Definite integral $\int_0^{\pi/12} \left( \frac{e^x \cos(x)}{\cos(x+\frac\pi4)} \right)^2 {\rm d}x= e^{\pi/6} \cos \left( \frac\pi6 \right) - \frac12$

$$\int_0^{\pi/12} \left( \frac{e^x \cos(x)}{\cos(x+\frac\pi4)} \right)^2 {\rm d}x= e^{\pi/6} \cos \left( \frac\pi6 \right) - \frac12$$

I've spent alot of time moving things around but I can't find a way to actually prove it... Would really appreciate some help.

$$\int_0^{\pi/12} \left( \frac{e^x \cos(x)}{\cos(x+\frac\pi4)} \right)^2 {\rm d}x \\= \int_0^{\pi/12} \frac{e^{2x}\cos^2(x)}{\frac12 (\cos(x)-\sin(x))^2}{\rm d}x \\= 2\int_0^{\pi/12}e^{2x} \left( \frac{\cos(x)}{ \cos(x)-\sin(x)} \right)^2{\rm d}x \\= \int_0^{\pi/12}e^{2x} \left( \frac{1 + \cos(2x)}{ 1-\sin(2x)} \right){\rm d}x \\= \frac12 \int_0^{\pi/6} e^u \frac{1+\cos(u)}{1-\sin(u)} {\rm d}u$$

I think integrating by parts works in most cases where there's a combination of $$e^x$$ and some other random function and of course we expect an elementary primitive. I will continue what you did: $$I=\frac12 \int_0^\frac{\pi}{6} e^x \frac{1+\cos x}{\color{red}{1-\sin x}}dx=\frac12 \int_0^\frac{\pi}{6} e^x (1+\cos x)\color{red}{\left(\frac{\cos x}{1-\sin x}\right)'}dx$$ $$=\frac12 e^x (1+\cos x)\left(\frac{\cos x}{1-\sin x}\right)\bigg|_0^\frac{\pi}{6}-\frac12 \int_0^\frac{\pi}{6} e^x \frac{1-\sin x+\cos x}{1-\sin x}\cos xdx$$ $$=e^{\frac{\pi}{6}}\left(\frac34+\frac{\sqrt 3}2\right)-1-\frac12\int_0^\frac{\pi}{6}e^x\left(\cos x +\frac{{\cos^2 x}}{1-\sin x}\right)dx$$ $$=e^{\frac{\pi}{6}}\left(\frac34+\frac{\sqrt 3}2\right)-1-\frac12\int_0^\frac{\pi}{6}e^x(\cos x+1+\sin x)dx$$ $$=e^{\frac{\pi}{6}}\left(\frac34+\frac{\sqrt 3}2\right)-1-\frac12 e^x(1+\sin x)\bigg|_0^\frac{\pi}{6}$$$$=e^{\frac{\pi}{6}}\left(\frac34+\frac{\sqrt 3}2\right)-1-\frac34 e^\frac{\pi}{6}+\frac12=e^{\large \frac{\pi}{6}}\frac{\sqrt 3}{2}-\frac12$$

Of course we also have: $$\int e^x\frac{1+\cos x}{1-\sin x}dx=e^x(1+\cos x)\frac{\cos x}{1-\sin x}-e^x(1+\sin x)+C=\frac{e^x\cos x}{1-\sin x}+C$$

• (+1) Why/how are you so damn good at integrating? Your work never ceases to amaze me. Commented Apr 28, 2019 at 0:32
• I'm supposed to keep an air of awesomeness and mystery about myself $\ddot \frown$, but I can tell you a little secret: Sometimes I type on the search button "user:583016 [integration]" and then I watch. Commented Apr 28, 2019 at 8:48

Yeet I found another way to solve it!!

$$\frac12 \int_0^{\pi/6} e^x \frac{1+\cos(x)}{1-\sin(x)} {\rm d}x \\ = \frac12 \int_0^{\pi/6} e^x \frac{(1+\cos(x))(1+\sin(x))}{1-\sin^2(x)} {\rm d}x \\ = \frac12 \int_0^{\pi/6} e^x \frac{1+\sin(x)+\cos(x)+\sin(x)\cos(x)}{\cos^2(x)} {\rm d}x \\ = \frac12 \int_0^{\pi/6} e^x( \tan(x) + \sec^2(x) + \sec(x) + \sec(x)\tan(x)) {\rm d}x \\ = \frac12 e^x \tan(x) \bigg|_0^{\pi/6} + \frac12 e^x \sec(x) \bigg|_0^{\pi/6} \\ = \frac12 e^{\pi/6} \frac{1}{\sqrt3} - 0 + \frac12 e^{\pi/6}\frac{2}{\sqrt3} - \frac12 \\ = \frac{\sqrt3}{2} e^{\pi/6} -\frac12$$

Using the fact that $$\int e^x (f(x) + f'(x)) {\rm d} x = e^x f(x) + C$$.