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

If I 10 = π / 2 ∫ 0 X 10 Sin X D X , Then the Value of I10 + 90i8 Is,9 ( π 2 ) 9,10 ( π 2 ) 9,( π 2 ) 9,9 ( π 2 ) 8 - Mathematics

Question

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

Options

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

MCQ

Solution

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

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Definite Integrals

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Q 27 Q 26 Q 28

Chapter 20: Definite Integrals - MCQ [Page 119]

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RD Sharma Mathematics [English] Class 12

Chapter 20 Definite Integrals
MCQ | Q 27 | Page 119

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