Integration Grandmaster - P13

$A = \int_{0}^{1} \frac{\cos{x}}{\sqrt{2 + \sin(2x)}} \ dx$ $A = \frac{1}{2} \int_{0}^{1} \frac{2\cos{x}}{\sqrt{2 + \sin(2x)}} \ dx$ $A = \frac{1}{2} \int_{0}^{1} \frac{(\cos{x}-\sin{x}) +(\cos{x}+\sin{x}) }{\sqrt{2 + \sin(2x)}} \ dx$ $A = \frac{1}{2} \int_{0}^{1} \frac{\cos{x}-\sin{x}}{\sqrt{2 + \sin(2x)}} \ dx + \frac{1}{2} \int_{0}^{1} \frac{\cos{x}+\sin{x}}{\sqrt{2 + \sin(2x)}} \ dx$ $A = \frac{1}{2} \int_{0}^{1} \frac{\cos{x}-\sin{x}}{\sqrt{1 + (1 + \sin(2x))}} \ dx + \frac{1}{2} \int_{0}^{1} \frac{\cos{x}+\sin{x}}{\sqrt{3 -(1 - \sin(2x))}} \ dx$ $A = \frac{1}{2} \int_{0}^{1} \frac{\cos{x}-\sin{x}}{\sqrt{1 + (\sin^2{x} + \cos^2{x} + 2\sin{x}\cos{x})}} \ dx + \frac{1}{2} \int_{0}^{1} \frac{\cos{x}+\sin{x}}{\sqrt{3 -(\sin^2{x} + \cos^2{x} - 2\sin{x}\cos{x})}} \ dx$ $A = \frac{1}{2} \int_{0}^{1} \frac{\cos{x}-\sin{x}}{\sqrt{1 + (\sin{x}+\cos{x})^2}} \ dx + \frac{1}{2} \int_{0}^{1} \frac{\cos{x}+\sin{x}}{\sqrt{3 -(\sin{x}-\cos{x})^2}} \ dx$ $A = A_1 + A_2$ $A_1 = \frac{1}{2} \int_{0}^{1} \frac{\cos{x}-\sin{x}}{\sqrt{1 + (\sin{x}+\cos{x})^2}} \ dx$ $A_2 = \frac{1}{2} \int_{0}^{1} \frac{\cos{x}+\sin{x}}{\sqrt{3 -(\sin{x}-\cos{x})^2}} \ dx$

Taking $\sin{x}+\cos{x}=z$ transforms $A_1$ to:

$A_1 =\frac{1}{2} \int_{1}^{\sin{1} + \cos{1}} \frac{dz}{\sqrt{1 +z^2}}$

Taking $\sin{x}-\cos{x}=z$ transforms $A_2$ to:

$A_2 =\frac{1}{2} \int_{-1}^{\sin{1} - \cos{1}} \frac{dz}{\sqrt{3 -z^2}}$

Now, $A_1$ and $A_2$ can be evaluated by using standard formulas or by trigonometric substitution. I have chosen to leave out these steps as they are doable, but tedious.

The result is:

$\lfloor 10000A \rfloor = 5181$

If we write $u \;= \; \frac{\cos x + \sin x}{\sqrt{2 + \sin2x}} \hspace{2cm} v \; = \; \frac{\cos x - \sin x}{\sqrt{2 + \sin2x}}$ then $\frac{1}{\sqrt{2 + \sin2x}} \; = \; \frac12\left[ \frac{1}{1-u^2}\frac{du}{dx} - \frac{1}{1+v^2}\frac{dv}{dx}\right]$ so that $\int \frac{1}{\sqrt{2 + \sin2x}}\,dx \; = \; \tfrac14\ln\left(\tfrac{1+u}{1-u}\right) - \tfrac12\tan^{-1}v + c$ Since $u(0) = v(0)= \tfrac{1}{\sqrt{2}} \hspace{2cm} u(1) = \frac{\cos1 + \sin1}{\sqrt{2 + \sin2}} \;,\; v(1) = \frac{\cos1 - \sin1}{\sqrt{2+\sin2}}$ we derive $A$ . Hence $\lfloor 10000A\rfloor = \boxed{5181}$ .