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Berkley Bee Rounds
3rd place
1

$$\int_{-2}^{-1}\tan^{-1}{\sqrt{\frac{x+1}{x-1}}}dx$$

$$\frac{\pi}{3}-\frac{1}{2}\ln{(2-\sqrt3)}$$
$$\int_{-2}^{-1}\tan^{-1}{\sqrt{\frac{x+1}{x-1}}}dx=x\tan^{-1}{\sqrt{\frac{x+1}{x-1}}}\Biggl|_{-2}^{-1}-\int_{-2}^{-1}\frac{x}{(1+\frac{x+1}{x-1})}\frac{1}{2}\sqrt{\frac{x-1}{x+1}}\frac{-2}{(x-1)^2}dx$$$$=\frac{\pi}{3}+\frac{1}{2}\int_{-2}^{-1}\frac{dx}{\sqrt{x^2-1}}=\frac{\pi}{3}+\frac{1}{2}\ln{(x+\sqrt{x^2-1})}\Big|_{-2}^{-1}$$$$\int_{-2}^{-1}\tan^{-1}{\sqrt{\frac{x+1}{x-1}}}dx=\frac{\pi}{3}-\frac{1}{2}\ln{(2-\sqrt3)}$$
2

$$\int_{-\infty}^{\infty}\frac{\cos{2021x}}{1+x^2}dx$$

$$\pi e^{-2021}$$
$$\int_{-\infty}^{\infty}\frac{\cos{2021x}}{1+x^2}dx=2\int_0^{\infty}\frac{\cos{2021x}}{1+x^2}dx\text{Let }I_0=\int_0^{\infty}\frac{\cos{tx}}{1+x^2}dx$$$$\text{Laplace transform for }I_0\text{is}\mathcal{L}\left\{I_0\right\}=\int_0^{\infty}e^{-gt}\int_0^{\infty}\frac{\cos{2021x}}{1+x^2}dx$$$$\text{Let's change integration order: }\mathcal{L}\left\{ I_0 \right\}=\int_0^{\infty}\frac{1}{1+x^2}\biggl(\int_0^{\infty}e^{-gt}\cos{tx}\,dt\biggl)\,dx$$$$\text{1st part of I is}I_1=\int_0^{\infty}e^{-gt}\cos{tx}\,dt=\frac{1}{x}e^{-gt}\sin{tx}\Big|_0^{\infty}+\frac{g}{x}\int_0^{\infty}e^{-gt}\sin{tx}dx=0-\frac{g}{x^2}e^{-gt}\cos{tx}\Big|_0^{\infty}-\frac{g^2}{x^2}\int_0^{\infty}e^{-gt}\cos{tx}dx$$$$I_1=\frac{g}{x^2}-\frac{g^2}{x^2}I_1\Rightarrow I_1=\frac{g}{g^2+x^2}$$$$\text{Hence }\mathcal{L}\left\{ I_0 \right\}=\int_0^{\infty}\frac{gdx}{(1+x^2)(g^2+x^2)}=\frac{g}{g^2-1}\int_0^{\infty}\frac{dx}{1+x^2}-\frac{g}{g^2-1}\int_0^{\infty}\frac{dx}{g^2+x^2}$$$$\mathcal{L}\left\{ I_0 \right\}=\frac{g}{g^2-1}\frac{\pi}{2}-\frac{1}{g^2-1}\frac{\pi}{2}$$$$I_0=\mathcal{L}^{-1}\left\{ \mathcal{L}\left\{ I_0 \right\} \right\}=\frac{\pi}{2}\mathcal{L}^{-1}\left\{ \frac{1}{g^2-1} \right\}-\frac{\pi}{2}\mathcal{L^{-1}}\left\{ \frac{1}{g^2-1} \right\}$$$$\text{But }\mathcal{L}^{-1}\left\{ \frac{g}{g^2-1} \right\}=\cosh{t}\,,\,\mathcal{L}^{-1}\left\{ \frac{1}{g^2-1} \right\}=\sinh{t}$$$$\Downarrow$$$$I_0=\frac{\pi}{2}(\cosh{t}-\sinh{t})=\frac{\pi}{2}\Big(\frac{e^t+e^{-t}}{2}-\frac{e^t-e^{-t}}{2}\Big)=\frac{\pi e^{-t}}{2}$$$$I_0(2021)=\frac{\pi e^{-2021}}{2}$$$$\text{Finally }\int_{-\infty}^{\infty}\frac{\cos{2021x}}{1+x^2}dx=2I_0(2021)=\pi e^{-2021}$$
3

