Subido por marlith betsaida simon santiago

# EJERCICIOS RESUELTOS

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```EJERCICIOS RESUELTOS
1) ∫(𝑐𝑜𝑠 4 𝑥 − 𝑠𝑒𝑛4 𝑥)𝑑𝑥 = ∫ 𝑐𝑜𝑠 4 𝑥𝑑𝑥 − ∫
= ∫(𝑐𝑜𝑠 2 𝑥 − 𝑠𝑒𝑛2 𝑥) (𝑐𝑜𝑠
⏟ 2 𝑥 + 𝑠𝑒𝑛2 𝑥) 𝑑𝑥 = ∫(𝑐𝑜𝑠 2 𝑥 − 𝑠𝑒𝑛2 𝑥) 𝑑𝑥
1
= ∫ 𝑐𝑜𝑠 2 𝑥𝑑𝑥 − ∫ 𝑠𝑒𝑛2 𝑥𝑑𝑥 = ∫
𝑥
𝑥
= 2−2+∫
𝑐𝑜𝑠2𝑥 𝑑𝑥.2
2
2
+ ∫
𝑥
= 0 + 4 ∫ 𝑐𝑜𝑠2𝑥. 𝑑(2𝑥) +
1
= 2 ∫ 𝑐𝑜𝑠2𝑥𝑑(2𝑥) =
2
𝑢2 −2
1 (√2+4𝑥)
1
8
1
𝑑𝑥
𝑑𝑥
2
∫ 𝑐𝑜𝑠2𝑥. (2𝑑𝑥)
4
+𝑐
𝑢2 −2
4
=𝑥
1 𝑢3
3
1
− 2√4 (2 + 𝑥)] + 𝐶 =
3
=√2 + 𝑥 [
2
− 2√2 + 4𝑥] + 𝐶
1
2
1
+𝑥
2
√ +𝑥[
2
1−𝑐𝑜𝑠2𝑥
3
3
1
𝑑𝑥 − ∫
∫(𝑢2 − 2)𝑑𝑢 = 8 [ 3 − 2𝑢] + 𝐶
8
(√4( +𝑥))
[
.
1
=∫ 4𝑥2 𝑢 . 𝑢𝑑𝑢 =
= 8[
2
𝑢2 = 2 + 4𝑥
2𝑢 𝑑𝑢 = 4𝑑𝑥
𝑑𝑢
𝑑𝑥 =
. 2𝑢
4
𝑢𝑑𝑢
𝑑𝑥 =
2
𝑢𝑑𝑢
.
1
2
𝑥
∫
2
𝑠𝑒𝑛 (2𝑥)
2) ∫
𝑑𝑥
√2+4𝑥
𝑢2 −2
4
√𝑢2
𝑐𝑜𝑠2𝑥
1𝑥𝑐𝑜𝑠2𝑥
3
]−
2+4𝑥−6
12
1
2
√ +𝑥
2
1
= √2 + 𝑥 [
1
] + 𝐶 = √2 + 𝑥 [
1
8
1+2𝑥
6
3
1
2
8(√ +𝑥)
[
1
− 2.2√2 + 𝑥]
3
1
− 2] + 𝐶
4(𝑥−1)
(𝑥−1)
12
3
]=
1
√ +𝑥+𝐶
2
8 𝐶𝑂𝑆√𝑥
3) ∫0
√𝑥 = 𝑡 ⇒ 𝑥 = 𝑡 2 ⟹ 𝑑𝑥 = 2𝑡𝑑6
8
𝑑𝑥 = 𝐹(𝑥) ∫0
√𝑥
⟹∫
8 𝑐𝑜𝑠√𝑥
=∫0 𝑥 𝑑𝑥
√
=
8 𝑐𝑜𝑠√𝑥
lim ∫0 𝑥 𝑑𝑥
√
𝑎→0
cos 𝑡
. 2𝑡𝑑𝑡
𝑡
= 2 ∫ 𝑐𝑜𝑠 𝑡𝑑𝑡 = 2𝑠𝑒𝑛 𝑡 + 𝐶
=⏟
2𝑠𝑒𝑛√𝑥 + C
𝐹(𝑥)
=lim ∫ 2𝑠𝑒𝑛√𝑥
𝑎→0
8
∫0
= lim (2𝑠𝑒𝑛√8 − 2𝑠𝑒𝑛√𝑎 )
𝑎→0
=2𝑠𝑒𝑛(2√2) − ⏟
lim 2𝑠𝑒𝑛√𝑎 = 2𝑠𝑒𝑛 (2√2)
𝑎→0
0
1
4) ∫ 𝑠𝑒𝑐𝑥𝑑𝑥 = ∫ 𝑐𝑜𝑠𝑥 𝑑𝑥
𝜋⁄
2
5) ∫0
𝑒 𝑠𝑒𝑛𝑥
𝑡𝑔𝑥 .
