Find each of the following.
(a) (3t^2 + t/2)dt
(b) x^(-5/4) dx
(c)(2e^x -3e^-2x)dx
(d) ((2/sqrt1-y^2) - (1/y^(1/4)))dy
(f) (csc theta/csc theta-sin theta) d theta
To find each of the given integrals, we will use basic integration techniques. Let's solve each of them step by step:
(a) ∫(3t^2 + t/2) dt:
To find the integral of (3t^2 + t/2) with respect to t, we apply the power rule and the constant multiple rule. The power rule states that the integral of tn dt is (tn+1)/(n+1), where n is any real number except -1. The constant multiple rule allows us to split the integral of a sum or difference of functions into the sum or difference of their integrals.
∫(3t^2 + t/2) dt = ∫3t^2 dt + ∫t/2 dt
= 3∫t^2 dt + 1/2 ∫t dt
= 3(t^3/3) + 1/2 (t^2/2) + C
= t^3 + t^2/4 + C
(b) ∫x^(-5/4) dx:
To find the integral of x^(-5/4) with respect to x, we apply the power rule in reverse. If we have x^(n), where n is not equal to -1, the integral is (x^(n+1))/(n+1) * 1/n.
∫x^(-5/4) dx = x^(-5/4 + 1)/(-5/4 + 1) * 1/(-5/4 + 1) + C
= x^(-1/4)/(-1/4) * 4/(-1)
= -4x^(-1/4) + C
(c) ∫(2e^x - 3e^(-2x)) dx:
To find the integral of (2e^x - 3e^(-2x)) with respect to x, we use the power rule for exponential functions. The integral of e^(ax) is (1/a)e^(ax).
∫(2e^x - 3e^(-2x)) dx = 2∫e^x dx - 3∫e^(-2x) dx
= 2e^x - 3(1/(-2))e^(-2x) + C
= 2e^x + 3/2e^(-2x) + C
(d) ∫((2/sqrt(1-y^2)) - (1/y^(1/4))) dy:
To find the integral of ((2/sqrt(1-y^2)) - (1/y^(1/4))) with respect to y, we use some algebraic manipulation.
∫((2/sqrt(1-y^2)) - (1/y^(1/4))) dy = 2∫(1/sqrt(1-y^2)) dy - ∫(1/y^(1/4)) dy
We can now evaluate each integral separately.
For the first term, if we let y = sin(theta), we can use the trigonometric substitution. Remember that dy = cos(theta) d(theta).
2∫(1/sqrt(1-y^2)) dy = 2∫(1/cos(theta)) cos(theta) d(theta)
= 2∫d(theta)
= 2theta + C
For the second term, we apply the power rule for integrals.
∫(1/y^(1/4)) dy = ∫y^(-1/4) dy = 4y^(3/4) + C
Putting it all together:
∫((2/sqrt(1-y^2)) - (1/y^(1/4))) dy = 2theta + 4y^(3/4) + C
(f) ∫(csc(theta)/csc(theta)-sin(theta)) d(theta):
To find the integral of (csc(theta)/csc(theta)-sin(theta)) with respect to theta, we use some trigonometric identities and algebraic manipulation.
Start by simplifying the expression using the reciprocal identities:
csc(theta)/(csc(theta)-sin(theta)) = 1/(1-sin(theta)/csc(theta))
Since csc(theta) = 1/sin(theta):
1/(1-sin(theta)/csc(theta)) = 1/(1-1) = 1
∫(csc(theta)/csc(theta)-sin(theta)) d(theta) = ∫1 d(theta) = theta + C
Therefore, the integral is simply theta + C.