$$\int_0^{\frac{\pi}{2}}\frac{x}{\tan{x}}dx$$

$$\frac{\pi\ln2}{2}$$
$$\int_0^{\frac{\pi}{2}}\frac{x}{\tan{x}}dx=x\ln{\sin{x}}\Big|_0^{\frac{\pi}{2}}-\int_0^{\frac{\pi}{2}}\ln{\sin{x}}dx=-\int_0^{\frac{\pi}{2}}\ln{\sin{x}}dx=-\int_0^{\frac{\pi}{2}}\ln{\cos{x}}dx$$$$\text{Hence}2I=-\int_0^{\frac{\pi}{2}}\ln{(\sin{x}\cos{x})}dx=-\int_0^{\frac{\pi}{2}}\ln{\frac{\sin{2x}}{2}}dx=-\int_0^{\frac{\pi}{2}}\ln{\sin{2x}}dx+\frac{\pi\ln2}{2}$$$$\Downarrow$$$$I=-\frac{1}{2}\int_0^{\frac{\pi}{2}}\ln{\sin{2x}}dx+\frac{\pi\ln2}{4}$$$$\text{But}-\frac{1}{2}\int_0^{\frac{\pi}{2}}\ln{\sin{2x}}dx=-\frac{1}{4}\int_0^{\pi}\ln{\sin{t}}dt\,\,,\,\small{t=2x}$$$$\text{Note that}-\frac{1}{4}\int_0^{\pi}\ln{\sin{t}}dt\equiv-\frac{1}{4}\int_0^{\pi}\ln{\sin{x}}dx$$$$\text{And}\,\,\,\sin{(\pi-x)}=\sin{x}\Rightarrow-\frac{1}{4}\int_0^{\pi}\ln{\sin{x}}dx=-\frac{1}{4}\biggl(2\int_0^{\frac{\pi}{2}}\ln{\sin{x}dx}\biggl)=\frac{1}{2}I$$$$\text{Let's rewrite first part: }I=\frac{1}{2}I+\frac{\pi\ln2}{4}$$$$\Downarrow$$$$\int_0^{\frac{\pi}{2}}\frac{x}{\tan{x}}dx=\frac{\pi\ln2}{2}$$
4

$$\int_0^{\frac{5}{3}}\sqrt{25-9x^2}dx$$

$$\frac{25\pi}{12}$$
$$\int\sqrt{a^2-b^2x^2}dx=\frac{x\sqrt{a^2-b^2x^2}}{2}+\frac{a^2}{2b}\sin^{-1}{\frac{bx}{a}}\Rightarrow \int_0^{\frac{5}{3}}\sqrt{25-9x^2}dx=\frac{25\pi}{12}$$
semifinal
1

$$\int_0^{\infty}\Big(e^{-x-xe^{-xe^{-xe^{-\dots}}}}+e^{-x-xe^x}\Big)dx$$

$$1$$
$$\text{Let }\int_0^{\infty}e^{-x-xe^{-xe^{-xe^{-\dots}}}}dx=I_1; \int_0^{\infty}e^{-x-xe^x}dx=I_2$$$$\text{Let }\lim_{N\to \infty}\underbrace{-xe^{-xe^{-xe^{-\dots}}}}_{N }=-a\Rightarrow -xe^{-a}=-a\text { or }x=ae^a$$$$dx=e^a(a+1)da\longrightarrow I_1=\int_0^{\infty}e^{-x-xe^{-xe^{-xe^{-\dots}}}}dx=\int_0^{\infty}e^{-ae^a-a}e^a(1+a)da\equiv\int_0^{\infty}e^{-x-xe^x}e^x(1+x)dx$$$$I=\int_0^{\infty}e^{-x-xe^x}e^x(1+x)dx+\int_0^{\infty}e^{-x-xe^x}dx=\int_0^{\infty}(xe^x+e^x+1)e^{-x-xe^x}dx$$$$\text{Note }\frac{d}{dx}(x+xe^x)=\frac{dt}{dx}=xe^x+e^x+1\Rightarrow I=\int_0^{\infty}e^{-t}dt=1$$$$\int_0^{\infty}\Big(e^{-x-xe^{-xe^{-xe^{-\dots}}}}+e^{-x-xe^x}\Big)dx=1$$
2

