-6.r.10 evaluate int with u v subst

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SUMMARY

The integral $\displaystyle\int_{0}^{\pi/8}\dfrac{\sec^2(2x)}{2+\tan\left({2x}\right)}$ can be evaluated using substitution methods. Two substitution approaches were discussed: first, using $u=\dfrac{\tan\left(2x\right)}{\sqrt{2}}$ leading to $du=\sqrt{2}\sec^2(2x) dx$, and second, using $u = 2+\tan(2x)$ resulting in $du = 2\sec^2(2x) \, dx$. The final evaluation yields $\dfrac{1}{2} \int_2^3 \dfrac{du}{u} = \ln\sqrt{\dfrac{3}{2}}$.

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Evaluate
$\displaystyle\int_{0}^{\pi/8}\dfrac{\sec^2(2x)}{2+\tan\left({2x}\right)}$
ok not sue if this u v is correct or maybe better...
$u=\dfrac{\tan\left(2x\right)}{\sqrt{2}}\therefore du=\sqrt{2}\sec^2(2x) dx$
 
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one can use straight up substitution …

let $u = 2+\tan(2x) \implies du = 2\sec^2(2x) \, dx$

$\displaystyle \dfrac{1}{2} \int_2^3 \dfrac{du}{u} = \ln\sqrt{\dfrac{3}{2}}$
 
https://dl.orangedox.com/GXEVNm73NxaGC9F7Cy
 

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