# Couple of Integration Questions

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1. ysin^3x dy/dx = cosx

I've done: INT y dy = INT cosx/sin^3x dx

y^2/2 = ???

How would yo go about intergrating the RHS?

2. Find the equation of the curve whose gradient function is (y+1)/(x^2-1) which passes through the point (-3,1)

Any help would be much appreciated

I've done: INT y dy = INT cosx/sin^3x dx

y^2/2 = ???

How would yo go about intergrating the RHS?

2. Find the equation of the curve whose gradient function is (y+1)/(x^2-1) which passes through the point (-3,1)

Any help would be much appreciated

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#2

1. Have you tried using a substitution? u=sinx should work fine.

2.

dy/dx = (y+1)/(x^2-1)

1/(y+1) dy = 1/(x^2-1) dx

ln|y+1| = 0.5ln|(x-1)/(x+1)| + C [use partial fractions for the RHS -- (x^2-1)=(x-1)(x+1)]

Sub in (-3, 1) to find C.

2.

dy/dx = (y+1)/(x^2-1)

1/(y+1) dy = 1/(x^2-1) dx

ln|y+1| = 0.5ln|(x-1)/(x+1)| + C [use partial fractions for the RHS -- (x^2-1)=(x-1)(x+1)]

Sub in (-3, 1) to find C.

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#3

(Original post by

1. Integrate ysin^3x dy/dx = cosx

2. Find the equation of the curve whose gradient function is (y+1)/(x^2-1) which passes through the point (-3,1)

**Alex H**)1. Integrate ysin^3x dy/dx = cosx

2. Find the equation of the curve whose gradient function is (y+1)/(x^2-1) which passes through the point (-3,1)

---> y dy = (cosx / sin^3x) dx

---> Int y dy = Int (cosx / sin^3x) dx

---> y^2/2 = Int (cosx / sin^3x) dx

To find Int (cosx / sin^3x) dx:

Let u = sinx ---> du/dx = cosx ---> dx = 1/cosx du

Hence: Int (cosx / sin^3x) dx = Int [cosx / (u^3.cosx)] du = u^-3 du = -(u^-2)/2 + k = -(1/sin^2x)/2 + k = -cosec^2x/2 + k

Hence:

y^2/2 = -cosec^2x/2 + k

---> y^2 = 2k - cosec^2x

Let 2k = c:

---> Solution:

**y = Sqrt[c - cosec^2x]**

2.) Find the equation of the curve whose gradient function is (y+1)/(x^2-1) which passes through the point (-3,1).

dy/dx = (y + 1)/(x^2 - 1)

---> 1/(y + 1) dy = 1/(x^2 - 1) dx = 1/[(x + 1)(x - 1)] dx = 1/[2(x - 1)] - 1/[2(x + 1)] dx = 1/2 [1/(x - 1) - 1/(x + 1)] dx

---> Int 1/(y + 1) dy = 1/2 Int 1/(x - 1) - 1/(x + 1) dx

---> ln|y + 1| = 1/2 [ln|x - 1| - ln|x + 1| + k]

---> ln|y + 1| = 1/2 {ln[|x - 1|/|x + 1|] + k]

---> y + 1 = e^{lnSqrt[|x - 1|/|x + 1|] + k/2}

---> y + 1 = e^(k/2). Sqrt[|x - 1|/|x + 1|]

---> y + 1 = Sqrt{[e^k|x - 1|]/|x + 1]}

---> y = Sqrt{[e^k|x - 1|]/|x + 1]} - 1

When x = -3, y = 1:

---> 1 = Sqrt{4e^k/2} - 1

---> Sqrt (2e^k) = 2

---> 2e^k = 4

---> e^k = 2

---> lne^k = ln2

---> klne = ln2

---> k = ln2

Hence Equation Of Curve Is:

**y = Sqrt{2|x - 1|]/|x + 1]} - 1**

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#4

(Original post by

1.) Integrate ysin^3x dy/dx = cosx

---> y dy = (cosx / sin^3x) dx

---> Int y dy = Int (cosx / sin^3x) dx

---> 1 = Int (cosx / sin^3x) dx

**Nima**)1.) Integrate ysin^3x dy/dx = cosx

---> y dy = (cosx / sin^3x) dx

---> Int y dy = Int (cosx / sin^3x) dx

---> 1 = Int (cosx / sin^3x) dx

To find Int (cosx / sin^3x) dx:

Let u = sin^3x ---> du/dx = 3cosx(sin^2x) ---> dx = 1/[3cosx(sin^2x)] du

Hence: Int (cosx / sin^3x) dx = Int [cosx / (u.3.cosx.u^(2/3))] du = Int 1/[3u^(5/3)] = 1/3 Int u^(-5/3) = 1/3 . [-3/2.u^(-2/3)] + k = -[u^(-2/3)]/2 + k = -[(sin^3x)^(-2/3)]/2 + k = -[1/(sin^2x)]/2 + k = -cosec^2x/2 + k

Let u = sin^3x ---> du/dx = 3cosx(sin^2x) ---> dx = 1/[3cosx(sin^2x)] du

Hence: Int (cosx / sin^3x) dx = Int [cosx / (u.3.cosx.u^(2/3))] du = Int 1/[3u^(5/3)] = 1/3 Int u^(-5/3) = 1/3 . [-3/2.u^(-2/3)] + k = -[u^(-2/3)]/2 + k = -[(sin^3x)^(-2/3)]/2 + k = -[1/(sin^2x)]/2 + k = -cosec^2x/2 + k

int (cosx/sin^3x) dx = int cosx/(sinx.sin^2x) dx = int cotxcosec^2x dx = - int cotx(-cosec^2x) dx = -0.5cot^2x + C, since the integrand became a function (cotx) and its derivative (-cosec^2x).

Another way to do it (and get the same answer you got) is to use the subs u=sinx and du=cosx dx:

int (cosx/sin^3x) dx = int (1/u^3) du = -1/(2u^2) = -0.5cosec^2x + K.

Both are correct answers (that are a constant apart).

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