$$\lim_{n\to\infty}\int_{-\infty}^{\infty}2n\sin{(x^{2n})}dx$$

$$\pi$$
$$\int_{-\infty}^{\infty}2n\sin{(x^{2n})dx=2\int_0^{\infty}}2n\sin{(x^{2n})}dx$$$$\text{Let }t=x^{2n}:\frac{dt}{dx}=\frac{dx^{2n}}{dx}=2n\frac{x^{2n}}{x}=\frac{2nt}{\sqrt[2n]{t}}$$$$I=2\int_0^{\infty}2n\sin{t}\frac{\sqrt[2n]{t}dt}{2nt}=2\int_0^{\infty}\frac{\sin{t}dt}{t}\sqrt[2n]{t}$$$$\text{For }t\in(0, \infty)\,\lim_{n\to\infty}\sqrt[2n]{t}=1\rightarrow I=2\int_0^{\infty}\frac{\sin{t}dt}{t}=2\times\frac{\pi}{2}=\pi-\text{ it's Dirichlet integral}$$
3

$$\int_0^{\infty}\frac{\ln{x}}{(x+\sqrt3)^2+1}dx$$

$$\frac{\pi\ln2}{6}$$
$$\text{Let }t=\arctan{(x+\sqrt3)}:\int_0^{\infty}\frac{\ln{x}}{(x+\sqrt3)^2+1}dx=\int_{\frac{\pi}{3}}^{\frac{\pi}{2}}\ln{(\tan{t}-\sqrt3)}dt=\int_{\frac{\pi}{3}}^{\frac{\pi}{2}}\ln{\Big[2\frac{\sin{(t-\frac{\pi}{3})}}{\cos{t}}\Big]}dt$$$$=\int_{\frac{\pi}{3}}^{\frac{\pi}{2}}\ln2dt+\int_{\frac{\pi}{3}}^{\frac{\pi}{2}}\ln{\sin{\big(t-\frac{\pi}{3}\big)}}dt-\int_{\frac{\pi}{3}}^{\frac{\pi}{2}}\ln{\cos{t}}dt$$$$\text{Note }\int_a^bf(x)dx=\int_a^bf(a+b-x)dx\Rightarrow\int_{\frac{\pi}{3}}^{\frac{\pi}{2}}\ln{\cos{t}}dt=\int_{\frac{\pi}{3}}^{\frac{\pi}{2}}\ln{\cos{\Big(\frac{\pi}{2}+\frac{\pi}{3}-t\Big)}}dt=\int_{\frac{\pi}{3}}^{\frac{\pi}{2}}\sin{\Big(t-\frac{\pi}{3}\Big)}dt$$$$\text{Hence }I=\int_{\frac{\pi}{3}}^{\frac{\pi}{2}}\ln2dt+\int_{\frac{\pi}{3}}^{\frac{\pi}{2}}\ln{\sin{\big(t-\frac{\pi}{3}\big)}}dt-\int_{\frac{\pi}{3}}^{\frac{\pi}{2}}\ln{\sin{\big(t-\frac{\pi}{3}\big)}}dt=\int_{\frac{\pi}{3}}^{\frac{\pi}{2}}\ln2dt=\frac{\pi\ln2}{6}$$
4

$$\lim_{n\to \infty}\int_n^{n+2}\frac{dx}{n^{\sin{\pi x}}+2021}$$

$$\frac{1}{2021}$$
$$$$
5

$$\int\frac{x^2-1}{x\sqrt{x^2+4x+1}\sqrt{x^2+6x+1}}dx$$

$$\cosh^{-1}{(x+\frac{1}{x}+5)}$$
$$\text{Substitute } t=x+\frac{1}{x}+5$$
6

$$\lim_{h\to 0^{+}}h\int_{-\infty}^{\infty}e^{-e^x-\pi h^2x^2}dx$$

$$\frac{1}{2}$$
$$$$
final
1

$$\int_{-\infty}^{\infty}\frac{\sin{2x}\cos{x^{-1}}-\sin{x^{-1}\cos{2x}}}{2x^3-x}dx$$