𝑒 𝑠𝑒𝑛𝑥
𝑒 𝑠𝑒𝑛𝑥
. 𝑑𝑥 ⇒ ∫ 𝑡𝑔𝑥.𝑐𝑜𝑠𝑒𝑐𝑥 𝑑𝑥 = ∫ 𝑠𝑒𝑛𝑥
𝑐𝑜𝑠𝑒𝑐𝑥
𝑐𝑜𝑠𝑥
.
1
𝑠𝑒𝑛𝑥
𝑑𝑥
= ∫ 𝑒 𝑠𝑒𝑛𝑥 . 𝑐𝑜𝑠𝑥𝑑𝑥 = ∫ 𝑒 𝑠𝑒𝑛𝑥 . 𝑑(𝑠𝑒𝑛𝑥) = 𝑒⏟𝑠𝑒𝑛𝑥 + 𝐶
𝜋⁄
2
=∫0
𝜋⁄
2
𝑒 𝑠𝑒𝑛𝑥
𝑑𝑥 = 𝑒 𝑠𝑒𝑛𝑥 ∫0
𝑡𝑔𝑥.𝑐𝑜𝑠𝑒𝑐𝑥
= 𝑒 𝑠𝑒𝑛(
𝜋⁄ )
2
− 𝑒 𝑠𝑒𝑛(0)
=𝑒 1 − 𝑒 0 = 𝑒 − 1
𝑒
6) ∫1 𝑥(𝑙𝑜𝑔𝑥)2 𝑑𝑥
𝑢 = (𝑙𝑜𝑔𝑥)2
𝑑𝑢 = 2𝑙𝑜𝑔𝑥
= ∫ 𝑥(𝑙𝑜𝑔𝑥)2 𝑑𝑥 = 𝑥(𝑙𝑜𝑔𝑥)2 − ∫ 𝑥. 2𝑙𝑜𝑔𝑥.
= 𝑥(𝑙𝑜𝑔𝑥)2 − 2 ∫ 𝑙𝑜𝑔𝑥𝑑𝑥
𝑑𝑣 = 𝑑𝑥
𝑑𝑥
𝑥
𝑑𝑥
𝑥
𝑣=𝑥
𝐼. 𝑃. 𝑢 = 𝑙𝑜𝑔𝑥
𝑑𝑥
𝑑𝑢 =
𝑥
= 𝑥(log(𝑥))2 − 2 [𝑥𝑙𝑜𝑔𝑥 − ∫
𝑥.𝑑𝑥
𝑥
]
2
= 𝑥(𝑙𝑜𝑔𝑥)
− 2𝑥𝑙𝑜𝑔𝑥 + 2𝑥 + 𝐶
⏟
𝑓(𝑥)
𝑒
𝑒
=∫1 𝑥(𝑙𝑜𝑔𝑥)2 𝑑𝑥 = 𝑥(𝑙𝑜𝑔𝑥)2 − 2𝑥𝑙𝑜𝑔𝑥 + 2𝑥 ∫1
=𝑒(𝑙𝑜𝑔𝑒)2 − 2𝑒𝑙𝑜𝑔𝑒 + 2𝑒 − 1(log(1))2 + 2(1) log(1) − 2(1)
=𝑒 − 2
𝑑𝑣 = 𝑥
𝑣=𝑥
7) ∫
5+4
1
√𝑥+𝑥
𝐶. 𝑉. √𝑋 = 𝑡 ⟹ 𝑥 = 𝑡 2 ⟹ 𝑑𝑥 = 2𝑡𝑑𝑡
𝑑𝑥
1
2𝑡𝑑𝑡
2𝑡+4−4
(2𝑡+4)𝑑𝑡
4𝑡𝑑
=∫ 5+4𝑡+𝑡 2 . 2𝑡𝑑𝑡 = ∫ 𝑡 2 +4𝑡+5 = ∫ 𝑡 2 +4𝑡+5 𝑑𝑡 = ∫ 𝑡 2 +4𝑡+5 − ∫ (𝑡+2)2 +1
=𝑙𝑛|𝑡 2 + 4𝑡 + 5| − 4𝑥 𝑎𝑟𝑐𝑡𝑔(𝑡 + 2) + 𝐶
8)
2
9) ∫0
