Hussain-Alqatari
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Here are some interesting integrals: (Will include more interesting integrals soon), also will include interesting things other than integrals.

$$\tag{1}\int_{-\infty}^{\infty}\frac{\mathrm{d}x}{x^2-2x\cot(x)+\csc^2(x)}=\pi$$

$$\tag{2}\int_{0}^{1}\bigg(\left\lfloor\frac{a}{x}\right\rfloor-a\left\lfloor\frac{1}{x}\right\rfloor\bigg)\mathrm{d}x=a\log(a),[\text{where }0

$$\tag{3}\int_{-1}^{1}\frac{1}{x}\sqrt{\frac{1+x}{1-x}}\log\bigg(\frac{2x^2+2x+1}{2x^2-2x+1}\bigg)\mathrm{d}x=4\pi\cot\big(\sqrt{\varphi}\big), \bigg[\text{where }\varphi=\text{golden ratio}=\frac{1+\sqrt{5}}{2}\bigg]$$

$$\tag{4}\int_{-1}^{1}\frac{1}{x}\sqrt{\frac{1+x}{1-x}}\log(1-ax)\mathrm{d}x=\pi\bigg(\log\bigg(\frac{1}{2}+\frac{1}{2}\sqrt{1-a^2}\bigg)-\sin^{-1}(a)\bigg),[\text{where }-1\le a\le1]$$

$$\tag{5}\int_{0}^{\pi}\bigg(\frac{\sin\big(2^nx\big)}{\sin(x)}\bigg)^2\mathrm{d}x=2^n\pi, \big[\text{where }n\in\mathbb{Z}_{\ge0}\big]$$

$$\tag{6}\int_{0}^{2\pi}\exp\bigg(\frac{7+5\cos(x)}{10+6\cos(x)}\bigg)\cos\bigg(\frac{\sin(x)}{10+6\cos(x)}\bigg)\mathrm{d}x=2\pi e^{2/3}$$

$$\tag{7}\int_{0}^{2\pi}\cos\big(\cos(x)+1\big)\sinh\big(\sin(x)\big)\mathrm{d}x=2\pi\cos(1)$$

$$\tag{8}\int_{0}^{\infty}\frac{\mathrm{d}x}{\big(1+x^\varphi\big)^\varphi}=1, \bigg[\text{where }\varphi=\text{golden ratio}=\frac{1+\sqrt{5}}{2}\bigg]$$

$$\tag{9}\int_{0}^{1/2}\frac{\mathrm{d}x}{\sqrt{x^2+1}}=\log(\varphi), \bigg[\text{where }\varphi=\text{golden ratio}=\frac{1+\sqrt{5}}{2}\bigg]$$

$$\tag{10}\int_{4}^{5}\bigg(\frac{3}{2}+\frac{1}{4\sqrt{x}}\bigg)\mathrm{d}x=\varphi, \bigg[\text{where }\varphi=\text{golden ratio}=\frac{1+\sqrt{5}}{2}\bigg]$$

$$\tag{11}\int_{0}^{\infty}\frac{x^2}{1+x^{10}}\mathrm{d}x=\frac{\pi}{5\varphi}, \bigg[\text{where }\varphi=\text{golden ratio}=\frac{1+\sqrt{5}}{2}\bigg]$$

$$\tag{12}\int_{0}^{\infty}\frac{\sin(x)\cos^2(x)}{x}\mathrm{d}x=\frac{\pi}{4}$$

$$\tag{13}\int_{0}^{\infty}\frac{x-\sinh(x)}{x^2\sinh(x)}\mathrm{d}x=-\log(2)$$

$$\tag{14}\int_{0}^{\infty}\frac{\cos(x)}{x^2+1}\mathrm{d}x=\frac{\pi}{2e}$$

$$\tag{15}\int_{0}^{1}\frac{\log(1+x)}{x}\mathrm{d}x=\frac{\pi^2}{12}$$ $$\tag{16}\int_{0}^{\pi/2}\frac{\sqrt[3]{\tan(x)}}{\big(\sin(x)+\cos(x)\big)^2}\mathrm{d}x=\frac{2\pi}{3\sqrt{3}}$$

$$\tag{17}\int_{1}^{\infty}\bigg(\frac{\sin\big(\log(x)\big)}{x\log(x)}\bigg)^2\mathrm{d}x=\tan^{-1}(2)-\frac{\log(5)}{4}$$

$$\tag{18}\int_{0}^{\pi/2}\frac{\sqrt{\tan(x)}}{\sin(x)\big(\sin(x)+\cos(x)\big)}\mathrm{d}x=\pi$$

$$\tag{19}\int_{0}^{\pi/2}\log\big(\tan^4(x)+\sec^2(x)\big)\mathrm{d}x=\pi\log(2+\sqrt{3})$$

$$\tag{20}\int_{0}^{\infty}\frac{\mathrm{d}x}{1+x^n}=\frac{\pi}{n}\csc\big(\frac{\pi}{n}\big), \big[\text{where }n\in\mathbb{N}_{\ge2}\big]$$

$$\tag{21}\int_{0}^{\pi/2}\frac{x}{\sec(x)+\csc(x)}\mathrm{d}x=\frac{\pi}{4\sqrt{2}}\big(\sqrt{2}+\log(-1+\sqrt{2})\big)=\frac{\pi}{4}\bigg(\frac{1}{\sqrt{2}}\tanh^{-1}\bigg(\frac{1}{\sqrt{2}}\bigg)\bigg)$$

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