( x ) = sec 2 ( x ) ⋅ ( sec 2 ( x ) ) k − 1 = sec 2 ( x ) ⋅ ( 1 + tan 2 ( x ) ) k − 1 = sec 2 ( x ) ∑ n = 0 k − 1 ( k − 1 n ) tan 2 n ( x ) .by binomial theory).

2) csc 2 k ( x ) = csc 2 ( x ) ⋅ ( csc 2 ( x ) ) k − 1 = csc 2 ( x ) ⋅ ( 1 + cot 2 ( x ) ) k − 1 = csc 2 ( x ) ∑ n = 0 k − 1 ( k − 1 n ) cot 2 n ( x ) .by binomial theory).

In this section, we find the integrations of power of function trigonometric

Theorem 2.1: For all k ∈ N . then:

1) ∫ sin 2 k − 1 ( x ) d x = ( − 1 ) k − 1 2 2 k − 2 ∑ n = 1 k ( − 1 ) n − 1 ( 2 k − 1 n − 1 ) 2 k − 2 n + 1 cos ( 2 k − 2 n + 1 ) x + c

2) ∫ sin 2 k ( x ) d x = 1 2 2 k ( 2 k k ) x + ( − 1 ) k 2 2 k − 1 ∑ n = 0 k − 1 ( − 1 ) n ( 2 k n ) 2 k − 2 n sin ( 2 k − 2 n ) x + c

3) ∫ cos 2 k − 1 ( x ) d x = 1 2 2 k − 2 ∑ n = 1 k ( 2 k − 1 n − 1 ) 2 k − 2 n + 1 sin ( 2 k − 2 n + 1 ) x + c .

4) ∫ cos 2 k ( x ) d x = 1 2 2 k ( 2 k k ) x + 1 2 2 k − 1 ∑ n = 0 k − 1 ( 2 k n ) 2 k − 2 n sin ( 2 k − 2 n ) x + c . #

Proof: Directed from Theorem (1.1). #

Theorem 2.2: For all k ∈ N . then:

1) ∫ tan 2 k ( x ) d x = ( − 1 ) k x + ∑ n = 1 k ( − 1 ) n + 1 2 k − 2 n + 1 tan 2 k − 2 n + 1 ( x ) + c .

2) ∫ tan 2 k + 1 ( x ) d x = ( − 1 ) k + 1 ln | cos ( x ) | + ∑ n = 1 k ( − 1 ) n + 1 2 k − 2 n + 2 tan 2 k − 2 n + 2 ( x ) + c

3) ∫ cot 2 k ( x ) d x = ( − 1 ) k x + ∑ n = 1 k ( − 1 ) n 2 k − 2 n + 1 cot 2 k − 2 n + 1 ( x ) + c .

4) ∫ cot 2 k + 1 ( x ) d x = ( − 1 ) k ln | sin ( x ) | + ∑ n = 1 k ( − 1 ) n 2 k − 2 n + 2 cot 2 k − 2 n + 2 ( x ) + c

5) ∫ sec 2 k ( x ) d x = ∑ n = 0 k − 1 ( k − 1 n ) 2 n + 1 tan 2 n + 1 ( x ) + c .

6) ∫ csc 2 k ( x ) d x = − ∑ n = 0 k − 1 ( k − 1 n ) 2 n + 1 cot 2 n + 1 ( x ) + c .

Proof: Directed from Theorem (1.2) and Theorem (1.3). #

Lemma 2.3: For all k ∈ N . then:

∫ sec 2 k + 1 ( x ) d x = 1 2 k [ sec 2 k − 1 ( x ) tan ( x ) + ( 2 k − 1 ) ∫ sec 2 k − 1 ( x ) d x ]

Proof: ∫ sec 2 k + 1 ( x ) d x = ∫ sec 2 k − 1 ( x ) sec 2 ( x ) d x

Using integration by parts:

Let u = sec 2 k − 1 ( x ) . then d u = ( 2 k − 1 ) sec 2 k − 1 ( x ) tan ( x ) d x

d v = sec 2 ( x ) d x . then v = tan (x)