$$\pi$$
$$\int_{-\infty}^{\infty}\frac{\sin{2x}\cos{x^{-1}}-\sin{x^{-1}\cos{2x}}}{2x^3-x}dx=\int_{-\infty}^{\infty}\frac{\sin{\Big(2x-\frac{1}{x}\Big)}}{(2x-\frac{1}{x})x^2}dx$$$$\text{But }\frac{dx}{x^2}=d\Big(-\frac{1}{x}\Big)=d\Big(2x-\frac{1}{x}\Big)-2dx$$$$\Downarrow$$$$\text{With }t=2x-\frac{1}{x}:\int_{-\infty}^{\infty}\frac{\sin{\Big(2x-\frac{1}{x}\Big)}}{(2x-\frac{1}{x})x^2}dx=2\int_{-\infty}^{\infty}\frac{\sin{\Big(2x-\frac{1}{x}\Big)}}{2x-\frac{1}{x}}d\Big(2x-\frac{1}{x}\Big)-\int_{-\infty}^{\infty}\frac{\sin{\Big(2x-\frac{1}{x}\Big)}}{2x-\frac{1}{x}}2dx$$$$=2\int_{-\infty}^{\infty}\frac{\sin{t}}{t}dt-2\int_{-\infty}^{\infty}\frac{\sin{\Big(2x-\frac{1}{x}\Big)}}{2x-\frac{1}{x}}dx,\,\text{ where }2\int_{-\infty}^{\infty}\frac{\sin{t}}{t}dt=2\pi$$$$\text{But with }v=\frac{1}{2x}:I=\int_{-\infty}^{\infty}\frac{\sin{\Big(2x-\frac{1}{x}\Big)}}{(2x-\frac{1}{x})x^2}dx=2\int_{0}^{\infty}\frac{\sin{\Big(2x-\frac{1}{x}\Big)}}{(2x-\frac{1}{x})x^2}dx=4\int_0^{\infty}\frac{\sin{\Big(\frac{1}{v}-2v\Big)}}{(\frac{1}{v}-2v)}dv$$$$=4\int_{0}^{\infty}\frac{\sin{\Big(2v-\frac{1}{v}\Big)}}{(2v-\frac{1}{v})}dv=2\int_{-\infty}^{\infty}\frac{\sin{\Big(2v-\frac{1}{v}\Big)}}{(2v-\frac{1}{v})}dv,\,\text{ which is equivalent to }2\int_{-\infty}^{\infty}\frac{\sin{\Big(2x-\frac{1}{x}\Big)}}{(2x-\frac{1}{x})}dx$$$$\text{Finally }I=2\int_{-\infty}^{\infty}\frac{\sin{\Big(2x-\frac{1}{x}\Big)}}{(2x-\frac{1}{x})}dx=2\pi-2\int_{-\infty}^{\infty}\frac{\sin{\Big(2x-\frac{1}{x}\Big)}}{(2x-\frac{1}{x})}dx$$$$\Downarrow$$$$2I=2\pi\longrightarrow \int_{-\infty}^{\infty}\frac{\sin{2x}\cos{x^{-1}}-\sin{x^{-1}\cos{2x}}}{2x^3-x}dx=\pi$$
2

$$\int_0^{2\pi}\sum_{n=0}^{\infty}\frac{2(-1)^n}{(n+\frac{1}{2})^2}\sin{\Big[\Big(n+\frac{1}{2}\Big)x\Big]}dx$$

$$\pi^3$$
$$\int_0^{2\pi}\sum_{n=0}^{\infty}\frac{2(-1)^n}{(n+\frac{1}{2})^2}\sin{\Big[\Big(n+\frac{1}{2}\Big)x\Big]}dx=\sum_{n=0}^{\infty}\frac{2(-1)^n}{(n+\frac{1}{2})^2}\int_0^{2\pi}\sin{\Big[\Big(n+\frac{1}{2}\Big)x\Big]}dx$$$$\int_0^{2\pi}\sin{\Big[\Big(n+\frac{1}{2}\Big)x\Big]}dx=\frac{1}{n+\frac{1}{2}}\times2$$$$\Downarrow$$$$I=\sum_{n=0}^{\infty}\frac{4(-1)^n}{(n+\frac{1}{2})^3}=32\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)^3}=32\frac{\pi^3}{32}=\pi^3$$
3

$$\int_{-1}^1\frac{\sqrt{1-x^2}dx}{(\cosh{\sqrt{x}}+\cos{\sqrt{x}})^{\cosh{\sqrt{x}}-\cos{\sqrt{x}}}+1}$$

$$\frac{\pi}{4}$$
$$$$
4

$$\lim_{n\to\infty}\frac{1}{n}\int_0^{\infty}\min{\Big(2, \max{(0, 4-\frac{x}{n}\big)}\Big)}dx$$

