NIMCET 2017 Question Paper with solutions

Let I = $\int {\mathit{x}}^3$ sin x dx
= ${\mathit{x}}^3$ [- cos x] - ∫ (- cos x) $3{\mathit{x}}^2$ dx
$-{\mathit{x}}^3\mathit{c}\mathit{o}\mathit{s}\mathit{x}$ + 3 ∫ ${\mathit{x}}^2$ cos x dx
= $-{\mathit{x}}^3$ cos x + 3 [${\mathit{x}}^2$ (sin x) - ∫ sinx . (2x) dx]
= $-{\mathit{x}}^3$ cos x + $3{\mathit{x}}^2$ sin x - 6 ∫ c sin x dx
= $-{\mathit{x}}^3$ cos x + $3{\mathit{x}}^2$ sin x - 6
[x (- cos x) - ∫ (- cos x) 1 dx]
= ${\mathit{x}}^3$ cos x + $3{\mathit{x}}^2$ sin x + 6x cos x - 6 sin x
∴ $\underset{0}{\overset{\mathit{\pi }}{\int }}{\mathit{x}}^3$ sin x dx = ${\left[-{\mathit{x}}^3\mathit{c}\mathit{o}\mathit{s}\mathit{x}+3{\mathit{x}}^2\mathit{s}\mathit{i}\mathit{n}\mathit{x}+6\mathit{x}\mathit{c}\mathit{o}\mathit{s}\mathit{x}-6\mathit{s}\mathit{i}\mathit{n}\mathit{x}\right]}_0^{\mathit{\pi }}$
= ($-{\mathit{\pi }}^3$ cos π + $3{\mathit{\pi }}^2$ sin π + 6π cos π - 6 sin π)
= 90 + 0 + 0 - 6 sin 0)
= ${\mathit{\pi }}^3$ + 0 - 6π + 0 + 0 = ${\mathit{\pi }}^3$ - 6π