Let ∫x3sin⁡xdx=g(x)+C, where C is the constant of integration. If 8gπ2+g′π2=απ3+βπ2+γ,α,β,γ∈Z,Then α+β−γ equals :

∫x3sin⁡xdx=−x3cos⁡x+∫3x2cos⁡xdx=−x3cos⁡x+3x2sin⁡x−∫6xsin⁡xdx=−x3cos⁡x+3x2sin⁡x+6xcos⁡x−6sin⁡x+cSo g(x)=−x3cos⁡x+3x2sin⁡x+6xcos⁡x−6sin⁡x gπ2=3π24−6 g′(x)=−3x2cos⁡x+x3sin⁡x+6cos⁡x−6cos⁡x g′π2=π38 8gπ2+g′π2=π3+6π2−48 So α+β−γ=55

Let ∫x3sin⁡xdx=g(x)+C, where C is the constant of integration. If 8gπ2+g′π2=απ3+βπ2+γ,α,β,γ∈Z,Then α+β−γ equals :

∫x3sin⁡xdx=−x3cos⁡x+∫3x2cos⁡xdx=−x3cos⁡x+3x2sin⁡x−∫6xsin⁡xdx=−x3cos⁡x+3x2sin⁡x+6xcos⁡x−6sin⁡x+cSo g(x)=−x3cos⁡x+3x2sin⁡x+6xcos⁡x−6sin⁡x gπ2=3π24−6 g′(x)=−3x2cos⁡x+x3sin⁡x+6cos⁡x−6cos⁡x g′π2=π38 8gπ2+g′π2=π3+6π2−48 So α+β−γ=55

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Let $\int x^{3} \sin xdx = g (x) + C$ , where C is the constant of integration. If $8 (g (\frac{π}{2}) + g^{'} (\frac{π}{2})) = {απ}^{3} + {βπ}^{2} + γ, α, β, γ \in Z,$
Then $α + β - γ$ equals :

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Detailed Solution

$\begin{array}{l} \int x^{3} \sin xdx = - x^{3} \cos x + \int 3 x^{2} \cos xdx \\ = - x^{3} \cos x + 3 x^{2} \sin x - \int 6 xsin xdx \\ = - x^{3} \cos x + 3 x^{2} \sin x + 6 xcos x - 6 \sin x + c \end{array}$
So $g (x) = - x^{3} \cos x + 3 x^{2} \sin x + 6 xcos x - 6 \sin x$
$\begin{array}{l} \begin{aligned} g (\frac{π}{2}) = \frac{3 π^{2}}{4} - 6 \\ g^{'} (x) = - 3 x^{2} \cos x + x^{3} \sin x + 6 \cos x - 6 \cos x \\ g^{'} (\frac{π}{2}) = \frac{π^{3}}{8} \\ 8 (g (\frac{π}{2}) + g^{'} (\frac{π}{2})) = π^{3} + 6 π^{2} - 48 \end{aligned} \\ So α + β - γ = 55 \end{array}$