integral por substituição

Prévia do material em texto

Integral por Substituição 
1) a) ∫ 2𝑥(𝑥2 + 1)23𝑑𝑥 = ∫(𝑢)23𝑑𝑢 = ∫
𝑢24
24
𝑑𝑢 =
(𝑥2+1)
24
24
+ 𝐶 
𝑢 = 𝑥2 + 1 
𝑢′ = 2𝑥 
𝑑𝑢
𝑑𝑥
= 2𝑥 
𝑑𝑢 = 2𝑥𝑑𝑥 
 
b) ∫ 𝑐𝑜𝑠3𝑥 𝑠𝑒𝑛𝑥 𝑑𝑥 = − ∫ 𝑢3𝑑𝑢 = − ∫
𝑢4
4
𝑑𝑢 =
−𝑐𝑜𝑠𝑥4
4
+ 𝐶 
𝑢 = 𝑐𝑜𝑠𝑥 
𝑢′ = −𝑠𝑒𝑛𝑥 
𝑑𝑢
𝑑𝑥
= −𝑠𝑒𝑛𝑥 
𝑑𝑢 = −𝑠𝑒𝑛𝑥 𝑑𝑥 
 
c)∫
1
√𝑥
𝑠𝑒𝑛√𝑥 𝑑𝑥 = ∫ 2𝑠𝑒𝑛(𝑢)𝑑𝑢 = 2 ∫ 𝑠𝑒𝑛(𝑢)𝑑𝑢 = 2 − cos(𝑢) = 2 −
cos(√𝑥) + 𝐶 
 
d) ∫
3𝑥𝑑𝑥
√4𝑥2+5 
= 3 ∫
𝑑𝑢
8√𝑢
= 3.
1
8
∫
1
√𝑢
𝑑𝑢 =
3
4
√4𝑥2 + 5 + 𝐶 
𝑢 = 4𝑥2 + 5 
𝑢′ = 8𝑥 
𝑑𝑢
𝑑𝑥
= 8𝑥 
𝑑𝑢 = 8𝑥𝑑𝑥 
 
2) a) ∫ 𝑠𝑒𝑐2(4𝑥 + 1)𝑑𝑥 = ∫ 𝑠𝑒𝑐2(𝑢) 
𝑑𝑢
4
=
1
4
∫ 𝑡𝑔(𝑢) =
1
4
𝑡𝑔(4𝑥 + 1) + 𝐶 
𝑢 = 4𝑥 + 1 
𝑢′ = 4 
 
𝑑𝑢
𝑑𝑥
= 4 
 𝑑𝑢 = 4𝑑𝑥 
b) ∫ 𝑦√1 + 2𝑦² 𝑑𝑦 = 
1
4
∫ 𝑢
1
2 𝑑𝑢 =
1
6
𝑢
3
2 =
1
6
(1 + 2𝑦2)
3
2 + 𝐶 
𝑢 = 1 + 2𝑦² 
𝑢′ = 4𝑦 
 
𝑑𝑢
𝑑𝑦
= 4𝑦 
 𝑑𝑢 = 4𝑦𝑑𝑦 
c) ∫(2𝑥 + 7)(𝑥2 + 7𝑥 + 3)
4
5𝑑𝑥 = ∫
(2𝑥+7)(𝑢)
4
5
2𝑥+7
𝑑𝑢 = ∫(𝑢)
4
5𝑑𝑢 =
5
9
(𝑥2 + 7𝑥 +
3)
9
5 + 𝐶 
 
𝑢 = 𝑥2 + 7𝑥 + 3 
𝑢′ = 2𝑥 + 7 
 
𝑑𝑢
𝑑𝑥
= 2𝑥 + 7 
 𝑑𝑢 = 2𝑥 + 7𝑑𝑥 
3) a) ∫(1 + 𝑠𝑒𝑛 𝑡)9 cos 𝑡 𝑑𝑡 = ∫ 𝑢9𝑑𝑢 = ∫
𝑢9+1
9+1
𝑑𝑢 = ∫
𝑢10
10
𝑑𝑢 =
1
10
𝑢10 =
1
10
(1 + 𝑠𝑒𝑛 𝑡)10 + 𝐶 
 
𝑢 = 1 + 𝑠𝑒𝑛 𝑡 
𝑢′ = cos 𝑡 
 
𝑑𝑢
𝑑𝑡
= cos 𝑡 
 𝑑𝑢 = cos 𝑡 𝑑𝑡 
b) ∫ cos 2𝑥 𝑑𝑥 = ∫
cos(𝑢)
2
𝑑𝑢 =
1
2
∫ cos(𝑢)𝑑𝑢 =
1
2
𝑠𝑒𝑛 (2𝑥) + 𝐶 
𝑢 = 2𝑥 
𝑢′ = 2 
 
