2. integrate each of the following with respect to x:
g)2sin(3x+30)^。
h)xcos(x^2)
3.find each of the following integrals:
f)∫ 1/(2t-t^2)^0.5 dt
6.find the following integrals:
a) ∫ ln(2-3x)dx
B) ∫ xsecxtanxdx
D) ∫ e^2x sinx dx
7. find the exact value of each of the following integrals:
A) ∫ dx/(4+9x^2) limit:from 0 to 2/(3^0.5).
C) ∫ xcos2xdx limit from -pi/4 to 0
Please help me.i need working solutions to understand.these are my practice qns which I didn't manage to get the ans.
Ans to the qns are:
2g)-120/pi • cospi/6((x/10) +1)+c
H)0.5sin(x^2)+c
3f)sin^(-1)(t-1)+c
6a)ln(2-3x)(x-(2/3))-x+c
B)xsecx-ln|secx+tanx|+c
D) (1/5)e^(2x) (2sinx-cosx)+c
7a)pi/18
C) (2-pi)/8
g)2sin(3x+30)^。
h)xcos(x^2)
3.find each of the following integrals:
f)∫ 1/(2t-t^2)^0.5 dt
6.find the following integrals:
a) ∫ ln(2-3x)dx
B) ∫ xsecxtanxdx
D) ∫ e^2x sinx dx
7. find the exact value of each of the following integrals:
A) ∫ dx/(4+9x^2) limit:from 0 to 2/(3^0.5).
C) ∫ xcos2xdx limit from -pi/4 to 0
Please help me.i need working solutions to understand.these are my practice qns which I didn't manage to get the ans.
Ans to the qns are:
2g)-120/pi • cospi/6((x/10) +1)+c
H)0.5sin(x^2)+c
3f)sin^(-1)(t-1)+c
6a)ln(2-3x)(x-(2/3))-x+c
B)xsecx-ln|secx+tanx|+c
D) (1/5)e^(2x) (2sinx-cosx)+c
7a)pi/18
C) (2-pi)/8
-
2g)
∫ 2 sin((3x+30)°) dx = ∫ 2 sin((3x+30)*π/180) dx
............................ = ∫ 2 sin(π/60(x+10)) dx
Use a substitution:
u = π/60(x+10)
du = π/60 dx
60/π du = dx
............................ = ∫ 2 sin(u) * 60/π du
............................ = ∫ 120/π sin(u) du
............................ = −120/π cos(u) + C
............................ = −120/π cos(π/60(x+10)) + C
............................ = −120/π cos(π/6(x/10+1)) + C
——————————————————————————————
2h)
u = x²
du = 2x dx
∫ x cos(x²) dx = ∫ 1/2 cos(x²) * 2x dx
.................. = ∫ 1/2 cos(u) du
.................. = 1/2 cos(u) + C
.................. = 1/2 cos(x²) + C
——————————————————————————————
3f)
∫ 1/√(2t−t²) dt = ∫ 1/√(1−(t²−2t+1)) dt
................... = ∫ 1/√(1−(t−1)²) dt
Use a substitution:
t − 1 = u
dt = du
................... = ∫ 1/√(1−u²) du
................... = sin⁻¹(u) + C
................... = sin⁻¹(t−1) + C
——————————————————————————————
6a)
Integrate by parts:
u = ln(2−3x) ............ dv = dx
∫ 2 sin((3x+30)°) dx = ∫ 2 sin((3x+30)*π/180) dx
............................ = ∫ 2 sin(π/60(x+10)) dx
Use a substitution:
u = π/60(x+10)
du = π/60 dx
60/π du = dx
............................ = ∫ 2 sin(u) * 60/π du
............................ = ∫ 120/π sin(u) du
............................ = −120/π cos(u) + C
............................ = −120/π cos(π/60(x+10)) + C
............................ = −120/π cos(π/6(x/10+1)) + C
——————————————————————————————
2h)
u = x²
du = 2x dx
∫ x cos(x²) dx = ∫ 1/2 cos(x²) * 2x dx
.................. = ∫ 1/2 cos(u) du
.................. = 1/2 cos(u) + C
.................. = 1/2 cos(x²) + C
——————————————————————————————
3f)
∫ 1/√(2t−t²) dt = ∫ 1/√(1−(t²−2t+1)) dt
................... = ∫ 1/√(1−(t−1)²) dt
Use a substitution:
t − 1 = u
dt = du
................... = ∫ 1/√(1−u²) du
................... = sin⁻¹(u) + C
................... = sin⁻¹(t−1) + C
——————————————————————————————
6a)
Integrate by parts:
u = ln(2−3x) ............ dv = dx
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