$$6$$
$$\lim_{n\to\infty}\frac{1}{n}\int_0^{\infty}\min{\Big(2, \max{(0, 4-\frac{x}{n}\big)}\Big)}dx=\lim_{n\to\infty}\biggl(\frac{1}{n}\int_0^{2n}2dx+\frac{1}{n}\int_{2n}^{4n}(4-\frac{x}{n})dx+\frac{1}{n}\int_{4n}^{\infty}0dx\biggl)$$$$=\lim_{n\to\infty}\Big(\frac{4n}{n}+\frac{8n}{n}-\frac{12n^2}{2n^2}\Big)=6$$
5

$$\int\frac{x}{x^2+x^3\tan^{-1}{x}+x^4+x^5\tan^{-1}{x}}dx$$

$$-\ln{\Big(\tan^{-1}{x}+\frac{1}{x}\Big)}$$
$$\int\frac{x}{x^2+x^3\tan^{-1}{x}+x^4+x^5\tan^{-1}{x}}dx=\int\frac{dx}{x^2(1+x^2)(\frac{1}{x}+\tan^{-1}{x})}$$$$=-\int\Big(\frac{1}{x^2+1}-\frac{1}{x^2}\Big)\frac{dx}{\frac{1}{x}+\tan^{-1}{x}}=-\int\frac{d(\tan^{-1}{x}+\frac{1}{x})}{\frac{1}{x}+\tan^{-1}{x}}=-\ln{\Big(\tan^{-1}{x}+\frac{1}{x}\Big)}$$
final
1

$$\int_{-\infty}^{\infty}\frac{2020^{-|x|}}{1+5^{\arcsin{(\sin^5{x})}}}dx$$

$$\frac{1}{\ln{2020}}$$
$$\int_{-\infty}^{\infty}\frac{2020^{-|x|}}{1+5^{\arcsin{(\sin^5{x})}}}dx=\int_0^{\infty}\frac{2020^{-x}}{1+5^{\arcsin{(\sin^5{x})}}}dx+\int_{-\infty}^0\frac{2020^{x}}{1+5^{\arcsin{(\sin^5{x})}}}dx$$$$\text{In }\int_{-\infty}^0\frac{2020^{x}}{1+5^{\arcsin{(\sin^5{x})}}}dx\,\,\text{ let's substitute }v=-x:$$$$\int_{\infty}^0\frac{2020^{-v}}{1+5^{-\arcsin{(\sin^5{v})}}}d(-v)=\int_0^{\infty}\frac{2020^{-v}5^{\arcsin{(\sin^5{v})}}}{1+5^{\arcsin{(\sin^5{v})}}}dv\equiv \int_0^{\infty}\frac{2020^{-x}5^{\arcsin{(\sin^5{x})}}}{1+5^{\arcsin{(\sin^5{x})}}}dv$$$$I=\int_0^{\infty}2020^{-x}dx=\frac{1}{\ln{2020}}$$
2

$$\int_1^2\biggl[\Big(e^{1-\frac{1}{(x-1)^2}}+1\Big)+\Big(1+\frac{1}{\sqrt{1-\ln{(x-1)}}}\Big)\biggl]dx$$

$$3$$
$$\text{By substituting }v=x-1 \text{ we get:}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$$$$\int_1^2\biggl[\Big(e^{1-\frac{1}{(x-1)^2}}+1\Big)+\Big(1+\frac{1}{\sqrt{1-\ln{(x-1)}}}\Big)\biggl]dx\equiv\int_0^1\biggl[\Big(e^{1-\frac{1}{v^2}}+1\Big)+\Big(1+\frac{1}{\sqrt{1-\ln{v}}}\Big)\biggl]dv$$$$I=\int_0^1e^{1-\frac{1}{v^2}}dv+2\int_0^1dv+\int_0^1\frac{1}{\sqrt{1-\ln{v}}}dv=I_1+I_2+I_3$$$$\text{In }I_3\text{ let's substitute }t=\frac{1}{\sqrt{1-\ln{v}}}:\begin{gather}v=e^{1-\frac{1}{t^2}}\\\,\,\,\,\,\,\,\,\,dv=\frac{2}{t^3}e^{1-t^2}dt\end{gather}\Rightarrow I_3=2\int_0^1\frac{e^{1-\frac{1}{t^2}}}{t^2}dt$$$$\text{But }I_1=\int_0^1e^{1-\frac{1}{v^2}}dv=ve^{1-\frac{1}{v^2}}\Big|_0^1-2\int_0^1\frac{e^{1-\frac{1}{v^2}}}{v^2}dv=1-2\int_0^1\frac{e^{1-\frac{1}{v^2}}}{v^2}dv$$$$\text{Hence }I=1-2\int_0^1\frac{e^{1-\frac{1}{v^2}}}{v^2}dv+2x\Big|_0^1+2\int_0^1\frac{e^{1-\frac{1}{t^2}}}{t^2}dt=3$$
3