2
𝑥
(𝑥−3)(𝑥+5)2
𝑑𝑥 = 𝐹(𝑥) ∫0
𝐻𝑎𝑙𝑙𝑎𝑚𝑜𝑠 𝐹(𝑥) =
∫
𝑥
𝐴
𝐵
𝐶
𝑑𝑥 = ∫ [
+
+
] 𝑑𝑥
2
(𝑥 − 3)(𝑥 + 5)
𝑥 − 3 𝑥 + 5 (𝑥 + 5)2
⟹
𝑥
𝐴
𝐵
𝐶
𝐴(𝑥 + 5)2 + 𝐵(𝑥 − 3)(𝑥 + 5) + 𝐶(𝑥 − 3)
=
+
+
=
(𝑥 − 3)(𝑥 + 5)2 𝑥 − 3 𝑥 + 5 (𝑥 + 5)2
(𝑥 + 3)(𝑥 + 5)2
⟹ 𝑥 = 𝐴(𝑥 + 5)2 + 𝐵(𝑥 − 3)(𝑥 + 5) + 𝐶(𝑥 − 3)
Si 𝑥 = 3 ⇒ 3 = 𝐴(8)2 ⇒ 𝐴 = 3⁄64
Si 𝑥 = −5 ⇒ −5 = 𝐶(−5 − 3) ⇒ 𝐶 = 5⁄8
3
5
Si 𝑥 = 0 ⇒ 0 = 64 (5)2 + 𝐵(−3)(5) + 8 (−3)
0=
⟹
3𝑋25
64
− 15𝐵 −
15
8
⟹−
15
8
75
+ 64 = 15𝐵 ⇒
15+8
64
75
+ 64 = 15𝐵
75 − 120
−45
−3
= 15𝐵 ⟹
= 15𝐵 ⇒ 𝐵 =
64
64
64
3⁄
−3⁄
5⁄
𝑥
64
64
8 ] 𝑑𝑥
∴∫
𝑑𝑥 = ∫ [
+
+
(𝑥 − 3)(𝑥 + 5)2
𝑥 − 3 𝑥 + 5 (𝑥 + 5)2
⟹
3
−3
5 −1
|𝑛|𝑥 − 3|
|𝑛|𝑥 + 5| +
+𝐶
64
64
8 𝑥+5
2
2
𝑥
3
−3
−5
|𝑛|𝑥 − 3|
|𝑛|𝑥 + 5|
⇒∫
𝑑𝑥 =
∫
2
64
64
8(𝑥 + 5) 0
0 (𝑥 − 3)(𝑥 + 5)
3
𝑥−3
3
−1
2
5
3
2−3
−5
3
−3
5
=64 𝑙𝑛 (𝑥+5) − 8(𝑥+5) ∫0 = 64 𝑙𝑛 (2+5) 8(2+5) − 64 𝑙𝑛 ( 5 ) + 8