𝑑𝑢
𝑑𝑥
= 2 
 𝑑𝑢 = 2𝑑𝑥 
4) a) ∫ 𝑥2√1 + 𝑥 𝑑𝑥 = ∫(𝑢 − 1)2 𝑢
1
2 𝑑𝑢 = ∫(𝑢2 − 2𝑢 − 12) 𝑢
1
2𝑑𝑢 =
∫ (𝑢
5
2 − 2𝑢
3
2 − 𝑢
5
2) 𝑑𝑢 =
𝑢
5
2
+1
5
2
+1
−
2𝑢
3
2
+1
3
2
+1
−
𝑢
5
2
+1
5
2
+1
=
𝑢
7
2
7
2
−
2𝑢
5
2
5
2
−
𝑢
7
2
7
2
=
2
7
𝑢
7
2 −
4
5
𝑢
5
2 −
2
7
𝑢
7
2 =
2
7
(1 + 𝑥)
7
2 −
4
5
(1 + 𝑥)
5
2 −
2
7
(1 + 𝑥)
7
2 + 𝐶 
 
𝑢 = 1 + 𝑥 → 𝑥 = 𝑢 − 1 
𝑢′ = 1 
 
𝑑𝑢
𝑑𝑥
= 1 
 𝑑𝑢 = 𝑑𝑥 
b) ∫ 𝑠𝑒𝑛 (𝑥 − 𝜋)𝑑𝑥 = ∫ 𝑠𝑒𝑛(𝑢)𝑑𝑢 = ∫ − cos(𝑢) 𝑑𝑢 = − cos(𝑥 − 𝜋) + 𝐶 
𝑢 = 𝑥 − 𝜋 
𝑢′ = 1 
 𝑑𝑢 = 𝑑𝑥 
c) ∫
5𝑥4
(𝑥5+1)2
𝑑𝑥 = ∫
𝑑𝑢
(𝑢)²
=
1
𝑢
=
1
(𝑥5+1)
+ 𝐶 
𝑢 = 𝑥5 + 1 
𝑢′ = 5𝑥4 
 
𝑑𝑢
𝑑𝑥
= 5𝑥4 
 𝑑𝑢 = 5𝑥4𝑑𝑥 
 
5) a) ∫
𝑑𝑥
𝑥𝑙𝑛𝑥
= ∫
1
𝑢
𝑑𝑢 = ln(𝑢) = ln(𝑙𝑛𝑥) + 𝐶 
𝑢 = 𝑙𝑛𝑥 
𝑢′ =
1
𝑥
 
 
𝑑𝑢
𝑑𝑥
=
1
𝑥
 
 𝑑𝑢 =
1
𝑥
𝑑𝑥 
 b) ∫ 𝑒−5𝑥𝑑𝑥 = ∫ 𝑒𝑢
𝑑𝑢
−5
= −
1
5
∫ 𝑒𝑢𝑑𝑢 = −
1
5
𝑒𝑢 = −
1
5
𝑒−5𝑥 + 𝐶 
 
