вправа 14.35 гдз 11 клас алгебра Істер 2019
Вправа 14.35
Обчисліть інтеграл:
Умова:
Відповідь ГДЗ:
π/3
✔ 1) ∫cos7х • cos5хdx
0
cos7х • cos5хdx = 1/2(cos2х + cos12х)
π/3
∫ 1/2(cos2х + cos12х)dx =
0
π/3 π/3
= 1/2∫cos2хdx + 1/2∫cos12хdx =
0 0
π/3 π/3
= 1/2 • 1/2 sin2х | + 1/2 • 1/12 sin 12х | =
0 0
π/3 π/3
= 1/4 sin2х | + 1/24 sin12х | =
0 0
= 1/4 sin 2 • π/3 - 0 + 1/24 sin12 • π/3 - 0 =
= 1/4 sin 2π/3 + 1/24 sin4π = 1/4 • √3/2 + 0 =
= √3/8;
π
✔ 2) ∫ sin23хdx
-π
sin23хdx = (1-cos6х) : 2
π π
∫ sin23хdx = 1/2 ∫ (1 - cos6х)dx =
-π -π
π π
= 1/2 ∫ dx - 1/2 ∫ cos6хdx =
-π -π
π π
= 1/2 х | - 1/2 • 1/6 sin6х | =
-π -π
π π
= 1/2 х | - 1/12 sin6х | =
-π -π
= 1/2 π - 1/2 • (-π) - 1/12 sin6π + 1/12 sin(-6π) =
= 1/2 π + 1/2 π - 0 + 0 = π;
0
✔ 3) ∫ (cos22х - sin22х)dx
π/8
cos22х - sin22х = cos4х
0 0
∫ (cos22х - sin22х)dx = ∫ cos4хdx,
π/8 π/8
так як π/8 > 0, робимо перестановку
0 π/8 π/8
∫cos4хdx = -∫cos4хdx = - 1/4 sin4х | =
π/8 0 0
= - 1/4 sin4 • π/8 + 1/4 sin4 • 0 =
= - 1/4 sin π/2 - 0 = - 1/4 • 1 = - 1/4;
3π
✔ 4) ∫ (1 - 2cos2 х/3)dx
0
cos2 х/3 = 1/2(1 + cos 2х/3)
(1 - 2 • 1/2(1 + cos 2х/3)) = 1 - 1 - cos 2х/3 =
= -cos 2х/3
3π/4 3π/4
∫ (1 - 2cos2 х/3)dx = -∫ cos 2х/3 dx =
0 0
3π/4
= - 1/2 • 3 • sin 2х/3 | =
0
3π/4
= - 3/2 sin 2х/3 | =
0
= - 3/2 sin ((2•3π/4)/3) + 0 =
= - 3/2 sin π/2 = - 3/2;
π/4
✔ 5) ∫ (1 + 2tg2х)dx
π/6
Пояснення:
1. 1 + 2tg2х = 1 + (2sin2х) : (cos2α) =
= (cos2х+sin2х+sin2х) : (cos2α) =
= (1+sin2х) : (cos2α) = 1/cos2α + sin2х/cos2α;
π/4 π/4
2. ∫ tg2хdx = ∫ sin2х:cos2х =
π/6 π/6
π/4
= ∫(1-cos2х):cos2х dx =
π/6
π/4 π/4
= ∫ 1:cos2х - ∫ dx = tgх - х
π/6 π/6
π/4 π/4
∫ 1/cos2х dx + ∫ tg2хdx =
π/6 π/6
π/4 π/4 π/4
= tgх | + tgх | - х | =
π/6 π/6 π/4
π/4 π/4
= 2tgх | - х | =
π/6 π/6
= 2tg π/4 - 2tg π/6 - (π/4 - π/6) =
= 2 • 1 - 2 • √3/3 - π/12 =
= 2 - 2√3/3 - 0,26 = 1,74 - 2√3/3 =
= 1,74 - 1,15 = 0,59;
π/2
✔ 6) ∫ (6cos22х + sin3х)dx =
π/6
π/2 π/2
= ∫ 6cos22хdx + ∫ sin3хdx =
π/6 π/6
π/2 π/2
= 6 ∫ (1+cos4х):2 dx + (-cos3х):3 | =
π/6 π/6
π/2 π/2
= 6 1/2 ∫ (1 + cos4х)dx - cos3х:3 | =
π/6 π/6
π/2 π/2 π/2
= 3 ∫ dx + 3 ∫ cos4хdx - cos3х:3 | =
π/6 π/6 π/6
π/2 π/2 π/2
= 3х | + sin4х:4 | - cos3х:3 | =
π/6 π/6 π/6
= 3 • π/2 - 3 • π/6 + 1/4 • sin 4 • π/2 -
- 1/4 sin 4 • π/6 - (cos3•π/2):3 + (cos•3•π/6):3 =
= 3π/2 - π/2 + 1/4 sin 2π - 1/4 • sin π/3 -
- 1/3 cos 3π/2 + (cosπ/20:3 = 3π/2 + 0 -
- 1/4 • √3/2 - 1/3 • 0 + 1/3 • 0 = π - √3/8.
