Calculate the indefinite integral of y = cos x / (sin x)^3.

hala718 | High School Teacher | (Level 1) Educator Emeritus

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y= cosx /(sin x)^3

Let t= sinx

==> dt = cosx dx

=> y=  (dt / t^3)

intg y = intg (dt/t^3)

= intg (t^-3) dt

= -t^-2/2  + c

= -1/2t^2 + C

Now substitute with t= sinx

==> intg y = -1/2(sin x)^2 + C

neela | High School Teacher | (Level 3) Valedictorian

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To find Integral {cosx/(sinx)^3} dx

Let us have a transformation  sinx = t. Then cosx dx = dt.

So Integral { cosxdx/ (sinx)^3} = Integral  dt/t^3

= t^(-3+1)/(-3+1) +C

= t^-2/((-2) + C

= (sinx)^(-2) / 2 +

= 1/[2(sinx)^2] + const

So Int { cosxdx/(sinx)^3 } = 1/{2(sinx)^2} + constant

giorgiana1976 | College Teacher | (Level 3) Valedictorian

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To calculate the indefinite integral, we'll use the substitution method.

We'll note y = f(x)

We'll calculate Integral of f(x) = y = cos x / (sin x)^3.

We notice that if we'll differentiate sin x, we'll get cos x.

So, we'll note sin x = t

(sin x)'dx = dt

(cos x)dx = dt

We'll re-write the integral in the variable t:

Int  (cos x)dx / (sin x)^3 = Int dt / t^3

Int dt / t^3 = Int [t^(-3)]dt

Int [t^(-3)]dt = t^(-3+1) / (-3+1) + C

Int [t^(-3)]dt = t^(-2)/-2 + C

Int [t^(-3)]dt = -1 / 2t^2 + C

But sin x = t.

Int  (cos x)dx / (sin x)^3 = -1 / 2(sin x)^2 + C, where C is a family of constants.