Question

If $I_{10}=\underset{0}{\overset{\mathrm{\pi }/2}{\int }}{x}^{10}\sin xdx,$ then the value of I₁₀ + 90I₈ is

(a) $9{\left(\frac{\mathrm{\pi }}{2}\right)}^9$

(b) $10{\left(\frac{\mathrm{\pi }}{2}\right)}^9$

(c) ${\left(\frac{\mathrm{\pi }}{2}\right)}^9$

(d) $9{\left(\frac{\mathrm{\pi }}{2}\right)}^8$

Answer 1

$$\begin{gathered} \left(\mathrm{b}\right)10{\left(\frac{\mathrm{\pi }}{2}\right)}^9 \\ \mathrm{We}\mathrm{have}, \\ {I}_{10}=\underset{0}{\overset{\mathrm{\pi }/2}{\int }}{x}^{10}\sin xdx \\ ={\left[x^{10}\left(-\cos x\right)\right]}_0^{\frac{\mathrm{\pi }}{2}}-\underset{0}{\overset{\mathrm{\pi }/2}{\int }}\left[10x^9\int \sin xdx\right]dx \\ ={\left[-x^{10}\cos x\right]}_0^{\frac{\mathrm{\pi }}{2}}-10\underset{0}{\overset{\mathrm{\pi }/2}{\int }}{x}^9\left(-\cos x\right)dx \\ =-{\left[x^{10}\cos x\right]}_0^{\frac{\mathrm{\pi }}{2}}+10\underset{0}{\overset{\mathrm{\pi }/2}{\int }}{x}^9\cos xdx \\ =-{\left[x^{10}\cos x\right]}_0^{\frac{\mathrm{\pi }}{2}}+10{\left[x^9\sin x\right]}_0^{\frac{\mathrm{\pi }}{2}}-10\underset{0}{\overset{\mathrm{\pi }/2}{\int }}9x^8\sin xdx \\ =-\left[{\left(\frac{\mathrm{\pi }}{2}\right)}^{10}\times 0-0^{10}\cos 0\right]+10\left[{\left(\frac{\mathrm{\pi }}{2}\right)}^9\times 1-0^9\times 0\right]-90\underset{0}{\overset{\mathrm{\pi }/2}{\int }}{x}^8\sin xdx \\ =10\left[{\left(\frac{\mathrm{\pi }}{2}\right)}^9\times 1\right]-90I_8 \\ =10{\left(\frac{\mathrm{\pi }}{2}\right)}^9-90I_8 \\ \therefore I_{10}+90I_8=10{\left(\frac{\mathrm{\pi }}{2}\right)}^9 \end{gathered}$$