solve the equation Integral sin^2xdx+Integral cos^2xdx=(x^3-4x)/(x-2)

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We have to solve the equation: Int[(sin x)^2 dx]+Int[(cos x)^2 dx] = (x^3-4x)/(x-2)

Now Int [ a dx ] + Int [ b  dx] = Int [ (a + b) dx]

Int[(sin x)^2 dx]+Int[(cos x)^2 dx] = (x^3 - 4x)/(x-2)

=> Int [ (sin x)^2 + (cos x)^2 dx] = (x^3 - 4x)/(x-2)

Use (sinx)^2 + (cos x)^2 = 1

=> Int [ 1 dx] = (x^3 - 4x)/(x-2)

=> x = (x^3 - 4x)/(x-2)

=> x = x(x^2 - 4)/(x-2)

=> x - 2 = x^2 - 4

=> x^2 - x - 2 = 0

=> x^2 - 2x + x - 2 = 0

=> x(x - 2) + 1(x - 2) = 0

=> (x + 1)(x - 2) = 0

=> x = -1 and x = 2

But x = 2 makes the left hand side undefined, so we do not consider it.

The required solution of the equation is x = -1.

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