Question

The value of ${\int }_a^{b}f\left(x\right)dx$

Answer 1

A) $I={\int }_1^{\sqrt{3}}\frac{\sqrt{1+x^2}}{x^2}dx$
Substituting $x=\tan t\Rightarrow dx={\sec }^{2}tdt$, we get
$I={\int }_{\frac{\pi }{4}}^{\frac{\pi }{3}}{\csc }^{2}t\sec tdt={\int }_{\frac{\pi }{4}}^{\frac{\pi }{3}}\sec tdt+{\int }_{\frac{\pi }{4}}^{\frac{\pi }{3}}\cot t\csc tdt={\left[\log (\sec t+\tan t)\right]}_{\frac{\pi }{4}}^{\frac{\pi }{3}}-{\left[\csc t\right]}_{\frac{\pi }{4}}^{\frac{\pi }{3}}=\log \left(\frac{2+\sqrt{3}}{\sqrt{2}+1}\right)-(\frac{2}{\sqrt{3}}-\sqrt{2})$
B) $J={\int }_1^2\frac{\sqrt{x^2-1}}{x}dx$
Substituting $x=\sec t\Rightarrow dx=\tan t\sec tdt$, we get
$J={\int }_0^{\frac{\pi }{3}}{\tan }^{2}tdt={\int }_0^{\frac{\pi }{3}}({\sec }^{2}t-1)dt={[\tan t-t]}_0^{\frac{\pi }{3}}=\sqrt{3}-\frac{\pi }{3}$
C) $K={\int }_0^1\sqrt{{(1-x^2)}^3}$
Substituting $x=\sin t\Rightarrow dx=\cos tdt$, we get
$K={\int }_0^{\frac{\pi }{2}}{\cos }^{4}tdt={\int }_0^{\frac{\pi }{2}}\left({\cos }^{2}t(1-{\sin }^{2}t)\right)dt={\int }_0^{\frac{\pi }{2}}{\cos }^{2}tdt-{\int }_0^{\frac{\pi }{2}}{\sin }^{2}t{\cos }^{2}tdt={[\frac{3t}{8}+\frac{\sin t{\cos }^{3}t}{4}+\frac{3\sin t\cos t}{8}]}_0^{\frac{\pi }{2}}=\frac{3\pi }{16}$
D) $L={\int }_{\sqrt{2}}^2\frac{1}{x^5\sqrt{x^2-1}}dx$
Substituting $x=\sec t\Rightarrow dx=\tan t\sec tdt$, we get
$L={\int }_{\frac{\pi }{4}}^{\frac{\pi }{3}}{\cos }^{4}tdt={\int }_{\frac{\pi }{4}}^{\frac{\pi }{3}}\left({\cos }^{2}t(1-{\sin }^{2}t)\right)dt={\int }_{\frac{\pi }{4}}^{\frac{\pi }{3}}{\cos }^{2}tdt-{\int }_{\frac{\pi }{4}}^{\frac{\pi }{3}}{\sin }^{2}t{\cos }^{2}tdt={[\frac{3t}{8}+\frac{\sin t{\cos }^{3}t}{4}+\frac{3\sin t\cos t}{8}]}_{\frac{\pi }{4}}^{\frac{\pi }{3}}=(\frac{\pi }{8}+\frac{\sqrt{3}}{64}+\frac{3\sqrt{3}}{32}-\frac{3\pi }{32}-\frac{1}{16}-\frac{3}{16})=\frac{1}{32}(\pi +\frac{7\sqrt{3}}{2}-8)$