1. Find the indefinite integral.
Indefinite integral tan^3(pix/7)sec^2(pix/7)dx
2. Find the indefinite integral by making the substitution x=3tan(theta).
Indefinite integral x*sqrt(9+x^2)dx
3. Find the indefinite integral.
Indefinite integral cos(x)sin^3(x)dx
recall that tan' = sec^2
If we let v = pi/7 x, then dv = pi/7 dx
and we have Int[tan^3(v) sec^2(v)] 7/pi dv
Now, let u = tan(v). du = sec^2(v) dv Our integral now becomes
7/pi Int[u^3 du] = 7/pi * 1/4 u^4 + C
= 7/4pi tan^4(pi/7 x) + C
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If x = 3tan(u) then 9+x^2 = 9 + 9tan^2(u) = 9 sec^2(u)
dx = 3sec^2(u) du
The integral now becomes
Int[3tan(u) * 3sec*u)* 3sec^2(u)] du = 9Int[sec(u) tan(u)*sec^2(u)]
Now, if v = sec(u), dv = sec(u) tan(u) du, and the integral now becomes
9 Int[v^2 dv] = 1/3 v^3 = 1/33 sec^3(u) = 1/3 sqrt(9+x^2)^(3/2) + C
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This could also have been obtained by letting u=9+x^2, so du = 2xdx, making the integral
Int[u^1/2 * du/2] = 1/3(9+x^2)^(3/2) + C
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Let u = sin(x), so du = cos(x) dx and the integral becomes
Int[u^3 du] = 1/4 u^4 = 1/4 sin^4(x) + C
To find the indefinite integral of these functions, we will apply a combination of integration techniques such as substitution and trigonometric identities. Let's tackle each problem step by step:
1. Find the indefinite integral of tan^3(pix/7)sec^2(pix/7)dx:
To solve this integral, we will use the substitution method.
Let u = tan(pix/7), then du = sec^2(pix/7)dx.
By substituting the expressions for u and du, we convert the given integral into a new integral:
∫ tan^3(pix/7)sec^2(pix/7)dx = ∫ u^3du.
Now, this becomes a straightforward power rule integration:
∫ u^3du = u^4/4 + C, where C is the constant of integration.
Substituting back the value of u:
∫ tan^3(pix/7)sec^2(pix/7)dx = (tan^4(pix/7))/4 + C.
2. Find the indefinite integral of x*sqrt(9+x^2)dx using the substitution x=3tan(theta):
To solve this integral, we will use the substitution method.
Let x = 3tan(theta), then dx = 3sec^2(theta)d(theta).
By substituting the expressions for x and dx, we convert the given integral into a new integral:
∫ x*sqrt(9+x^2)dx = ∫ (3tan(theta))*sqrt(9+(3tan(theta))^2)*(3sec^2(theta))d(theta).
Simplifying the expression further, we have:
= 27∫ tan(theta)*sec(theta)*sqrt(9tan^2(theta)+9) d(theta).
Now, using the trigonometric identity sec^2(theta) = 1 + tan^2(theta), we can simplify the expression inside the square root:
= 27∫ tan(theta)*sec(theta)*3sec(theta) d(theta).
= 81∫ tan(theta)sec^2(theta)d(theta).
This integral simplifies back to the original integral of tan(theta)sec^2(theta), which is simply sec(theta) + C.
Substituting back the value of x, we have:
∫ x*sqrt(9+x^2)dx = 81(sec(theta)) + C.
Now, we need to convert the answer back to x form. We can use the right triangle relationship from the substitution:
tan(theta) = x/3, so sec(theta) = sqrt(1 + tan^2(theta)) = sqrt(1 + (x/3)^2) = sqrt((x^2+9)/9).
Finally, substituting sec(theta) = sqrt((x^2+9)/9) back into the answer gives us:
∫ x*sqrt(9+x^2)dx = 81(sqrt((x^2+9)/9)) + C.
3. Find the indefinite integral of cos(x)sin^3(x)dx:
To solve this integral, we will use the integration by parts method.
Let u = sin(x), then du = cos(x)dx.
And dv = sin^2(x)dx, then v = ∫ sin^2(x)dx = (1/2) ∫ (1 - cos(2x))dx = (1/2) (x - (1/2)sin(2x)).
Now, using the integration by parts formula:
∫ u dv = uv - ∫ v du.
We have:
∫ cos(x)sin^3(x)dx = (1/2)sin^2(x) - (1/2) ∫ (x - (1/2)sin(2x)) cos(x)dx.
Simplifying the expression further, we have:
= (1/2)sin^2(x) - (1/2) ∫ (x cos(x) - (1/2)sin(2x)cos(x))dx.
= (1/2)sin^2(x) - (1/2) ∫ x cos(x)dx + (1/4) ∫ sin(2x)cos(x)dx.
The first integral, ∫ x cos(x)dx, can be integrated using integration by parts or direct integration to give x sin(x) - ∫ sin(x)dx = x sin(x) + cos(x) + C1.
The second integral, ∫ sin(2x)cos(x)dx, can be rewritten using the double angle identity sin(2x) = 2sin(x)cos(x):
= (1/4) ∫ 2sin(x)cos^2(x)dx.
Now, we let u = cos(x), then du = -sin(x)dx.
And dv = 2sin(x)cos^2(x)dx, then v = ∫ 2sin(x)cos^2(x)dx = 2/3 cos^3(x).
Therefore, applying the integration by parts formula to the second integral:
= (1/4) (-2/3 cos^3(x)) + C2 = -1/6 cos^3(x) + C2.
Putting everything together, the final solution is:
∫ cos(x)sin^3(x)dx = (1/2)sin^2(x) - (1/2)(x sin(x) + cos(x)) - 1/6 cos^3(x) + C, where C = C1 + C2 is the constant of integration.