✔ 1) ∫cos7х • cos5хdx
0
cos7х • cos5хdx = 1/2(cos2х + cos12х)
π/3
∫ 1/2(cos2х + cos12х)dx =
0
π/3 π/3
= 1/2∫cos2хdx + 1/2∫cos12хdx =
0 0
π/3 π/3
= 1/2 • 1/2 sin2х | + 1/2 • 1/12 sin 12х | =
0 0
π/3 π/3
= 1/4 sin2х | + 1/24 sin12х | =
0 0
= 1/4 sin 2 • π/3 - 0 + 1/24 sin12 • π/3 - 0 =
= 1/4 sin 2π/3 + 1/24 sin4π = 1/4 • √3/2 + 0 =
= √3/8;
π
✔ 2) ∫ sin23хdx
-π
sin23хdx = (1-cos6х) : 2
π π
∫ sin23хdx = 1/2 ∫ (1 - cos6х)dx =
-π -π
π π
= 1/2 ∫ dx - 1/2 ∫ cos6хdx =
-π -π
π π
= 1/2 х | - 1/2 • 1/6 sin6х | =
-π -π
π π
= 1/2 х | - 1/12 sin6х | =
-π -π
= 1/2 π - 1/2 • (-π) - 1/12 sin6π + 1/12 sin(-6π) =
= 1/2 π + 1/2 π - 0 + 0 = π;
0
✔ 3) ∫ (cos22х - sin22х)dx
π/8
cos22х - sin22х = cos4х
0 0
∫ (cos22х - sin22х)dx = ∫ cos4хdx,
π/8 π/8
так як π/8 > 0, робимо перестановку
0 π/8 π/8
∫cos4хdx = -∫cos4хdx = - 1/4 sin4х | =
π/8 0 0
= - 1/4 sin4 • π/8 + 1/4 sin4 • 0 =
= - 1/4 sin π/2 - 0 = - 1/4 • 1 = - 1/4;
3π
✔ 4) ∫ (1 - 2cos2 х/3)dx
0
cos2 х/3 = 1/2(1 + cos 2х/3)
(1 - 2 • 1/2(1 + cos 2х/3)) = 1 - 1 - cos 2х/3 =
= -cos 2х/3
3π/4 3π/4
∫ (1 - 2cos2 х/3)dx = -∫ cos 2х/3 dx =
0 0
3π/4
= - 1/2 • 3 • sin 2х/3 | =
0
3π/4
= - 3/2 sin 2х/3 | =
0
= - 3/2 sin ((2•3π/4)/3) + 0 =
= - 3/2 sin π/2 = - 3/2;
π/4
✔ 5) ∫ (1 + 2tg2х)dx
π/6
Пояснення:
1. 1 + 2tg2х = 1 + (2sin2х) : (cos2α) =
= (cos2х+sin2х+sin2х) : (cos2α) =
= (1+sin2х) : (cos2α) = 1/cos2α + sin2х/cos2α;
π/4 π/4
2. ∫ tg2хdx = ∫ sin2х:cos2х =
π/6 π/6
π/4
= ∫(1-cos2х):cos2х dx =
π/6
π/4 π/4
= ∫ 1:cos2х - ∫ dx = tgх - х
π/6 π/6
π/4 π/4
∫ 1/cos2х dx + ∫ tg2хdx =
π/6 π/6
π/4 π/4 π/4
= tgх | + tgх | - х | =
π/6 π/6 π/4
π/4 π/4
= 2tgх | - х | =
π/6 π/6
= 2tg π/4 - 2tg π/6 - (π/4 - π/6) =
= 2 • 1 - 2 • √3/3 - π/12 =
= 2 - 2√3/3 - 0,26 = 1,74 - 2√3/3 =
= 1,74 - 1,15 = 0,59;
π/2
✔ 6) ∫ (6cos22х + sin3х)dx =
π/6
π/2 π/2
= ∫ 6cos22хdx + ∫ sin3хdx =
π/6 π/6
π/2 π/2
= 6 ∫ (1+cos4х):2 dx + (-cos3х):3 | =
π/6 π/6
π/2 π/2
= 6 1/2 ∫ (1 + cos4х)dx - cos3х:3 | =
π/6 π/6
π/2 π/2 π/2
= 3 ∫ dx + 3 ∫ cos4хdx - cos3х:3 | =
π/6 π/6 π/6
π/2 π/2 π/2
= 3х | + sin4х:4 | - cos3х:3 | =
π/6 π/6 π/6
= 3 • π/2 - 3 • π/6 + 1/4 • sin 4 • π/2 -
- 1/4 sin 4 • π/6 - (cos3•π/2):3 + (cos•3•π/6):3 =
= 3π/2 - π/2 + 1/4 sin 2π - 1/4 • sin π/3 -
- 1/3 cos 3π/2 + (cosπ/20:3 = 3π/2 + 0 -
- 1/4 • √3/2 - 1/3 • 0 + 1/3 • 0 = π - √3/8.
