1.integral cos(x)sqrt(sin(x))dx. upper limit pi and lower limit 0
2.integral 1/sqrt(x)(sqrt(x) +1)^2. Upper limit is 4 and lower limit is 1.
1. let u=sinx
then du = cosx dx
and you have ∫ u^(1/2) du
2. let u = √x+1
then du = 1/(2√x)
and you have ∫2u^2 du
To solve the integrals:
1. Integral of cos(x)sqrt(sin(x)) dx, with upper limit π and lower limit 0:
Step 1: We can use the substitution method for this integral. Let's substitute u = sqrt(sin(x)), then du = (1/2)sqrt(cos(x)) dx.
Step 2: Rearrange the equation to express dx in terms of du: dx = 2du/sqrt(cos(x)).
Step 3: Substitute the new variables in the integral: ∫[0 to π] cos(x)sqrt(sin(x)) dx = ∫[u(0) to u(π)] 2cos(x) du/sqrt(cos(x)) = 2∫[u(0) to u(π)] du.
Step 4: Evaluate the integral: 2 * (u(π) - u(0)) = 2 * [sqrt(sin(π)) - sqrt(sin(0))].
Step 5: Simplify the result: 2 * [sqrt(0) - sqrt(0)] = 2 * (0 - 0) = 0.
Therefore, the value of the integral is 0.
2. Integral of 1/sqrt(x)(sqrt(x) +1)^2, with upper limit 4 and lower limit 1:
Step 1: Let's rewrite the integral as follows: ∫[1 to 4] (1/x)(sqrt(x) + 1)^2) dx.
Step 2: Expand the expression within the parentheses: ∫[1 to 4] (1/x)(x + 2√x + 1) dx.
Step 3: Distribute and simplify the integral: ∫[1 to 4] (1/x)(x) dx + ∫[1 to 4] (1/x)(2√x) dx + ∫[1 to 4] (1/x)(1) dx.
Step 4: Evaluate each integral separately:
- For the first integral, we have ∫[1 to 4] dx, which is simply x evaluated from 1 to 4. So, it becomes 4 - 1 = 3.
- For the second integral, we have 2∫[1 to 4] √x dx. By applying the power rule, it simplifies to 2(2/3)x^(3/2) evaluated from 1 to 4. So, it becomes 2(2/3)(4)^(3/2) - 2(2/3)(1)^(3/2).
- For the third integral, we have ∫[1 to 4] (1/x) dx, which is ln|x| evaluated from 1 to 4. So, it becomes ln(4) - ln(1) = ln(4).
Step 5: Add up the results from each integral: 3 + 2(2/3)(4)^(3/2) - 2(2/3)(1)^(3/2) + ln(4).
Therefore, the value of the integral is 3 + 2(2/3)(4)^(3/2) - 2(2/3)(1)^(3/2) + ln(4).
1. To evaluate the integral ∫cos(x)√(sin(x))dx with upper limit π and lower limit 0, you can follow these steps:
Step 1: Start by applying the substitution method. Let u = sin(x), which implies du = cos(x)dx.
Step 2: Rewrite the integral in terms of u. Since u = sin(x), we have cos(x) = √(1 - sin^2(x)) = √(1 - u^2).
∫cos(x)√(sin(x))dx = ∫√(1 - u^2)du
Step 3: Plug in the limits of integration. When x = 0, u = sin(0) = 0, and when x = π, u = sin(π) = 0 as well.
∫[0,π] √(1 - u^2)du
Step 4: Simplify the integral. The integral of √(1 - u^2) is known as the definite integral of the cosine function.
∫[0,π] cos^2(t)dt
Step 5: Apply the trigonometric identity. Recall that cos^2(t) = (1 + cos(2t))/2.
∫[0,π] (1 + cos(2t))/2 dt
Step 6: Evaluate the integral of the first term. The integral of 1/2 with respect to t is (1/2)t.
∫[0,π] (1/2 + (1/2)cos(2t)) dt
Step 7: Evaluate the integral of the second term using the formula ∫cos(Ax)dx = sin(Ax)/A.
= (1/2)t + (1/4)sin(2t) + C
Step 8: Plug in the limits of integration and simplify.
= (1/2)π + (1/4)sin(2π) - (1/4)sin(0)
= (1/2)π
Therefore, the value of the integral from 0 to π of cos(x)√(sin(x))dx is π/2.
2. To evaluate the integral ∫(1/√(x))(√(x) + 1)^2dx with upper limit 4 and lower limit 1, follow these steps:
Step 1: Expand (√(x) + 1)^2 using the formula (a + b)^2 = a^2 + 2ab + b^2.
(√(x) + 1)^2 = (√(x))^2 + 2(√(x))(1) + (1)^2 = x + 2√(x) + 1.
Step 2: Rewrite the integral as a sum of integrals.
∫(1/√(x))(x + 2√(x) + 1)dx
Step 3: Split the integral into three separate integrals.
∫(1/√(x))xdx + ∫(1/√(x))(2√(x))dx + ∫(1/√(x))dx
Step 4: Evaluate each integral separately.
∫(1/√(x))xdx = 2√(x)dx (using the substitution method with u = √(x))
∫2√(x)dx = 2(2/3)x^(3/2) + C = (4/3)x^(3/2) + C
∫(1/√(x))(2√(x))dx = ∫2dx = 2x + C
∫(1/√(x))dx = 2√(x)dx (using the substitution method with u = √(x))
∫2√(x)dx = 2(2/3)x^(3/2) + C = (4/3)x^(3/2) + C
Step 5: Plug in the limits of integration and subtract the lower limit from the upper limit.
[∫(1/√(x))xdx + ∫(1/√(x))(2√(x))dx + ∫(1/√(x))dx] from 1 to 4
[(4/3)x^(3/2) + 2x + (4/3)x^(3/2)] evaluated from 1 to 4
[(4/3)(4)^(3/2) + 2(4) + (4/3)(4)^(3/2)] - [(4/3)(1)^(3/2) + 2(1) + (4/3)(1)^(3/2)]
[(4/3)(8) + 8 + (4/3)(8)] - [(4/3)(1) + 2(1) + (4/3)(1)]
[(32/3) + 8 + (32/3)] - [(4/3) + 2 + (4/3)]
= (88/3) - (10/3) = 78/3 = 26
Therefore, the value of the integral from 1 to 4 of (1/√(x))(√(x) + 1)^2dx is 26.