# X 4 ln x 1 16x 4 c g e2 x sin x dx1 5 e 2 x cos x 2 5

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x4ln(x)−116x4+Cg) 2xsin(x)dx=−15e2xcos(x)+25e2xsin(x)+C(by parts twice –boomarangs)Here are the details:u = e2x du = 2e2xdv = sin(x)dx v = -cos(x)e2xsin(x)dx=−e2xcos(x)−cos(x)2e2xdxe2xsin(x)dx=−e2xcos(x)+2cos(x)e2xdx(expression 1)
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Now integrate cos(x)2xdxby parts: u = e2x du = 2e2xdv = cos(x)dx v = sin(x)cos(x)e2xdx=e2xsin(x)−sin(x)2e2xdxcos(x)e2xdx=e2xsin(x)−2e2xsin(x)dx<< substitute this expression for the integral on the right hand side of expression 1, to obtain:e2xsin(x)dx=−e2xcos(x)+2(e2xsin(x)−2e2xsin(x)dx)distribute the 2e2xsin(x)dx=−e2xcos(x)+2e2xsin(x)−4e2xsin(x)dxreplace the integral with II=−e2xcos(x)+2e2xsin(x)−4Isolve for I to obtain:5I=−e2xcos(x)+2e2xsin(x)I=15(e2xcos(x)+2e2xsin(x))which is the answer. h) 01xex=12e(by parts, with u = x, dv = e-xdx) F(x) = -e-xx – e-xi) xex2dx=12ex2+C(substitution u = x2)j) 3x43x331x26x70x2x12dx¿ ¿x10¿¿3x43x331x26x70x2x12dx=3x2+5+x10x2x12dx=x3+5xx+10x2x12dxFor x+10x2x12dxusepartial fractions.x+10x2x12=Ax4+Bx+3x + 10 = A(x + 3) + B(x – 4) 8
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x + 10 = (A + B)x + 3A – 4B A + B = 1 and 3A – 4B = 10 A = 2 and B= -1So x+10x2x12dx=2x4+1x+3dx=2ln(x4)−ln(x+3)3x43x3