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\(u_{10}=integration lim 0 to pi/2 x^10sinxdx , then the value of u_{10}+90u_{8}\) is equal to

Note by Anirudha Nayak 3 years, 10 months ago

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Let \(\displaystyle u_{10} = \int_{0}^{\frac{\pi}{2}}x^{10}\cdot \sin xdx\)

Using Integration by parts, we get

\(\displaystyle u_{10} =0+ 10\int_{0}^{\frac{\pi}{2}}x^9\cos xdx\)

\(\displaystyle u_{10} = 10\cdot \frac{\pi^9}{2^9} - 90u_{8}\)

\(\displaystyle u_{10}+90u_{8} = 10\cdot \frac{\pi^9}{2^9}\)

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TopNewestLet \(\displaystyle u_{10} = \int_{0}^{\frac{\pi}{2}}x^{10}\cdot \sin xdx\)

Using Integration by parts, we get

\(\displaystyle u_{10} =0+ 10\int_{0}^{\frac{\pi}{2}}x^9\cos xdx\)

\(\displaystyle u_{10} = 10\cdot \frac{\pi^9}{2^9} - 90u_{8}\)

\(\displaystyle u_{10}+90u_{8} = 10\cdot \frac{\pi^9}{2^9}\)

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