∴ ∫ sec 2 k + 1 ( x ) d x = sec 2 k − 1 ( x ) tan ( x ) − ( 2 k − 1 ) ∫ sec 2 k − 1 ( x ) tan 2 ( x ) d x = sec 2 k − 1 ( x ) tan ( x ) − ( 2 k − 1 ) ∫ sec 2 k − 1 ( x ) ( sec 2 ( x ) − 1 ) d x = sec 2 k − 1 ( x ) tan ( x ) − ( 2 k − 1 ) ∫ sec 2 k + 1 ( x ) d x + ( 2 k − 1 ) ∫ sec 2 k − 1 ( x ) d x

∴ 2 k ∫ sec 2 k + 1 ( x ) d x = sec 2 k − 1 ( x ) tan ( x ) + ( 2 k − 1 ) ∫ sec 2 k − 1 ( x ) d x .

Hence

∫ sec 2 k + 1 ( x ) d x = 1 2 k [ sec 2 k − 1 ( x ) tan ( x ) + ( 2 k − 1 ) ∫ sec 2 k − 1 ( x ) d x ] .#

Note: From clearly that:

∫ sec 3 ( x ) d x = 1 2 [ sec ( x ) tan ( x ) + ln | sec ( x ) + tan ( x ) | ] + c .

Theorem 2.4: For all k ∈ N . then:

∫ sec 2 k + 1 ( x ) d x = ∏ i = 1 k − 1 ( 2 i + 1 ) ∏ i = 1 k ( 2 i ) ln | sec ( x ) + tan ( x ) | + 1 2 k sec 2 k − 1 ( x ) tan ( x )     + tan ( x ) ∑ j = 1 k − 1 ∏ i = j k − 1 ( 2 i + 1 ) ∏ i = j k ( 2 i ) sec 2 i − 1 ( x ) + c

Proof: Since

∫ sec 2 k + 1 ( x ) d x = 1 2 k [ sec 2 k − 1 ( x ) tan ( x ) + ( 2 k − 1 ) ∫ sec 2 k − 1 ( x ) d x ]

(by Lemma 2.3). Then

∫ sec 2 k − 1 ( x ) d x = 1 2 k − 2 [ sec 2 k − 3 ( x ) tan ( x ) + ( 2 k − 3 ) ∫ sec 2 k − 3 ( x ) d x ]

and ∫ sec 2 k − 3 ( x ) d x = 1 2 k − 4 [ sec 2 k − 5 ( x ) tan ( x ) + ( 2 k − 5 ) ∫ sec 2 k − 5 ( x ) d x ] .

…

and so on

∫ sec 5 ( x ) d x = 1 4 [ sec 3 ( x ) tan ( x ) + 3 ∫ sec 3 ( x ) d x ]

and ∫ sec 3 ( x ) d x = 1 2 [ sec ( x ) tan ( x ) + ln | sec ( x ) + tan ( x ) | ] + c .

∴ ∫ sec 2 k + 1 ( x ) d x = 1 2 k sec 2 k − 1 ( x ) tan ( x ) + 2 k − 1 2 k ( 2 k − 2 ) sec 2 k − 3 ( x ) tan ( x )       + ( 2 k − 1 ) ( 2 k − 3 ) 2 k ( 2 k − 2 ) ( 2 k − 4 ) sec 2 k − 5 ( x ) tan ( x )       + ( 2 k − 1 ) ( 2 k − 3 ) ( 2 k − 5 ) 2 k ( 2 k − 2 ) ( 2 k − 4 ) ( 2 k − 6 ) sec 2 k − 7 ( x ) tan ( x )       + ( 2 k − 1 ) ( 2 k − 3 ) ( 2 k − 5 ) ( 2 k − 7 ) 2 k ( 2 k − 2 ) ( 2 k − 4 ) ( 2 k − 6 ) ( 2 k − 8 ) sec 2 k − 9 ( x ) tan (x)

