题目
:5计算下列不定积分.-|||-(1)int dfrac (dx)(xsqrt {{x)^2-1}}(xgt 1).-|||-(2) int dfrac (sqrt {{x)^2-1}}({x)^2}dx.-|||-(3)int dfrac (dx)(1+sqrt {1-{x)^2}}-|||-(4) int dfrac (dx)(sqrt {{({x)^2+1)}^3}}
题目解答
答案
解析
步骤 1：(1) 令 $x=\sec t$ ,则
$\dfrac {dx}{x\sqrt {{x}^{2}-1}}=\int \dfrac {\sec x+\sin xt}{\sec t}\arctan \dfrac {t}{\sec t}=-\dfrac {1}{t}-\cot =\arccos \dfrac {1}{x}+$ C.
步骤 2：(2) 令 $x=\sec t$ ,则
$\int \dfrac {\sqrt {{x}^{2}-1}}{{x}^{2}}dx=\int \dfrac {\tan t}{\sec {t}^{2}t}\cdot \sec xdx=\int \dfrac {{\tan }^{2}t}{\sec t}dt=\int \dfrac {\sec {t}^{2}t-1}{\sec t}dt$
$=|\sec \sec t+dt-|\cos \alpha =\ln |\sec t+\tan \alpha |-\sin t+C$
$=\ln |x+\sqrt {{x}^{2}-1}|-\dfrac {\sqrt {{x}^{2}-1}}{x}+C$
步骤 3：(3) 令 $x=\sin t$ ,则
$=t-\int {c}^{2}tdt+\int \dfrac {d(\sin t)}{{\sin }^{2}t}=t+ctt-\dfrac {1}{\sin t}-C$
$=\arcsin x+\dfrac {\sqrt {1-{x}^{2}}}{x}-\dfrac {1}{x}+C$
步骤 4：(4) 令 $x=\tan t$ ,则
$\int \dfrac {dx}{{(\sqrt {a})}^{2}+{{D}_{1}}^{2}}=\int \dfrac {{s}^{2}{t}^{2}t}{5{t}^{2}t}=at=\dfrac {2}{{t}^{2}+{t}^{2}}+c$
$\dfrac {dx}{x\sqrt {{x}^{2}-1}}=\int \dfrac {\sec x+\sin xt}{\sec t}\arctan \dfrac {t}{\sec t}=-\dfrac {1}{t}-\cot =\arccos \dfrac {1}{x}+$ C.
步骤 2：(2) 令 $x=\sec t$ ,则
$\int \dfrac {\sqrt {{x}^{2}-1}}{{x}^{2}}dx=\int \dfrac {\tan t}{\sec {t}^{2}t}\cdot \sec xdx=\int \dfrac {{\tan }^{2}t}{\sec t}dt=\int \dfrac {\sec {t}^{2}t-1}{\sec t}dt$
$=|\sec \sec t+dt-|\cos \alpha =\ln |\sec t+\tan \alpha |-\sin t+C$
$=\ln |x+\sqrt {{x}^{2}-1}|-\dfrac {\sqrt {{x}^{2}-1}}{x}+C$
步骤 3：(3) 令 $x=\sin t$ ,则
$=t-\int {c}^{2}tdt+\int \dfrac {d(\sin t)}{{\sin }^{2}t}=t+ctt-\dfrac {1}{\sin t}-C$
$=\arcsin x+\dfrac {\sqrt {1-{x}^{2}}}{x}-\dfrac {1}{x}+C$
步骤 4：(4) 令 $x=\tan t$ ,则
$\int \dfrac {dx}{{(\sqrt {a})}^{2}+{{D}_{1}}^{2}}=\int \dfrac {{s}^{2}{t}^{2}t}{5{t}^{2}t}=at=\dfrac {2}{{t}^{2}+{t}^{2}}+c$