5
3
−3
1
1
7
5
3
−1⁄
=64 𝑙𝑛 ( 7 ) − 56 − 64 𝑙𝑛 ( 5 ) + 8 = 8 𝑥 7 − 56 + 64 𝑙𝑛 (−3⁄7)
5
=
2
56
10) ∫
+
3
64
5
1
21
28
𝑙𝑛 ( ) =
log(1+𝑙𝑜𝑔𝑥)
𝑥
−
𝐵
64
21
𝑙𝑛 ( )
5
𝑑𝑥
𝐶. 𝑉. 1 + 𝑙𝑜𝑔𝑥 = 𝑡
𝑑𝑥
⟹ ∫ 𝑙𝑜𝑔𝑡. 𝑑𝑡 = 𝑡𝑙𝑜𝑔(𝑡) − ∫ 𝑑𝑡
𝑥
= 𝑑𝑡
=𝑡𝑙𝑜𝑔(𝑡) − 𝑡 + 𝐶
=(1 + 𝑙𝑜𝑔𝑥) log(1 + 𝑙𝑜𝑔𝑥) − 1 − 𝑙𝑜𝑔𝑥 + 𝐶
=log(1 + 𝑙𝑜𝑔𝑥) + 𝑙𝑜𝑔𝑥. log(1 + 𝑙𝑜𝑔𝑥) − 1 − 𝑙𝑜𝑔𝑥 + 𝐶
=log(1 + 𝑙𝑜𝑔𝑥) + 𝑙𝑜𝑔𝑥[log(1 + 𝑙𝑜𝑔𝑥) − 1] + 𝐶1
1−𝑠𝑒𝑛2 𝑥
1
11) ∫ √𝑐𝑜𝑠𝑐𝑥 − 𝑠𝑒𝑛𝑥 𝑑𝑥 = ∫ √𝑠𝑒𝑛𝑥 − 𝑠𝑒𝑛𝑥 𝑑𝑥 = ∫ √
𝑠𝑒𝑛𝑥
𝑑𝑥
1 − 𝑠𝑒𝑛2 𝑥
𝑐𝑜𝑠𝑥 𝑑𝑥
𝑑(𝑠𝑒𝑛𝑥)
√
∫
𝑑𝑥 = ∫
=∫
= √𝑠𝑒𝑛𝑥 + 𝐶
𝑠𝑒𝑛𝑥
√𝑠𝑒𝑛𝑥
√𝑠𝑒𝑛𝑥
1
12) ∫ √𝑥 2
+25
𝑑𝑥
Sustituci&oacute;n trigonom&eacute;trico
𝑥 = 5𝑡𝑔𝑡
𝑑𝑥 = 5𝑠𝑒𝑐 2 𝑡𝑑𝑡
√𝑥 2 + 25
= √25𝑡𝑔2 𝑡 + 25
= 5√𝑡𝑔2 𝑡 + 1
= 5𝑠𝑒𝑐𝑡
⟹∫
5𝑠𝑒𝑐 2 𝑡. 𝑑𝑡
5𝑠𝑒𝑐𝑡
⟹ ∫ 𝑠𝑒𝑐𝑡𝑑𝑡 = ∫
𝑑𝑡
𝑐𝑜𝑠𝑡
𝑒 𝑙𝑜𝑔2 𝑥−1
13) ∫2
∫
𝑥𝑙𝑜𝑔2 𝑥
𝑙𝑜𝑔2 𝑥
.
𝑥
⇒∫
𝑙𝑜𝑔2 𝑥−1 𝑑𝑥
𝑙𝑜𝑔2 𝑥
.