+ ⋯ + ( 2 k − 1 ) ( 2 k − 3 ) ( 2 k − 5 ) ( 2 k − 7 ) ⋯ ( 5 ) 2 k ( 2 k − 2 ) ( 2 k − 4 ) ( 2 k − 6 ) ( 2 k − 8 ) ⋯ ( 4 ) sec 3 ( x ) tan ( x ) + ( 2 k − 1 ) ( 2 k − 3 ) ( 2 k − 5 ) ( 2 k − 7 ) ⋯ ( 5 ) ( 3 ) 2 k ( 2 k − 2 ) ( 2 k − 4 ) ( 2 k − 6 ) ( 2 k − 8 ) ⋯ ( 4 ) ( 2 ) sec ( x ) tan ( x ) + ( 2 k − 1 ) ( 2 k − 3 ) ( 2 k − 5 ) ( 2 k − 7 ) ⋯ ( 5 ) ( 3 ) 2 k ( 2 k − 2 ) ( 2 k − 4 ) ( 2 k − 6 ) ( 2 k − 8 ) ⋯ ( 4 ) ( 2 ) ln | sec ( x ) + tan ( x ) | + c

∴ ∫ sec 2 k + 1 ( x ) d x = ∏ i = 1 k − 1 ( 2 i + 1 ) ∏ i = 1 k ( 2 i ) ln | sec ( x ) + tan ( x ) | + 1 2 k sec 2 k − 1 ( x ) tan ( x )                                                     + tan ( x ) ∑ j = 1 k − 1 ∏ i = j k − 1 ( 2 i + 1 ) ∏ i = j k ( 2 i ) sec 2 i − 1 ( x ) + c .

Lemma 2.5: For all k ∈ N . then:

∫ csc 2 k + 1 ( x ) d x = 1 2 k [ − csc 2 k − 1 ( x ) cot ( x ) + ( 2 k − 1 ) ∫ csc 2 k − 1 ( x ) d x ]

Proof: ∫ csc 2 k + 1 ( x ) d x = ∫ csc 2 k − 1 ( x ) csc 2 ( x ) d x

Using integration by parts:

Let u = csc 2 k − 1 ( x ) . then d u = − ( 2 k − 1 ) csc 2 k − 1 ( x ) cot ( x ) d x

d v = csc 2 ( x ) d x . then v = − cot (x)

∴ ∫ csc 2 k + 1 ( x ) d x = − csc 2 k − 1 ( x ) cot ( x ) − ( 2 k − 1 ) ∫ csc 2 k − 1 ( x ) cot 2 ( x ) d x = − csc 2 k − 1 ( x ) cot ( x ) − ( 2 k − 1 ) ∫ csc 2 k − 1 ( x ) ( csc 2 ( x ) − 1 ) d x = − csc 2 k − 1 ( x ) cot ( x ) − ( 2 k − 1 ) ∫ csc 2 k + 1 ( x ) d x + ( 2 k − 1 ) ∫ csc 2 k − 1 ( x ) d x

∴ 2 k ∫ csc 2 k + 1 ( x ) d x = − csc 2 k − 1 ( x ) cot ( x ) + ( 2 k − 1 ) ∫ csc 2 k − 1 ( x ) d x

∴ ∫ csc 2 k + 1 ( x ) d x = 1 2 k [ − csc 2 k − 1 ( x ) tan ( x ) + ( 2 k − 1 ) ∫ csc 2 k − 1 ( x ) d x ] . .

Note: From clearly that: ∫ csc 3 ( x ) d x = − 1 2 [ csc ( x ) cot ( x ) + ln | csc ( x ) + cot ( x ) | ] + c .

