Question

If \[I_{10} = \int\limits_0^{\pi/2} x^{10} \sin x\ dx,\]  then the value of I₁₀ + 90I₈ is

 

Options

• \[9 \left( \frac{\pi}{2} \right)^9\]
• \[10 \left( \frac{\pi}{2} \right)^9\]
• \[\left( \frac{\pi}{2} \right)^9\]
• \[9 \left( \frac{\pi}{2} \right)^8\]

Answer 1

\[10 \left( \frac{\pi}{2} \right)^9 \]
\[\text{We have}, \]
\[ I_{10} = \int\limits_0^{\pi/2} x^{10} \sin x\ dx\]
\[ = \left[ x^{10} \left( - \cos x \right) \right]_0^\frac{\pi}{2} - \int\limits_0^{\pi/2} \left[ 10 x^9 \int\sin x dx \right]dx\]
\[ = \left[ - x^{10} \cos x \right]_0^\frac{\pi}{2} - 10 \int\limits_0^{\pi/2} x^9 \left( - \cos x \right) dx\]
\[ = - \left[ x^{10} \cos x \right]_0^\frac{\pi}{2} + 10 \int\limits_0^{\pi/2} x^9 \cos x\ dx\]
\[ = - \left[ x^{10} \cos x \right]_0^\frac{\pi}{2} + 10 \left[ x^9 \sin x \right]_0^\frac{\pi}{2} - 10 \int\limits_0^{\pi/2} 9 x^8 \sin x dx\]
\[ = - \left[ \left( \frac{\pi}{2} \right)^{10} \times 0 - 0^{10} \cos 0 \right] + 10\left[ \left( \frac{\pi}{2} \right)^9 \times 1 - 0^9 \times 0 \right] - 90 \int\limits_0^{\pi/2} x^8 \sin x dx\]
\[ = 10\left[ \left( \frac{\pi}{2} \right)^9 \times 1 \right] - 90 I_8 \]
\[ = 10 \left( \frac{\pi}{2} \right)^9 - 90 I_8 \]
\[ \therefore I_{10} + 90 I_8 = 10 \left( \frac{\pi}{2} \right)^9\]