𝑥
𝑙𝑜𝑔𝑥 = 𝑡
𝑑𝑡
= 𝑑𝑡
𝑥
𝑡2 − 1
1
1
. 𝑑𝑡 = ∫(1 − 𝑡 −2 )𝑑𝑡 = 𝑡 + + 𝐶 = 𝑙𝑜𝑔𝑥 +
+𝐶
2
𝑡
𝑡
𝑙𝑜𝑔𝑥
(𝑙𝑜𝑔𝑥)2 +1
=
𝑒 𝑙𝑜𝑔2 𝑥−1 𝑑𝑥
𝑑𝑥 = ∫2
𝑙𝑜𝑔𝑥
𝑒
⟹∫
2
+𝐶
𝑙𝑜𝑔2 𝑥 − 1
𝑙𝑜𝑔2 𝑥 + 1 𝑒 𝑙𝑛2 𝑒 + 1 𝑙𝑜𝑔2 2 + 1
𝑑𝑥
=
∫ =
−
𝑥𝑙𝑜𝑔2 𝑥
𝑙𝑜𝑔𝑥
𝑙𝑜𝑔𝑒
𝑙𝑜𝑔2
2
𝑙𝑜𝑔2 2 + 1 2𝑙𝑜𝑔2 − 𝑙𝑜𝑔2 2 + 1 −(𝑙𝑜𝑔2 2 − 2𝑙𝑜𝑔2 + 1)
=2−
=
=
𝑙𝑜𝑔2
𝑙𝑜𝑔2
𝑙𝑜𝑔2
=
−(𝑙𝑜𝑔2 − 1)2
𝑙𝑜𝑔2
14) ∫ 𝑒 3𝑥 . 𝑎𝑟𝑡𝑔𝑒 𝑥 3𝑑𝑥
𝑒 𝑥 = 𝑡 ⟹ 𝑒 𝑥 𝑑𝑥 = 𝑑𝑡
∫(𝑒 𝑥 )3 . 𝑎𝑟𝑡𝑔𝑒 𝑥 . 𝑑𝑥 = ∫(𝑒 𝑥 )2 𝑎𝑟𝑡𝑔(𝑒 𝑥 ). 𝑒 𝑥 𝑑𝑥
=∫ 𝑡 2 . 𝑎𝑟𝑡𝑔𝑡. 𝑑𝑡
𝑢 = 𝑎𝑟𝑐𝑡𝑔𝑡
𝐼 = ∫ 𝑡 2 𝑎𝑟𝑡𝑔𝑡. 𝑑𝑡
𝑑𝑢 =
𝑑𝑡
1 + 𝑡2
𝑑𝑣 = 𝑡 2 𝑑𝑡
𝑣=
𝑡3
3
𝑡3
𝑡 3 𝑑𝑡
𝑡3
1
𝑡 2 . 𝑡𝑑𝑡
𝐼 = 𝑎𝑟𝑐𝑡𝑔𝑡 − ∫ .
= 𝑎𝑟𝑐𝑡𝑔𝑡. ∫
3
3 1 + 𝑡2
3
3
1 + 𝑡2
𝑡3
1
𝑡2 + 1 − 1
= 𝑎𝑟𝑐𝑡𝑔𝑡 − ∫ ( 2
) 𝑡𝑑𝑡
3
3
𝑡 +1
=
=
𝑡3
1
1 2𝑡𝑑𝑡
𝑡3
𝑡2
1
𝑎𝑟𝑐𝑡𝑔𝑡 − [∫ 𝑡𝑑𝑡 − ∫ 2
] = 𝑎𝑟𝑐𝑡𝑔𝑡 −
+ ln(𝑡 2 + 1) + 𝐶
3
3
2 𝑡 +1
3
3𝑥2 6
𝑒 3𝑥
3
𝑎𝑟𝑐𝑡𝑔(𝑒 𝑥 ) −
𝑒 2𝑥
6
1
+ 6 𝑙𝑜𝑔(𝑒 2𝑥 + 1) + 𝐶
|𝑥−1|
4
15) ∫0
|𝑥−2|+|𝑥−3|
|𝑥 − 1| = {
𝑑𝑥
|𝑥 − 2| = {
𝑥−1
−𝑥 + 1
𝑠𝑖 𝑥 ≥ 1
𝑠𝑖 𝑥 &lt; 0
𝑥−2
−𝑥 + 2
𝑠𝑖 𝑥 ≥ 2
𝑠𝑖 𝑥 &lt; 2
|𝑥 − 3| = {
𝑥−3
−𝑥 + 3
𝑠𝑖 𝑥 ≥ 3