Theorem 2.6: For all k ∈ N . then:

∫ csc 2 k + 1 ( x ) d x = − ∏ i = 1 k − 1 ( 2 i + 1 ) ∏ i = 1 k ( 2 i ) ln | csc ( x ) + cot ( x ) | − 1 2 k csc 2 k − 1 ( x ) cot ( x )                                               − cot ( x ) ∑ j = 1 k − 1 ∏ i = j k − 1 ( 2 i + 1 ) ∏ i = j k ( 2 i ) csc 2 i − 1 ( x ) + c

Proof: Since ∫ csc 2 k + 1 ( x ) d x = 1 2 k [ − csc 2 k − 1 ( x ) cot ( x ) + ( 2 k − 1 ) ∫ csc 2 k − 1 ( x ) d x ] .by Lemma 2.3) Then

∫ csc 2 k − 1 ( x ) d x = 1 2 k − 2 [ − csc 2 k − 3 ( x ) cot ( x ) + ( 2 k − 3 ) ∫ csc 2 k − 3 ( x ) d x ]

and ∫ csc 2 k − 3 ( x ) d x = 1 2 k − 4 [ − csc 2 k − 5 ( x ) cot ( x ) + ( 2 k − 5 ) ∫ csc 2 k − 5 ( x ) d x ] .

...

and so on

∫ csc 5 ( x ) d x = 1 4 [ − csc 3 ( x ) cot ( x ) + 3 ∫ csc 3 ( x ) d x ]

and ∫ csc 3 ( x ) d x = 1 2 [ − csc ( x ) cot ( x ) − ln | csc ( x ) + cot ( x ) | ] + c .

∴ ∫ csc 2 k + 1 ( x ) d x = − 1 2 k csc 2 k − 1 ( x ) cot ( x ) − 2 k − 1 2 k ( 2 k − 2 ) csc 2 k − 3 ( x ) cot ( x )       − ( 2 k − 1 ) ( 2 k − 3 ) 2 k ( 2 k − 2 ) ( 2 k − 4 ) csc 2 k − 5 ( x ) cot ( x )       − ( 2 k − 1 ) ( 2 k − 3 ) ( 2 k − 5 ) 2 k ( 2 k − 2 ) ( 2 k − 4 ) ( 2 k − 6 ) csc 2 k − 7 ( x ) cot ( x )       − ( 2 k − 1 ) ( 2 k − 3 ) ( 2 k − 5 ) ( 2 k − 7 ) 2 k ( 2 k − 2 ) ( 2 k − 4 ) ( 2 k − 6 ) ( 2 k − 8 ) csc 2 k − 9 ( x ) cot (x)

− ⋯ − ( 2 k − 1 ) ( 2 k − 3 ) ( 2 k − 5 ) ( 2 k − 7 ) ⋯ ( 5 ) 2 k ( 2 k − 2 ) ( 2 k − 4 ) ( 2 k − 6 ) ( 2 k − 8 ) ⋯ ( 4 ) csc 3 ( x ) cot ( x ) − ( 2 k − 1 ) ( 2 k − 3 ) ( 2 k − 5 ) ( 2 k − 7 ) ⋯ ( 5 ) ( 3 ) 2 k ( 2 k − 2 ) ( 2 k − 4 ) ( 2 k − 6 ) ( 2 k − 8 ) ⋯ ( 4 ) ( 2 ) csc ( x ) cot ( x ) − ( 2 k − 1 ) ( 2 k − 3 ) ( 2 k − 5 ) ( 2 k − 7 ) ⋯ ( 5 ) ( 3 ) 2 k ( 2 k − 2 ) ( 2 k − 4 ) ( 2 k − 6 ) ( 2 k − 8 ) ⋯ ( 4 ) ( 2 ) ln | csc ( x ) + cot ( x ) | + c

∴ ∫ csc 2 k + 1 ( x ) d x = − ∏ i = 1 k − 1 ( 2 i + 1 ) ∏ i = 1 k ( 2 i ) ln | csc ( x ) + cot ( x ) | − 1 2 k csc 2 k − 1 ( x ) cot ( x )                                                   − cot ( x ) ∑ j = 1 k − 1 ∏ i = j k − 1 ( 2 i + 1 ) ∏ i = j k ( 2 i ) csc 2 i − 1 ( x ) + c .

Theorem 3.1: For all n ∈ N . then:

1) ∫ sin n ( x ) cos ( x ) d x = { 1 n + 1 sin n + 1 ( x ) + c ,         n ≠ − 1 ln | sin ( x ) | + c ,                         n = − 1 .

