SolutionIntegrals - Penn State Altoona Math 141 Solutions to the integral problems 3 x4 ln x dx 1 1 Solution Integration by parts with u = ln x dv = x4

SolutionIntegrals - Penn State Altoona Math 141 Solutions...

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Penn State Altoona Math 141 Solutions to the integral problems 1. 3 1 x 4 ln x dx Solution : Integration by parts with u = ln x , dv = x 4 dx : 3 1 x 4 ln x dx = ln( x ) x 5 5 3 1 - 3 1 x 4 5 dx = 3 5 5 ln(3) - x 5 25 3 1 = 3 5 5 ln(3) + 1 - 3 5 25 . 2. 2 / 2 0 x 2 1 - x 2 dx Solution : Substitute x = sin( θ ), so dx = cos( θ ) . If x = 0 then θ = 0, if x = 2 / 2 then θ = arcsin( 2 / 2) = π/ 4. Thus 2 / 2 0 x 2 1 - x 2 dx = π/ 4 0 sin 2 ( θ ) cos( θ ) cos( θ ) = π/ 4 0 1 2 (1 - cos(2 θ )) = 1 2 θ - 1 2 sin(2 θ ) π/ 4 0 = π 8 - 1 4 . 3. x sin 2 x dx Solution : Integration by parts with u = x and dv = sin 2 ( x ) dx , so du = dx and v = 1 2 ( x - 1 2 sin(2 x ) ) . x sin 2 x dx = x 2 ( x - 1 2 sin(2 x ) ) - 1 2 ( x - 1 2 sin(2 x ) ) dx = x 2 ( x - 1 2 sin(2 x ) ) - x 2 4 - 1 8 cos(2 x )+ C. 4. ln( x 2 - 1) dx Solution : Integration by parts with u = ln( x 2 - 1) and dv = dx , so du = 2 x x 2 - 1 dx and v = x . ln( x 2 - 1) dx = x ln( x 2 - 1) - 2 x 2 x 2 - 1 dx = x ln( x 2 - 1) - 2 + 2 x 2 - 1 dx = x ln( x 2 - 1) - 2 x - 1 x - 1 - 1 x + 1 dx = x ln( x 2 - 1) - 2 x - ln | x - 1 | + ln | x + 1 | + C.
5. sin 2 x dx Solution : Substitute u = 2 x , so du = 2 2 2 x dx = 1 u dx .

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