If I_n = int x^n sin x , dx and I_6 - 360 I_2 | vedclass

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(A) Using integration by parts, $I_n = \int x^n \sin x \, dx = -x^n \cos x + n \int x^{n-1} \cos x \, dx = -x^n \cos x + n(x^{n-1} \sin x - (n-1) \int

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$-85$

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(A) Using integration by parts, $I_n = \int x^n \sin x \, dx = -x^n \cos x + n \int x^{n-1} \cos x \, dx = -x^n \cos x + n(x^{n-1} \sin x - (n-1) \int x^{n-2} \sin x \, dx)$.So, $I_n = -x^n \cos x + n x^{n-1} \sin x - n(n-1) I_{n-2}$.This gives the recurrence relation: $I_n + n(n-1) I_{n-2} = -x^n \cos x + n x^{n-1} \sin x$.For $n=6$: $I_6 + 30 I_4 = -x^6 \cos x + 6 x^5 \sin x$.For $n=4$: $I_4 + 12 I_2 = -x^4 \cos x + 4 x^3 \sin x$.Multiplying the second equation by $-30$: $-30 I_4 - 360 I_2 = 30 x^4 \cos x - 120 x^3 \sin x$.Adding the two equations: $I_6 - 360 I_2 = (30 x^4 - x^6) \cos x + (6 x^5 - 120 x^3) \sin x$.Comparing with $f(x) \cos x + g(x) \sin x$, we have $f(x) = 30 x^4 - x^6$ and $g(x) = 6 x^5 - 120 x^3$.Then $f(1) = 30(1)^4 - (1)^6 = 29$ and $g(1) = 6(1)^5 - 120(1)^3 = -114$.Thus, $f(1) + g(1) = 29 - 114 = -85$.

If $I_n = \int x^n \sin x \, dx$ and $I_6 - 360 I_2 = f(x) \cos x + g(x) \sin x$, then $f(1) + g(1) =$

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