$$\int_0^{2\pi}\frac{\min{(\sin{x}, \cos{x})}}{\max{(e^{\sin{x}}, e^{\cos{x}})}}dx$$

$$-4\sinh{\frac{1}{\sqrt2}}$$
$$\int_0^{2\pi}\frac{\min{(\sin{x}, \cos{x})}}{\max{(e^{\sin{x}}, e^{\cos{x}})}}dx=\int_{\frac{\pi}{4}}^{\frac{5\pi}{4}}\frac{\cos{x}dx}{e^{\sin{x}}}+\int_{-\frac{3\pi}{4}}^{\frac{\pi}{4}}\frac{\sin{x}dx}{e^{\cos{x}}}$$$$=\int_{\frac{\sqrt2}{2}}^{-\frac{\sqrt2}{2}}\frac{d\sin{x}}{e^{\sin{x}}}-\int_{-\frac{\sqrt2}{2}}^{\frac{\sqrt2}{2}}\frac{d\cos{x}}{e^{\cos{x}}}\equiv\int_{\frac{\sqrt2}{2}}^{-\frac{\sqrt2}{2}}\frac{dt}{e^t}-\int_{-\frac{\sqrt2}{2}}^{\frac{\sqrt2}{2}}\frac{dt}{e^t}=-2\int_{-\frac{\sqrt2}{2}}^{\frac{\sqrt2}{2}}\frac{dt}{e^t}$$$$=\frac{2}{e^t}\Big|_{-\frac{\sqrt2}{2}}^{\frac{\sqrt2}{2}}=2(e^{-\frac{\sqrt2}{2}}-e^{\frac{\sqrt2}{2}})=-4\sinh{\frac{1}{\sqrt2}}$$
4

$$\int_0^1\ln{x}\sin{(\ln{x})}dx$$

$$\frac{1}{2}$$
$$\text{Let }t=-\ln{x}:\int_0^1\ln{x}\sin{(\ln{x})}dx=\int^{\infty}_0te^{-t}\sin{t}\,dt$$$$=-te^{-t}\sin{t}\Big|_0^{\infty}+\int_0^{\infty}e^{-t}(\sin{t}+t\cos{t})dt=e^{-t}(\sin{t}+t\cos{t})\Big|_0^{\infty}+\int_0^{\infty}e^{-t}(2\cos{t}-t\sin{t})dt$$$$I=2\int_0^{\infty}e^{-t}\cos{t}dt-I\Rightarrow I=\int_0^{\infty}e^{-t}\cos{t}dt$$$$=e^{-t}\sin{t}\Big|_0^{\infty}+\int_0^{\infty}e^{-t}\sin{t}dt=0-e^{-t}\cos{t}\Big|_0^{\infty}-\int_0^{\infty}e^{-t}\cos{t}dt$$$$2\int_0^{\infty}e^{-t}\cos{t}dt=1\Rightarrow \int_0^{\infty}e^{-t}\cos{t}dt=\frac{1}{2}=I$$$$\int_0^1\ln{x}\sin{(\ln{x})}dx=\frac{1}{2}$$
5

$$\int_0^1\ln^{2020}{x}\,dx$$

$$2020!$$
$$t=-\ln{x}:\int_0^1\ln^{2020}{x}\,dx=\int_0^{\infty}t^{2020}e^{-t}dt$$$$k\in\left\{ \mathbb{Z_{+}} ,0\right\}:I_k=\int_0^{\infty}t^{k}e^{-t}dt=-t^ke^{-t}\Big|_0^{\infty}+k\int_0^{\infty}t^{k-1}e^{-t}dt=kI_{k-1}$$$$\Downarrow$$$$I_k=k!\times I_0=k!\int_0^{\infty}e^{-t}dt=k!\Big(-e^{-t}\Big|_0^{\infty}\Big)=k!$$$$I_{2020}=2020!$$