1.) upper bound: pi
lower bound: 0
8sinx / sqrt(5-4cosx) dx
2.) upper bound: sqrt 3
lower bound: 0
4x / sqrt(x^2 + 1) dx
3.) upper bound: 1
lower bound: -1
(dx) / (3x-4)
To find the definite integral for each of the given functions, we can start by finding the antiderivative of the function and then evaluate it at the upper and lower bounds. Here's how you can solve each problem:
1.) The integral of 8sinx / sqrt(5-4cosx) dx:
To solve this integral, we can use the substitution method.
Let u = 5 - 4cosx
Differentiate both sides: du = 4sinx dx
Rearrange the equation: dx = du / (4sinx)
Now substitute the values into the integral:
∫ (8sinx / sqrt(5-4cosx)) dx
= ∫ (8sinx / sqrt(u)) (du / (4sinx))
= 2∫ du / sqrt(u)
= 2√u + C, where C is the constant of integration
Substitute the value of u back in:
= 2√(5-4cosx) + C
Now evaluate the integral at the upper and lower bounds:
= 2√(5-4cos(pi)) - 2√(5-4cos(0))
= 2√(5 + 4) - 2√(5 - 4)
= 2√9 - 2√1
= 2(3) - 2(1)
= 6 - 2
= 4
Therefore, the definite integral is 4.
2.) The integral of 4x / sqrt(x^2 + 1) dx:
To solve this integral, we can use another substitution method.
Let u = x^2 + 1
Differentiate both sides: du = 2x dx
Rearrange the equation: dx = du / (2x)
Now substitute the values into the integral:
∫ (4x / sqrt(x^2 + 1)) dx
= ∫ (4x / sqrt(u)) (du / (2x))
= 2∫ du / sqrt(u)
= 2√u + C, where C is the constant of integration
Substitute the value of u back in:
= 2√(x^2 + 1) + C
Now evaluate the integral at the upper and lower bounds:
= 2√(sqrt(3)^2 + 1) - 2√(0^2 + 1)
= 2√4 - 2√1
= 2(2) - 2(1)
= 4 - 2
= 2
Therefore, the definite integral is 2.
3.) The integral of dx / (3x-4):
For this integral, we do not need any substitution method.
∫ (dx / (3x-4))
To solve this integral, we can use a logarithmic substitution:
Let u = 3x - 4
Differentiate both sides: du = 3 dx
Rearrange the equation: dx = du / 3
Now substitute the values into the integral:
∫ (dx / (3x-4))
= ∫ (du / (3u))
= (1/3) ∫ du / u
= (1/3) ln|u| + C, where C is the constant of integration
Substitute the value of u back in:
= (1/3) ln|3x-4| + C
Now evaluate the integral at the upper and lower bounds:
= (1/3) ln|3(1)-4| - (1/3) ln|3(-1)-4|
= (1/3) ln|-1| - (1/3) ln|-7|
= (1/3) ln(1) - (1/3) ln(7)
= 0 - ln(7)/3
= -ln(7)/3
Therefore, the definite integral is -ln(7)/3.