Integral Cake Delivery IIIII
If my delivery equals , a,b,c,d,e are positive integers , find .
The answer is 8.0.
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1 solution
\(\begin{align*} I &= \int_0^\frac{\pi}{4} \frac{\sin^2(x)}{\cos^3(x)} dx = \int_0^\frac{\pi}{4} \sec(x) \tan^2(x) dx \\
&= \int_0^\frac{\pi}{4} \sec^3(x) - \sec(x) dx & {\color{blue} \tan^2(x) =\sec^2(x) - 1 } \\
& {\color{red} \int \sec^3(x) dx = \sec(x)\cdot \tan(x) - \int \sec(x) \tan^2(x) dx = \sec(x)\cdot \tan(x) - I } & {\color{blue} u=\sec(x) \ dv=\sec^2(x) } \\
& {\color{red} \int \sec(x) dx = \int \sec(x) \cdot \frac{\sec(x)+\tan(x)}{\sec(x)+\tan(x)} dx = \int \frac{\sec^2(x)+\sec(x) \tan(x)}{\sec(x)+\tan(x)} dx } \\
& {\color{red} \int \sec(x) dx = \ln(\sec(x)+\tan(x)) } \\
I &= \sec(x)\cdot \tan(x) - I - \ln(\sec(x)+\tan(x)) \\ 2I &= \sec(x)\cdot \tan(x) - \ln(\sec(x)+\tan(x)) \\ I &= \frac{1}{2}\left (\left .\sec(x)\cdot \tan(x) - \ln(\sec(x)+\tan(x)) \right |_0^\frac{\pi}{4} \right ) \\
I &= \frac{1}{2}\left ( \sqrt(2) - \ln(\sqrt(2)+1)\right ) \\ \end{align*}\)