𝑠𝑖 𝑥 &lt; 3
|𝑥 − 1| = −𝑥 + 1
|𝑥 − 2| = −𝑥 + 2
|𝑥 − 3| = −𝑥 + 3
0
1
=∫0
1 −𝑥+1
=∫0
2
−𝑥+1
−𝑥+2−𝑥+3
𝑑𝑥 + ∫1
𝑑𝑥 + ∫1
1 −2𝑥+5−3
=∫0
1 𝑑𝑥
2
𝑥
1
1
− 3 ∫0
3
= ∫0 −
2
1
2
1
3
=2 − 2 ∫0
−3
1
∫0
𝑑𝑥
𝑥
−2𝑥+5
2𝑥−5
2
2
2
3
𝑑𝑥
+
−2𝑥+5
𝑑𝑥
4
1
2
3
1
4
−2𝑑𝑥
−2𝑥+5
2(2𝑥−5)
3
∫2 + ∫3
4
3
4
2
1
3
3
1
2𝑥−5
3
3
3
−3
4
4
9
3
1
3
3
3
1
3
3
4
2
2
4
4
9
3
𝑙𝑛5 + ln(3) + + + ln(3) − (0)
3
3
1
=4 ln(3) − 4 ln(5) + 2 + 2 = 2 + 4 ln(3) − 4 ln(5)
16)
4
2𝑑𝑥
2(2𝑥−5)
4 𝑑(2𝑥−5)
2𝑥−5
3
4
ln(−2𝑥 + 5) ∫1 + 2 + 2 + 4 ln(2𝑥 − 5) ∫3
3
3𝑑𝑥
2(2𝑥−5)
4 𝑑𝑥
=4 [𝑙𝑛3 − 𝑙𝑛5] − 4 [ln(1) − ln(3)] + 2 + 2 + 4 [ln(3) − ln(1)]
= 𝑙𝑛3
4
+ ∫3
+ 2 + 2 − 2 + 2 ∫3
+ 2 + 2 + 4 ∫3
2
𝑑𝑥
4 𝑑𝑥
2
3
3
𝑑𝑥
𝑑𝑥
(𝑥−1)2
𝑥
−2(−2𝑥+5)
2 𝑑(−2𝑥+5)
3
+ −4 ∫1
−𝑥+1
−𝑥+2−𝑥+3
4 2𝑥−5+3
+ 2 ∫1
− 1 + 2 + 2 ∫1
1 −3
=−4 . ln(−2𝑥 + 5) ∫0
3
4 𝑥−1
𝑑𝑥 + ∫3
− ∫1 +
+ − + ∫3 + ∫3
2
−2𝑥+5
2
2
2
2
−2𝑥+5
−3
𝑑𝑥 + ∫3
3
2 𝑑𝑥
− ∫1
2(−2𝑥+5)
1 𝑑(−2𝑥+5)
1
1
3
4
−𝑥+1
−𝑥+2−𝑥+3
𝑑𝑥 + ∫2 (𝑥 − 1)𝑑𝑥 + ∫3
2(−2𝑥+5)
𝑑𝑥
−2𝑑𝑥
(−2)(−2𝑥+5)