𝑢 = −5𝑥 
𝑢′ = −5 
 
𝑑𝑢
𝑑𝑥
= −5 
 𝑑𝑢 = −5𝑑𝑥 
 c) ∫
𝑠𝑒𝑛3𝜃
1+𝑐𝑜𝑠3𝜃
𝑑𝜃 = ∫
𝑠𝑒𝑛3𝜃
𝑢
 
𝑑𝑢
𝑠𝑒𝑛3𝜃
= −
1
3
∫
1
𝑢
𝑑𝑢 = −
1
3
𝑙𝑛|𝑢| = −
1
3
𝑙𝑛|1 + 𝑐𝑜𝑠3𝜃| +
𝐶 
 
𝑢 = 1 + 𝑐𝑜𝑠3𝜃 
𝑢′ = −𝑠𝑒𝑛3𝜃 
 
𝑑𝑢
𝑑𝑥
= −𝑠𝑒𝑛3𝜃 
 𝑑𝑢 = −𝑠𝑒𝑛3𝜃𝑑𝜃 
 d) ∫
𝑒𝑥
1+𝑒𝑥
𝑑𝑥 = ∫
𝑒𝑥
𝑢
 
𝑑𝑢
𝑒𝑥
= ∫
1
𝑢
𝑑𝑢 = 𝑙𝑛|𝑢| = 𝑙𝑛|1 + 𝑒𝑥| + 𝐶 
 
𝑢 = 1 + 𝑒𝑥 
𝑢′ = 𝑒𝑥 
 
𝑑𝑢
𝑑𝑥
= 𝑒𝑥 
 𝑑𝑢 = 𝑒𝑥𝑑𝑥 
6) a) ∫
𝑑𝑥
𝑥√1−(𝑙𝑛𝑥)²
= ∫ 𝑥√1 − 𝑢² 
𝑑𝑢
𝑥
= ∫
1
√1−𝑢²
𝑑𝑢 = 𝑎𝑟𝑐𝑠𝑒𝑛 𝑢 = 𝑎𝑟𝑐𝑠𝑒𝑛(𝑙𝑛|𝑥|) +
𝐶 
𝑢 = ln 𝑥 
𝑢′ =
1
𝑥
 
 
𝑑𝑢
𝑑𝑥
=
1
𝑥
 
 𝑑𝑢 =
𝑑𝑥
𝑥
 
 b) ∫
𝑑𝑥
𝑥 √9𝑥2−1
= ∫ 𝑢 − 3 √𝑢2 − 1 
𝑑𝑢
3𝑥
= ∫
𝑑𝑢
𝑢√𝑢2−1
= 𝑎𝑟𝑐𝑠𝑒𝑛(𝑢) = 𝑎𝑟𝑐𝑠𝑒𝑛 (3𝑥) +
𝐶 
𝑢² = 9𝑥² 
𝑢 = 3𝑥 
 
𝑑𝑢
𝑑𝑥
= 3𝑥 
 𝑑𝑢 = 3𝑥𝑑𝑥 
 c) ∫
𝑑𝑥
√𝑥 (1+𝑥)
= ∫ 𝑢(1 + 𝑢) 2𝑑𝑢 = 2 ∫
𝑑𝑢
(1+𝑢2)
= 2𝑎𝑟𝑐𝑡𝑔(𝑢) = 2𝑎𝑟𝑐𝑡𝑔 (√𝑥 ) + 𝐶 
𝑢 = √𝑥 
𝑢′ =
1
2
𝑥−
1
2 
 
𝑑𝑢
𝑑𝑥
=
𝑥
−
1
2
2
 
 𝑑𝑢 =
𝑥
−
1
2
2
𝑑𝑥 
7) ∫(4𝑥 − 3)9𝑑𝑥 = ∫
𝑢9+1
9+1
 
𝑑𝑢
4
=
1
4
∫
𝑢10
10
𝑑𝑢 =
1
4
.
1
10
 𝑢10 =
1
40
(4𝑥 − 3)10 + 𝐶 
𝑢 = 4𝑥 − 3 
𝑢′ = 4 
 
𝑑𝑢
𝑑𝑥
= 4 
 𝑑𝑢 = 4𝑑𝑥 
8) ∫ 𝑥3√5 + 𝑥4𝑑𝑥 = ∫ 𝑥3𝑢
1
2 
𝑑𝑢
4𝑥³
=
1
4
. 2 ∫ 𝑢
1
2𝑑𝑢 =
1
6
𝑢
3
2 =
1
6
(5 + 𝑥4)
3
2 + 𝐶 
𝑢 = 5 + 𝑥4 
𝑢′ = 4𝑥³ 
 
𝑑𝑢
𝑑𝑥
= 4𝑥³ 
 𝑑𝑢 = 4𝑥³𝑑𝑥 
9) ∫ 𝑠𝑒𝑛 7𝑥 𝑑𝑥 = ∫ − cos(𝑢)
𝑑𝑢
7
=
1
7
∫ − cos(𝑢) 𝑑𝑢 =
1
7
− cos(𝑢) = −
1
7
cos(7𝑥) +
𝐶 
𝑢 = 7𝑥 
𝑢′ = 7 
 
𝑑𝑢
𝑑𝑥
= 7 
 𝑑𝑢 = 7𝑑𝑥 
10) ∫ sec 4𝑥 𝑡𝑔 4𝑥 𝑑𝑥 = ∫ sec(𝑢) 𝑡𝑔 (𝑢)
𝑑𝑢
4
=
1
4
∫ sec(𝑢) =
1
4
sec(4𝑥) + 𝐶 
𝑢 = 4𝑥 
𝑢′ = 4 
 
𝑑𝑢
𝑑𝑥
= 4 
 𝑑𝑢 = 4𝑑𝑥