‹ Back to the paper
Solve any one of the following.
Prove: ∫_(π/4)^((3π)/4) (x dx)/(1 + sin x) = (√2 - 1)π
Prove: $\int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \frac{x\,dx}{1 + \sin x} = (\sqrt{2} - 1)\pi$
Given: $\int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \frac{x\,dx}{1 + \sin x}$
To show: $(\sqrt{2} - 1)\pi$
Answer
Answer
Let $I = \int_{\pi/4}^{3\pi/4}\dfrac{x\,dx}{1 + \sin x}$. Using the property $\int_a^b f(x)dx = \int_a^b f(a + b - x)dx$ with $a + b = \pi$ gives $I = \int_{\pi/4}^{3\pi/4}\dfrac{\pi - x}{1 + \sin x}dx$; adding the two forms, $2I = \pi\int_{\pi/4}^{3\pi/4}\dfrac{dx}{1 + \sin x}$, which works out to $2I = 2\pi(\sqrt{2} - 1)$.
- Let $I = \int_{\pi/4}^{3\pi/4}\dfrac{x\,dx}{1 + \sin x}$.
- Using $\int_a^b f(x)\,dx = \int_a^b f(a + b - x)\,dx$ with $a + b = \pi$, and $\sin(\pi - x) = \sin x$: $I = \int_{\pi/4}^{3\pi/4}\dfrac{\pi - x}{1 + \sin x}dx$.
- Adding the two forms: $2I = \pi\int_{\pi/4}^{3\pi/4}\dfrac{dx}{1 + \sin x}$.
- $\dfrac{1}{1 + \sin x} = \dfrac{1 - \sin x}{\cos^2 x} = \sec^2 x - \sec x\tan x$, whose antiderivative is $\tan x - \sec x$.
- At $x = \frac{3\pi}{4}$: $\tan x - \sec x = -1 + \sqrt{2}$; at $x = \frac{\pi}{4}$: $1 - \sqrt{2}$.
- So $\int_{\pi/4}^{3\pi/4}\dfrac{dx}{1 + \sin x} = (\sqrt{2} - 1) - (1 - \sqrt{2}) = 2(\sqrt{2} - 1)$.
- $2I = 2\pi(\sqrt{2} - 1)$, hence $I = (\sqrt{2} - 1)\pi$.