2) ∫ cos n ( x ) sin ( x ) d x = { − 1 n + 1 cos n + 1 ( x ) + c ,         n ≠ − 1 − ln | cos ( x ) | + c ,                     n = − 1 .

3) ∫ tan n ( x ) sec 2 ( x ) d x = { 1 n + 1 tan n + 1 ( x ) + c ,         n ≠ − 1 ln | tan ( x ) | + c ,                             n = − 1 .

4) ∫ cot n ( x ) csc 2 ( x ) d x = { − 1 n + 1 cot n + 1 ( x ) + c ,         n ≠ − 1 − ln | cot ( x ) | + c ,                     n = − 1 .

5) ∫ sec n ( x ) tan ( x ) d x = { 1 n sec n ( x ) + c ,               n ≠ − 1 − cos ( x ) + c ,                     n = − 1 .

6) ∫ csc n ( x ) cot ( x ) d x = { − 1 n csc n ( x ) + c ,               n ≠ − 1 sin ( x ) + c ,                                   n = − 1 .

where c is integral constant.

Proof: Directed proof. #

Theorem 3.2: For all n , m ∈ N .nd m is an odd number, then: ∫ sin m ( x ) cos n ( x ) d x = ∑ i = 0 k ( − 1 ) i + 1 n + 1 + 2 i ( k i ) cos n + 1 + 2 i ( x ) + c .

Proof: Since m is an odd number, then ∃ k ∈ N .uch that m = 2 k + 1 .

∫ sin m ( x ) cos n ( x ) d x = ∫ sin 2 k + 1 ( x ) cos n ( x ) d x = ∫ ( sin 2 ( x ) ) k sin ( x ) cos n ( x ) d x = ∫ ( 1 − cos 2 ( x ) ) k sin ( x ) cos n ( x ) d x = ∫ ∑ i = 0 k ( − 1 ) i ( k i ) cos 2 i ( x ) sin ( x ) cos n ( x ) d x

= ∑ i = 0 k ( − 1 ) i ( k i ) ∫ cos n + 2 i ( x ) sin ( x ) d x = ∑ i = 0 k ( − 1 ) i + 1 n + 1 + 2 i ( k i ) cos n + 1 + 2 i ( x ) + c .

Theorem 3.3: For all n , m ∈ N .nd m is an even number, then:

∫ sin m ( x ) cos n ( x ) d x = { ∑ i = 0 k ( − 1 ) i ( k i ) [ 1 2 α ( α α 2 ) x + 1 2 α − 1 ∑ j = 0 α 2 − 1 ( α j ) α − 2 j sin ( α − 2 j ) x ] + c                                                                                                         if   α = 2 i + n ,     n   is   even ∑ i = 0 k ( − 1 ) i ( k i ) [ 1 2 α − 1 ∑ j = 1 α + 1 2 ( α − 1 j − 1 ) α − 2 j + 1 sin ( α − 2 j + 1 ) x ] + c                                                                                                           if   α = 2 i + n ,     n   is   odd .span class="bracketMark">(3.3)

Proof: Since m is an even number, then ∃ k ∈ N .uch that m = 2 k .

∫ sin m ( x ) cos n ( x ) d x = ∫ sin 2 k ( x ) cos n ( x ) d x = ∫ ( sin 2 ( x ) ) k cos n ( x ) d x = ∫ ( 1 − cos 2 ( x ) ) k cos n ( x ) d x = ∫ ∑ i = 0 k ( − 1 ) i ( k i ) cos 2 i + n ( x ) d x

By Theorem 2.1, we have (3.3).

Theorem 3.4: For all n , m ∈ N .nd m is an odd number, then:

∫ tan m ( x ) sec n ( x ) d x = ∑ i = 0 k ( − 1 ) k + i n + 2 i ( k i ) sec n + 2 i ( x ) + c .

Proof: Since m is an odd number, then ∃ k ∈ N .uch that m = 2 k + 1 .

