Question Number 200464 by Frix last updated on 19/Nov/23
$$\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\frac{{dx}}{\mathrm{1}+\mathrm{tan}^{\mathrm{2023}} \:{x}}=??????? \\ $$
Answered by som(math1967) last updated on 19/Nov/23
![I=∫_0 ^(π/2) ((cos^(2023) xdx)/(sin^(2023) x+cos^(2023) x)) 2I=∫_0 ^(π/2) ((cos^(2023) ((π/2)−x)+cos^(2023) xdx)/(sin^(2023) ((π/2)−x)+cos^(2023) ((π/2)−x))) 2I=[x]_0 ^(π/2) ⇒I=(π/4)](https://www.tinkutara.com/question/Q200467.png)
$$\:{I}=\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\frac{{cos}^{\mathrm{2023}} {xdx}}{{sin}^{\mathrm{2023}} {x}+{cos}^{\mathrm{2023}} {x}}\: \\ $$$$\mathrm{2}{I}=\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\frac{{cos}^{\mathrm{2023}} \left(\frac{\pi}{\mathrm{2}}−{x}\right)+{cos}^{\mathrm{2023}} {xdx}}{{sin}^{\mathrm{2023}} \left(\frac{\pi}{\mathrm{2}}−{x}\right)+{cos}^{\mathrm{2023}} \left(\frac{\pi}{\mathrm{2}}−{x}\right)} \\ $$$$\mathrm{2}{I}=\left[{x}\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \Rightarrow{I}=\frac{\pi}{\mathrm{4}}\: \\ $$
Commented by Frix last updated on 21/Nov/23
Yes! Thank you!