3.∫8/√(6t-t^2)dt4.∫tan(x)In(cos(x))dx5.∫tsin(t)dt6.∫cos(x)e^7xdx7.∫In(x)x^5dxa.byintegrationbypartsb.byrewritingtheintegralas1/6∫6·In(x)·x^5dxthenusingthebasicpropertiesofnaturallogstofurtherrewritetheintegra

Question

3.∫8/√(6t-t^2)dt
4.∫tan(x)In(cos(x))dx
5.∫t sin(t)dt
6.∫cos(x)e^7x dx
7.∫In(x)x^5 dx
a.by integration by parts
b.by rewriting the integral as 1/6∫6·In(x)·x^5 dx then using the basic properties of natural logs to further rewrite the integral so that you can use the u-subsitution u=x^6
8.∫x√(x-8) dx
10.∫cos^6 (z) dz - by using the trig identity cos^2 (z) = （1+cos(2z))/2
11.∫cos^7 (x) sin^3 (x) dx

Answer 1

3.∫8/√(6t-t^2) dt = ∫8/√[9-(t-3)²] dt = ∫8/√[1-(t/3 - 1)²] d(t/3 - 1) = 8arcsin(t/3 - 1) + C 4.∫tan(x)In(cos(x))dx = ∫sin(x)In(cos(x))/cosx(x) dx = -∫In(cos(x))/cosx(x) dcos(x) = -∫In(cos(x)) dln[cos(x)] = -1/2 [ln(cosx)]² + C 5.∫t sin(t)dt = -∫t dcos(t) = -tcost + ∫costdt = sint - tcost + C 6.∫cos(x)e^7x dx = ∫e^7x dsinx = e^7x sinx - 7∫sinx e^7x dx = e^7x sinx + 7∫e^7x dcosx = e^7x sinx + 7e^7x cosx - 49∫cosx e^7x dx 所以∫cosx e^7x dx = (e^7x sinx + 7e^7x cosx)/50 + C 7.∫In(x)x^5 dx = 1/6 * ∫In(x) dx^6 = 1/6 * lnx x^6 - 1/6 * ∫x^5 dx = 1/6 * lnx x^6 - (x^6)/36 + C 8.∫x√(x-8) dx = ∫[(x-8)√(x-8) + 8√(x-8)] dx = 2/5 * (x-8)²√(x-8) + 16/3 * (x-8)√(x-8) + C 10.∫cos^6 (z) dz = 1/8 * ∫[1 + cos(2z)]³ dz = z/8 + 1/8 * ∫[3cos(2z) + 3cos²(2z) + cos³(2z)] dz = z/8 + 3sin(2z)/16 + 3/16 ∫[1+cos(4z)]dz + 1/16 ∫[1-sin²(2z)]dsin(2z) = z/8 + 3sin(2z)/16 + 3z/16 + 3sin(4z)/64 + sin(2z)/16 - sin³(2z)/48 + C = 5z/16 + sin(2z)/4 + 3sin(4z)/64 - sin³(2z)/48 + C