∫ tan m ( x ) sec n ( x ) d x = ∫ tan 2 k + 1 ( x ) sec n ( x ) d x = ∫ ( tan 2 ( x ) ) k tan ( x ) sec n ( x ) d x = ∫ ( sec 2 ( x ) − 1 ) k tan ( x ) sec n ( x ) d x

= ∫ ( − 1 ) k ( 1 − sec 2 ( x ) ) k tan ( x ) sec n ( x ) d x = ∫ ( − 1 ) k ∑ i = 0 k ( − 1 ) i ( k i ) sec 2 i ( x ) tan ( x ) sec n ( x ) d x = ∑ i = 0 k ( − 1 ) k + i ( k i ) ∫ sec n + 2 i − 1 ( x ) sec ( x ) tan ( x ) d x = ∑ i = 0 k ( − 1 ) k + i n + 2 i ( k i ) sec n + 2 i ( x ) + c .

Theorem 3.5: For all n , m ∈ N .nd m is an even number, then: ∫ tan m ( x )   sec n ( x )   d x =

∫ tan m ( x ) sec n ( x ) d x = { ∑ i = 0 k ( − 1 ) k + i ( k i ) [ ∑ j = 0 α 2 − 1 ( α 2 − 1 j ) 2 j + 1 tan 2 j + 1 ( x ) ] + c ,       if   α = 2 i + n ,     n   is   even ∑ i = 0 k ( − 1 ) k + i ( k i ) [ 1 α − 1 sec α − 2 ( x ) tan ( x ) + ∏ m = 1 α − 3 2 ( 2 m + 1 ) ∏ m = 1 α − 1 2 ( 2 m ) ln | sec ( x ) + tan ( x ) |     + tan ( x ) ∑ j = 1 α − 3 2 ∏ m = j α − 3 2 ( 2 m + 1 ) ∏ m = j α − 1 2 ( 2 m ) sec 2 m − 1 ( x ) + c ] ,         if   α = 2 i + n ,     n   is   odd .span class="bracketMark">(3.5)

Proof: Since m is an even number, then ∃ k ∈ N .uch that m = 2 k .

∫ tan m ( x ) sec n ( x ) d x = ∫ tan 2 k ( x ) sec n ( x ) d x = ∫ ( tan 2 ( x ) ) k sec n ( x ) d x

= ∫ ( sec 2 ( x ) − 1 ) k sec n ( x ) d x = ∫ ( − 1 ) k ( 1 − sec 2 ( x ) ) k sec n ( x ) d x = ∫ ∑ i = 0 k ( − 1 ) k + i ( k i ) sec 2 i + n ( x ) d x

By Theorem 2.2, we have (3.5). #

Theorem 3.6: For all n , m ∈ N .nd m is an odd number, then: ∫ cot m ( x ) csc n ( x ) d x = ∑ i = 0 k ( − 1 ) k + i + 1 n + 2 i ( k i ) csc n + 2 i ( x ) + c .

Proof: Since m is an odd number, then ∃ k ∈ N .uch that m = 2 k + 1 .

∫ cot m ( x ) csc n ( x ) d x = ∫ cot 2 k + 1 ( x ) csc n ( x ) d x = ∫ cot 2 k ( x ) cot ( x ) csc n ( x ) d x = ∫ ( cot 2 ( x ) ) k cot ( x ) csc n ( x ) d x = ∫ ( csc 2 ( x ) − 1 ) k cot ( x ) csc n ( x ) d x = ∫ ( − 1 ) k ( 1 − csc 2 ( x ) ) k cot ( x ) csc n ( x ) d x = ∫ ( − 1 ) k ∑ i = 0 k ( − 1 ) i ( k i ) csc 2 i ( x ) cot ( x ) csc n ( x ) d x = ∑ i = 0 k − ( − 1 ) k + i ( k i ) ∫ csc n + 2 i − 1 ( x ) ( − csc ( x ) ) cot ( x ) d x = ∑ i = 0 k ( − 1 ) k + i + 1 n + 2 i ( k i ) csc n + 2 i ( x ) + c .

