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Question Number 197766 by universe last updated on 28/Sep/23

$${\int }_0^{\pi /2}\left(\underset{n\to \mathrm{\infty }}{\lim }n{\sin }^{2n+1}x\cos x\right)dx=?$$

Commented by witcher3 last updated on 28/Sep/23

$$\mathrm{dominate}\mathrm{cv}\mathrm{Theorem}$$

Answered by witcher3 last updated on 28/Sep/23

$$={\int }_0^1{\underset{\mathrm{n}\to \mathrm{\infty }}{\lim x}}^{2\mathrm{n}+1}\mathrm{dx},\sin \left(\mathrm{x}\right)\to \mathrm{x}$$$$\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }{\int }_0^1{\mathrm{nx}}^{2\mathrm{n}+1}\mathrm{dx}=\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\frac{\mathrm{n}}{2\mathrm{n}+2}=\frac{1}{2}$$$${\mathrm{x}}^{\mathrm{n}}=\mathrm{u}\Rightarrow \mathrm{dx}={\mathrm{nx}}^{\mathrm{n}-1}\mathrm{dx}=\mathrm{du}$$$$\mathrm{x}={\mathrm{u}}^{\frac{1}{\mathrm{n}}}$$$$=\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }{\int }_0^1\mathrm{u}.{\mathrm{u}}^{\frac{2}{\mathrm{n}}}\mathrm{du}=\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }{\int }_0^1{\mathrm{u}}^{1+\frac{2}{\mathrm{n}}}⩽\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }{\int }_0^1\mathrm{du}=1$$$$\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }{\int }_0^1{\mathrm{nsin}}^{2\mathrm{n}+1}\left(\mathrm{x}\right)\cos \left(\mathrm{x}\right)\mathrm{dx}=\frac{1}{2}={\int }_0^1{\underset{\mathrm{n}\to \mathrm{\infty }}{\lim nsin}}^{2\mathrm{n}+1}\left(\mathrm{x}\right)\cos \left(\mathrm{x}\right)\mathrm{dx}$$$$$$

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