= 2 . (−2) ∫0
𝑑𝑥 + ∫2
2 −2𝑥+5−3
𝑑𝑥 + ∫1
2(−2𝑥+5)
=∫0
𝑑𝑥 + ∫2
3 𝑥−1
−2𝑥+5
2
3
−𝑥+1
−𝑥+2−𝑥+3
2 𝑥−1
−2𝑥+5
1
4
1+𝑒 𝑥
17) ∫
𝑑𝑥 = ∫
1−𝑒 𝑥
1+𝑒𝑥
𝑒𝑥
1−𝑒𝑥
𝑒𝑥
𝑒 −𝑥 −1
𝑑𝑥 = ∫
𝑒 𝑥 +1
𝑒 −𝑥 −1
𝑑𝑥 = ∫
2𝑑𝑥
=∫ −𝑥 𝑑𝑥 + ∫ −𝑥 = 𝑥 + 2 ∫
𝑒 −1
𝑒 −1
=𝑥 − 2 ∫
−𝑒 𝑥 𝑑𝑥
1−𝑒 𝑥
𝑑𝑥
1−𝑒𝑥
𝑒𝑥
𝑒 −𝑥 +1+2
𝑒 −1−1
𝑑𝑥
= 𝑥 + 2∫
𝑒 𝑥 𝑑𝑥
1−𝑒 𝑥
= 𝑥 − 2 log(1 − 𝑒 𝑥 ) + 𝐶
18) ∫ 𝑡𝑔4 𝑥. 𝑑𝑥 = ∫ 𝑡𝑔2 𝑥. 𝑡𝑔2 𝑥𝑑𝑥 = ∫(𝑠𝑒𝑐 2 𝑥 − 1)2 𝑑𝑥
2
=∫ 𝑠𝑒𝑐 4 𝑥. 𝑑𝑥 − ∫ 2𝑠𝑒𝑐 𝑥. 𝑑𝑥 + ∫ 𝑑𝑥
=∫ 𝑠𝑒𝑐 2 𝑥. 𝑠𝑒𝑐 2 𝑥. 𝑑𝑥 − 2𝑡𝑔𝑥 + 𝑥 = ∫(𝑡𝑔2 𝑥 + 1)𝑠𝑒𝑐 2 𝑥. 𝑑𝑥 − 2𝑡𝑔𝑥 + 𝑥
=∫ 𝑡𝑔2 𝑥. 𝑠𝑒𝑐 2 𝑥. 𝑑𝑥 + ∫ 𝑠𝑒𝑐 2 𝑥. 𝑑𝑡 − 2𝑡𝑔𝑥 + 𝑥
𝑡𝑔3 𝑥
=
3
+ 𝑡𝑔𝑥 − 2𝑡𝑔𝑥 + 𝑥 + 𝐶 =
𝑡𝑔3 𝑥
3
− 𝑡𝑔𝑥 + 𝑥 + 𝐶
𝑠𝑒𝑛2 𝑥
𝑠𝑒𝑛2 𝑥
19) ∫ 𝑠𝑒𝑛𝑥. 𝑡𝑔2 𝑥. 𝑑𝑥 = ∫ 𝑠𝑒𝑛𝑥. 2 𝑑𝑥 = ∫
. 𝑠𝑒𝑛𝑥. 𝑑𝑥
𝑐𝑜𝑠 𝑥
1−𝑠𝑒𝑛2 𝑥
=∫
1−𝑐𝑜𝑠 2 𝑥
𝑐𝑜𝑠 2 𝑥
𝑠𝑒𝑛𝑥. 𝑑𝑥 = ∫
=∫ 𝑑(𝑐𝑜𝑠𝑥) − ∫
𝑥+1
𝑑(𝑐𝑜𝑠𝑥)
𝑐𝑜𝑠 2 𝑥
20) ∫ 2
𝑑𝑥 = ∫
𝑥 +2𝑥+3
1
𝑐𝑜𝑠 2 𝑥
= 𝑐𝑜𝑠𝑥 +
𝑥+1
2
(𝑥+1)2 +(√2)
= log(𝑥 2 + 2𝑥 + 3) + 𝐶
2
(𝑐𝑜𝑠 2 𝑥−1)
− 𝑠𝑒𝑛𝑥. 𝑑𝑥 = ∫
1
𝑐𝑜𝑠𝑥
𝑑𝑥 = ∫
+𝐶
2𝑥+2
𝑥 2 2𝑥+3
𝑐𝑜𝑠 2 𝑥−1
𝑐𝑜𝑠 2 𝑥
. 𝑑(𝑐𝑜𝑠𝑥)
```