# soln2 - R Balan Homework#2 Solutions MATH 464 1 0 1 dx = 2...

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R. Balan Homework #2 Solutions MATH 464 1. Z 0 1 x 2 + 1 dx = Z π/ 2 0 1 tan 2 ( θ ) + 1 sec 2 ( θ ) = Z π/ 2 0 sec 2 ( θ ) sec 2 ( θ ) = Z π/ 2 0 = π/ 2 2. Z -∞ 1 x 2 + a 2 dx = 2 Z 0 1 x 2 + a 2 dx = 2 a 2 Z 0 1 x 2 /a 2 + 1 dx = 2 | a | Z 0 1 y 2 + 1 dy = 2 | a | · π/ 2 = π a 3. If a 6 = 1, we have that 1 ( x 2 + 1)( x 2 + a 2 ) = 1 a 2 - 1 ± 1 x 2 + 1 - 1 x 2 + a 2 ² , and the integral is then π a ( a +1) . If a = 1, the integral is π 2 . 4. If a > 0, we have Z 0 e - ax dx = 1 a Z 0 -∞ e y dy = 1 a lim y →-∞ (1 - e y ) = 1 a . 5. Using integration by parts, the integral becomes lim x →∞ - ( x 2 + 2 x + 2) e - x + 2 = 2 . 6. Denote the value of these integrals by I n for n > 0. Integration by parts yields the recursion I n = n · I n - 1 , so we conclude that I n = n !. 7. If ω = 0, then the integral is 1. Otherwise, integration by parts yields Z t 0 e - x cos( ωx ) dx = - e - x cos( ωx ) - ω Z e - x sin( ωx ) dx ³ ³ ³ ³ t 0 = - e - x cos( ωx ) - ω ± - e - x sin( ωx ) + ω Z e - x cos( ωx ) dx ²³ ³ ³ ³ t 0 . Thus, (1 + ω 2 ) Z t 0 e - x cos( ωx ) dx = - e - x cos( ωx ) + e - x ω sin( ωx ) ³ ³ t 0 , and hence Z 0 e - x cos( ωx ) dx = lim x →∞ e - x 1 + ω 2 [ ω sin(
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