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# web-7.1 - Web 7.1 In w1w4 integrate by parts w1 x3 ex dx =...

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Web 7.1 In w1–w4, integrate by parts. w1. Z x 3 e x 2 dx = Z [ x 2 ] e x 2 x dx = x 2 1 2 e x 2 - Z 1 2 e x 2 2 x dx by parts = - 1 2 Z e x 2 2 x dx = - 1 2 e x 2 + C by B3 w2. Do 35 in Ex 7.1 in the book without first substituting. Solution. Z π π 2 θ 3 cos θ 2 = Z π π 2 [ θ 2 ] (cos θ 2 θ dθ ) = θ 2 1 2 sin θ 2 - Z 1 2 sin θ 2 2 θ dθ π π 2 = · · · w3. Do 30 in Ex 7.1 in the book. Solution. Z 1 0 r 3 4 + r 2 dr = Z 1 0 [ r 2 ] (4 + r 2 ) - 1 2 r dr = r 2 (4 + r 2 ) 1 2 - Z (4 + r 2 ) 1 2 2 r dr 1 0 = · · · w4. Z x 3 dx 5 9 + x 2 = Z [ x 2 ] (9 + x 2 ) - 1 5 x dx = x 2 1 2 5 4 (9 + x 2 ) 4 5 - Z 5 8 (9 + x 2 ) 4 5 2 x dx = 5 8 x 2 (9 + x 2 ) 4 5 - 25 72 (9 + x 2 ) 9 5 + C In w5 and w6, integrate by parts twice. Then transpose the desired unknown integral from the right side to the left side of the equation. Then solve the equation for this unknown integral. w5. Z [ e ax ] (cos bx dx ) = e ax 1 b sin bx - a b Z [ e ax ] (sin bx dx )

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= 1 b e ax sin bx - a b e ax - cos bx b - Z - cos bx b e ax a dx ! = 1 b e ax sin bx + a b 2 e ax cos bx - a 2 b 2 Z e ax cos bx dx Now we write the integral at the beginning of this ”string” equal to the expression after the last ”equal sign” and we get:
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