This integral is from the 2010 MIT Integration Bee. My attempt is below. $$I=\int_0^{\frac{\pi}{2}}\frac{dx}{\sin x+\sec x}$$ We can rewrite this as the following: $$I=\int_0^{\frac{\pi}{2}}\frac{\cos x}{1+\sin x\cos x}dx=\int_0^{\frac{\pi}{2}}\frac{2\cos x}{2+\sin(2x)}dx$$ From here I attempt to calculate the definite integral. $$\int\frac{2\cos x}{2+\sin(2x)}dx=\frac{\cos x+\sin x+\cos x-\sin x}{2+\sin(2x)}dx=\int\frac{\cos x+\sin x}{2+\sin(2x)}dx+\int\frac{\cos x-\sin x}{2+\sin(2x)}dx$$ After implementing the double angle formula and the substitutions $t=x-\cos x$ and $u=\sin x+\cos x$ for the left and right sides respectively we are left with: $$\int\frac{dt}{3-t^2}+\int\frac{du}{1+u^2}=\frac{1}{2\sqrt{3}}\ln\left| \frac{\sqrt{3}+t}{\sqrt{3}-t} \right|+\arctan u+C$$ $$=\frac{1}{2\sqrt{3}}\ln\left| \frac{\sqrt{3}+\sin x-\cos x}{\sqrt{3}-\sin x+\cos x} \right|+\arctan(\sin x+\cos x)+C$$ Now we evaluate all of that from $0$ to $\frac{\pi}{2}$ and after some tedious computation: $$I=\frac{1}{\sqrt{3}}\ln(2+\sqrt{3})$$ I want to learn other solutions that are not so computationally intensive. There is a lot of room to make errors in the attempt I made.
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Are you sure $t=x-\cos(x)$? – Тyma Gaidash Sep 10 '24 at 19:02
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$$\int_0^{\frac{\pi}{2}}\frac{1}{\cos x+\sec x}dx=\int_0^{\frac{\pi}{2}} \frac{\sin x}{1+\sin x\cos x}dx\ =\frac12 \int_0^{\frac{\pi}{2}} \frac{\sin x+\cos x}{1+\sin x\cos x}dx =\frac12 \int_0^{\frac{\pi}{2}} \frac{d(\sin x-\cos x)}{1+\frac12[1-(\sin x-\cos x)^2]} $$ – Quanto Sep 10 '24 at 19:31
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Why did you have $dx$ written twice? – Dan Sep 10 '24 at 19:37
4 Answers
It is much quicker and less painful if you use the King Property. $$I=\int_0^{\frac{\pi}{2}}\frac{dx}{\sin x+\sec x}dx=\int_0^{\frac{\pi}{2}}\frac{\cos(\frac{\pi}{2}-x)}{1+\sin(\frac{\pi}{2}-x)\cos(\frac{\pi}{2}-x)}dx=\int_0^{\frac{\pi}{2}} \frac{\sin x}{1+\sin x\cos x}dx$$ After substitutions, we arrive at a much more manageable computation: $$\frac{1}{\sqrt{3}}\left[ \ln\left| \frac{\sqrt{3}+t}{\sqrt{3}-t} \right| \right]^1_{-1}=\color{blue}{\frac{1}{\sqrt{3}}\ln(2+\sqrt{3})}$$
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$$\begin{align*} & \int_0^\tfrac\pi2 \frac{dx}{\sin x+\sec x} = \int_0^\tfrac\pi2 \frac{\cos x}{\frac12\sin(2x)+1} \, dx \\ &= \sqrt2 \int_{-\tfrac\pi4}^\tfrac\pi4 \frac{\cos x-\sin x}{\cos(2x)+2} \, dx & x\to x+\frac\pi4 \\ &= 2\sqrt2 \int_0^\tfrac\pi4 \frac{d(\sin x)}{3-2\sin^2x} & \rm symmetry \\ &= \frac2{\sqrt3} \operatorname{artanh}\left(\sqrt{\frac23}\sin \frac\pi4\right) = \boxed{\frac2{\sqrt3} \operatorname{artanh}\frac1{\sqrt3}} \end{align*}$$
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$$ I =\int_0^{\frac{\pi}{2}} \frac{\cos x}{\sin x \cos x+1} dx=\int_0^{\frac{\pi}{2}} \frac{\sin x}{\cos x \sin x+1} d x $$ Averaging them yields $$ \begin{aligned} I & =\frac{1}{2} \int_0^{\frac{\pi}{2}} \frac{\cos x+\sin x}{\cos x \sin x+1} d x \\&= \int_0^{\frac{\pi}{2}} \frac{d(\sin x-\cos x)}{3-(\sin x-\cos x)^2}\\ & =\frac{1}{2 \sqrt{3}}\left[\ln \left(\frac{\sqrt{3}+u}{\sqrt{3}-u}\right)\right]_{-1}^1, \quad \textrm{ where }u=\sin x-\cos x \\ & =\frac{1}{\sqrt{3}} \ln (2+\sqrt{3}) \end{aligned} $$
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If,a you did, you use the tangent half-angle substitution
$$I=\int_0^1 \frac{2-2 t^2}{t^4-2 t^3+2 t^2+2 t+1}\,dt$$ $$t^4-2 t^3+2 t^2+2 t+1= $$ $$\big(t^2+\left(\sqrt{3}-1\right) t+\left(2-\sqrt{3}\right)\big)\big(t^2-\left(\sqrt{3}+1\right) t+\left(2+\sqrt{3}\right)\big)$$ Using partial frcation dcomposition, the integrate is $$\frac{1}{\sqrt{3}}\Bigg(\frac {t+1}{t^2+\left(\sqrt{3}-1\right) t+\left(2-\sqrt{3}\right) }- \frac {t+1}{t^2-\left(\sqrt{3}+1\right) t+\left(2+\sqrt{3}\right) }\Bigg)$$ which are not very difficult each of them leading to a logarithm and an arctangent (as ususal).
Using the bounds, simplifying and using the logarithmic representation of $\cosh ^{-1}(t)$ $$I=\frac{\cosh ^{-1}(2)}{\sqrt{3}}$$
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