Test công thức toán

$\sin ^4x+\cos ^4x = \left(\sin ^2x+\cos ^2x\right)^2-2 \ \text{s}in^2x.\cos ^2x = 1-\dfrac{\left(2 \ \text{s}inx\cos x\right)^2}{2} = 1-\dfrac{\sin ^2 2x}{2}
\int_{0}^{\dfrac{\pi }{2}}{\cos _2x\left(\sin ^4x+\cos ^4x\right)dx} = \int_{0}^{\dfrac{\pi }{2}}{\cos _2x\left(1-\dfrac{\sin ^2 2x}{2}\right)dx} = \int_{0}^{\dfrac{\pi }{2}}{\cos _2xdx}-\dfrac{1}{4}\int_{0}^{\dfrac{\pi }{2}}{\sin ^2 2xd\left(\sin 2x\right)}
$
 
$\int x\tan xdx
\int x\cot xdx
\int \dfrac{\sin x}{x}dx
\int \dfrac{\cos x}{x}dx
\int e^{x^2}dx
\int e^xlnxdx
\int \sqrt{lnx}dx
\int \dfrac{dx}{lnx}
\int ln\left(\cos x\right)dx
\int ln\left(\sin x\right)dx
\int \cos \left(x^2\right)dx
\int \sin \left(x^2\right)dx
\int_{}^{}e^{\sin x}dx
\int_{}^{}e^{\cos x}dx
\int_{}^{}e^{\tan x}dx
\int_{}^{}e^{\cot x}dx
\int_{}^{}\sqrt{x^3+1}dx
\int_{}^{}\dfrac{dx}{\left(e^x+1\right)\left(x^2+1\right)}
\int_{}^{}\sqrt{\cos x}dx
\int_{}^{}\sqrt{\sin x}dx$
 
$$3\sqrt[9] {\dfrac{9\sqrt{3}\left(\sqrt{x^3+4}+\sqrt{3}\right)}{14+2x^3+2x^2-2x+4\sqrt{\left(x^2-2x\right)\left(x^3+4\right)}+4\sqrt{3x^2-6x}+4\sqrt{3x^3+12}}}+\sqrt[3] {\dfrac{6\sqrt{\left(x^2-2x\right)\left(x^3+4\right)}{\left(\sqrt{x^3+4}+\sqrt{3}\right)\left(\sqrt{3}\sqrt{x^3+4}+\sqrt{3}\right)}}=|1-2x|+|2x+3|.$$
 
$$U_{AN}+U_{NB}=\dfrac{U}{\cos \dfrac{\pi }{5}} \left[ \sin \left( \dfrac{\pi }{2}-\varphi \right)+\sin \left( \dfrac{\pi }{5}-\varphi \right) \right].$$

$$=\dfrac{2U}{\cos \dfrac{\pi }{5}}\sin \dfrac{3\pi }{10}\cos \left( \dfrac{7\pi }{20}-\varphi \right) \leq \dfrac{2U}{\cos \dfrac{\pi }{5}}\sin \dfrac{3\pi }{10}.$$
 
$$U_{AN}+U_{NB} =\dfrac{U}{\cos \dfrac{\pi }{5}}\left( \sin \left( \dfrac{\pi }{2}-\varphi \right)+\sin \left( \dfrac{\pi }{5}+\varphi \right) \right) .$$

$$=2\dfrac{U}{\cos \dfrac{\pi }{5}}\sin \left( \dfrac{7\pi }{20} \right)\cos \left( \dfrac{3\pi }{20}-\varphi \right).$$

$$ \leq 2\dfrac{U}{\cos \dfrac{\pi }{5}}\sin \dfrac{7\pi }{20}.$$
 

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