A calculus problem by Kishore S. Shenoy

I hope my solution would be the one you guys may be looking for.

$\displaystyle \int _{ 0 }^{ \frac { \pi }{ 4 } }{ \frac { \sec { x } }{ 1+2\sin ^{ 2 }{ x } } dx } =\int _{ 0 }^{ \frac { \pi }{ 4 } }{ \frac { 1 }{ (1+2\sin ^{ 2 }{ x } )\cos { x } } dx }$

Now, I call it "partial fraction"

$\displaystyle \int _{ 0 }^{ \frac { \pi }{ 4 } }{ \frac { 1 }{ (1+2\sin ^{ 2 }{ x } )\cos { x } } dx } =\frac { 1 }{ 3 } \int _{ 0 }^{ \frac { \pi }{ 4 } }{ \left( \frac { 1 }{ \cos { x } } +\frac { 2\cos { x } }{ 1+2\sin ^{ 2 }{ x } } \right) dx }$

$\displaystyle \frac { 1 }{ 3 } \int _{ 0 }^{ \frac { \pi }{ 4 } }{ \left( \frac { 1 }{ \cos { x } } +\frac { 2\cos { x } }{ 1+2\sin ^{ 2 }{ x } } \right) dx } =\frac { 1 }{ 3 } \int _{ 0 }^{ \frac { \pi }{ 4 } }{ \left( \frac { 1 }{ \cos { x } } \right) dx } +\frac { 1 }{ 3 } \int _{ 0 }^{ \frac { \pi }{ 4 } }{ \left( \frac { 2\cos { x } }{ 1+2\sin ^{ 2 }{ x } } \right) dx }$

$\displaystyle =\frac { 1 }{ 3 } \int _{ 0 }^{ \frac { \pi }{ 4 } }{ \sec { x } dx } +\frac { 1 }{ 3 } \int _{ x=0 }^{ \frac { \pi }{ 4 } }{ \left( \frac { 2 }{ 1+2\sin ^{ 2 }{ x } } \right) d\sin { x } }$

$\displaystyle { = }\frac { \ln { \left| \sec { x } +\tan { x } \right| } }{ 3 } +\frac { \sqrt { 2 } \arctan { \left( \sqrt { 2 } \sin { x } \right) } }{ 3 }$

$\displaystyle =\frac { \ln { \left| 1+\sqrt { 2 } \right| } }{ 3 } +\frac { \sqrt { 2 } \pi }{ 12 }$

$\displaystyle =\frac { \ln { \left| 17+12\sqrt { 2 } \right| } }{ 12 } +\frac { \sqrt { 2 } \pi }{ 12 }$

$\int_{0}^{\pi/4} \dfrac{\sec{x}}{1+2 \sin^2{x}} \; dx=\int_{0}^{\pi/4} \dfrac{\cos{x}}{\cos^2{x}+2 \sin^2{x} \cos^2{x}} \; dx =\int_{0}^{\pi/4} \dfrac{\cos{x}}{-2 \sin^4{x}+\sin^2{x}+1} \; dx$

$u=\sin{x} \Rightarrow \dfrac{du}{\cos{x}}=dx$

$\cdots=\int_{0}^{\frac{1}{\sqrt{2}}} \dfrac{1}{-2u^4+u^2+1} \; du=\int_{0}^{\frac{1}{\sqrt{2}}} \dfrac{2}{3(2u^2+1)}+\dfrac{1}{6(u+1)}-\dfrac{1}{6(u-1)} \; du$

$\cdots=\dfrac{2}{3} \int_{0}^{\frac{1}{\sqrt{2}}} \dfrac{1}{2u^2+1} \; du+ \dfrac{1}{6} \int_{0}^{\frac{1}{\sqrt{2}}} \dfrac{1}{u+1} \; du-\dfrac{1}{6} \int_{0}^{\frac{1}{\sqrt{2}}} \dfrac{1}{u-1} \; du$

$w=\sqrt{2} u \Rightarrow \dfrac{dw}{\sqrt{2}}=du$

$\dfrac{2}{3} \int_{0}^{\frac{1}{\sqrt{2}}} \dfrac{1}{2u^2+1} \; du =\dfrac{\sqrt{2}}{3} \int_{0}^{1} \dfrac{1}{w^2+1} \; dw= \dfrac{\sqrt{2}}{3} \left [\arctan{w} \right]_{0}^{1}=\dfrac{\sqrt{2}}{12} \pi$

$\dfrac{1}{6} \int_{0}^{\frac{1}{\sqrt{2}}} \dfrac{1}{u+1} \; du=\dfrac{1}{6} \left [\log{u+1} \right]_{0}^{\frac{1}{\sqrt{2}}} =\dfrac{1}{6} \log{\left (1+\dfrac{1}{\sqrt{2}} \right )}$

$\dfrac{1}{6} \int_{0}^{\frac{1}{\sqrt{2}}} \dfrac{1}{u-1} \; du=\dfrac{1}{6} \left [\log{u-1} \right]_{0}^{\frac{1}{\sqrt{2}}} =\dfrac{1}{6} \log{\left (1-\dfrac{1}{\sqrt{2}} \right )}$

$\dfrac{\sqrt{2}}{12} \pi+\dfrac{1}{6} \log{\left (1+\dfrac{1}{\sqrt{2}} \right )}-\dfrac{1}{6} \log{\left (1-\dfrac{1}{\sqrt{2}} \right )}=\dfrac{\sqrt{2}}{12} \pi+ \dfrac{1}{6} \left ( \log{\left ( \dfrac{\frac{\sqrt{2}+1}{\sqrt{2}}}{\frac{\sqrt{2}-1}{\sqrt{2}}} \right)} \right)$

$\cdots=\dfrac{\sqrt{2}}{12} \pi+ \dfrac{1}{6} \left ( \log{(3+2 \sqrt{2})} \right) =\dfrac{\sqrt{2}}{12} \pi+ \dfrac{1}{12} \left ( \log{(3+2 \sqrt{2})^2} \right)=\dfrac{\sqrt{2}}{12} \pi+ \dfrac{1}{12} \left ( \log{(17+12 \sqrt{2})} \right)$

$A=2,B=12,C=1,D=12,E=17 \Rightarrow A+B+C+D+E=\boxed{\boxed{44}}$