Theorem 3.7: For all n , m ∈ N .nd m is an even number, then: ∫ cot m ( x )   csc n ( x )   d x = ∫ cot m ( x ) csc n ( x ) d x = { − ∑ i = 0 k ( − 1 ) k + i ( k i ) [ ∑ j = 0 α 2 − 1 ( α 2 − 1 j ) 2 j + 1 cot 2 j + 1 ( x ) ] + c ,       if   α = 2 i + n ,     n   is     even − ∑ i = 0 k ( − 1 ) k + i ( k i ) [ 1 α − 1 csc α − 2 ( x ) cot ( x ) + ∏ m = 1 α − 3 2 ( 2 m + 1 ) ∏ m = 1 α − 1 2 ( 2 m ) ln | csc ( x ) + cot ( x ) |     + cot ( x ) ∑ j = 1 α − 3 2 ∏ m = j α − 3 2 ( 2 m + 1 ) ∏ m = j α − 1 2 ( 2 m ) csc 2 m − 1 ( x ) ] + c ,         if   α = 2 i + n ,     n   is   odd .span class="bracketMark">(3.7)

Proof: Since m is an even number, then ∃ k ∈ N .uch that m = 2 k

∫ cot m ( x ) csc n ( x ) d x = ∫ cot 2 k ( x ) csc n ( x ) d x = ∫ ( cot 2 ( x ) ) k csc n ( x ) d x = ∫ ( csc 2 ( x ) − 1 ) k csc n ( x ) d x

= ∫ ( − 1 ) k ∑ i = 0 k ( − 1 ) i ( k i ) csc 2 i + n ( x ) d x = ∫ ∑ i = 0 k ( − 1 ) k + i ( k i ) csc 2 i + n ( x ) d x = ∑ i = 0 k ( − 1 ) k + i ( k i ) ∫ csc 2 i + n ( x ) d x

Using Theorems 2.2 and 2.5, we have (3.7). #

Remark: There are many integration formulas, we can be finding it by previous results which obtained it from this paper. For example:

1) ∫ sin m ( x ) f ( x ) d x . ∫ cos m ( x ) f ( x ) d x .

where f ( x ) ∈ { tan n ( x ) , cot n ( x ) , sec n ( x ) , csc n ( x ) } . ∀ n , m ∈ N .

2) ∫ tan m ( x ) g ( x ) d x . where g ( x ) ∈ { cot n ( x ) , sec n ( x ) , csc n ( x ) } . ∀ n , m ∈ N .

3) ∫ cot m ( x ) sec n ( x ) d x .

The combinations of the exponential functions exp ( x ) .nd exp ( − x ) .alled hyperbolic functions. These functions which arise in various engineering applications, have many properties in common with the trigonometric functions. The hyperbolic functions have resulted from vibratory motions inside elastic solid and more general in many problems where mechanical energy is gradually absorbed by a surrounding medium. There are importance applications on hyperbolic functions and power integration of hyperbolic functions in physics, mathematics transformations and numerical analysis.

Remark: The proofs for this following result are similarly of the results for previous sections.

There are many integration formulas, we can find it by previous results which obtained it from this paper. For example:

1) ∫ sinh m ( x ) f ( x ) d x . ∫ cosh m ( x ) f ( x ) d x .

where f ( x ) ∈ { tanh n ( x ) , coth n ( x ) , sech n ( x ) , csch n ( x ) } . ∀ n , m ∈ N .

2) ∫ tanh m ( x ) g ( x ) d x . where

g ( x ) ∈ { coth n ( x ) , sech n ( x ) , csc h n ( x ) } . ∀ n , m ∈ N .

3) ∫ coth m ( x ) sech n ( x ) d x .

The authors declare no conflicts of interest regarding the publication of this paper.

Ali, A.M. and Abd-Alkanee, O.M. (2019) The Power Integrations of Trigonometric and Hyperbolic Functions. Open Access Library Journal, 6: e5618. https://doi.org/10.4236/oalib.1105618