Integral of cos^2(x)/sin^6(x) (substitution)

[@IntegralsForYou]

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[@mamy.dont.trogay]

why u put a minus between "du=" and "(sinx*sinx - cosx*cosx)/sin^2x"?

[@IntegralsForYou]

Hi! I do it in order to have sin^2(x)+cos^2(x) which is equal to 1: (- sin(x)*sin(x) - cos(x)*cos(x))/sin^2(x) = = (- sin^2(x) - cos^2(x))/sin^2(x) = = - (sin^2(x) + cos^2(x))/sin^2(x) = = -1/sin^2(x) Hope it helped! ;-D

[@sephyrias883]

I don't get how you made the "du = .... dx" - part. What even is "du" here?

[@IntegralsForYou]

When we do a substitution we write its expression and then we calculate the derivative on both sides and we put "du" and "dx". Some examples: u = x+1 du = dx u^2 = x^3 2u du = 3x^2 dx u = cos(x) du = -sin(x) dx

[@sephyrias883]

​@@IntegralsForYou So "u * du = (derivative of u) * dx" ? Then I still don't follow with what you wrote here: 0:45. Did you divide u in advance, making it "du=(derivative of u)/u * dx" ? Why would we do that?

[@IntegralsForYou]

Just let "u" on the left and "cos(x)/sin(x)" on the right. Derivate "u" with respect to "u" and derivate "cos(x)/sin(x)" with respect to "x" and add "du" on the left and "dx" on the right: u = cos(x)/sin(x) (derivative of u) du = (derivative of cos(x)/sin(x)) dx 1 du = -1/sin^2(x) dx du = -1/sin^2(x) dx (last step is only because 1*du=du)

[@sephyrias883]

@@IntegralsForYou But you wrote du = (-(sin x)^2 - (cos x)^2)/(sin x)^2 Is that the derivative of the entire (cosx)^2/(sinx)^6 ?

[@IntegralsForYou]

The first step splits cos^2(x)/sin^6(x) dx into [cos^2(x)/sin^2(x)]*[1/sin^2(x)]*[1/sin^2(x)]dx = [cot^2(x)]*[1/sin^2(x)]*[1/sin^2(x)]dx Since Integral of f(x) dx = -Integral of -f(x) dx, we put a minus inside and outside the integral. Now we have: [cot^2(x)]*[1/sin^2(x)]*[-1/sin^2(x)]dx We do: u = cot(x) ==> cot^2(x) = u^2 ==> 1/sin^2(x) = 1+u^2 ==> -1/sin^2(x) dx = du Then: [cot^2(x)]*[1/sin^2(x)]*[-1/sin^2(x)]dx = (u^2)(1+u^2)du

[@СнежныйБарс-г2я]

169

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