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素材/整合素材/高数素材/不定积分大集合——技巧篇.md

@@ -28,15 +28,19 @@ $$\begin{align*}
 = \int \frac{\sec t \tan t}{(\sec^2 t+1)\tan t} dt 
 = \int \frac{\cos t}{\cos^2 t+1} dt \\
 \end{align*}$$
-令 $u = sin t$
+令$u = \sin t$，则$du = \cos t dt$ ，所以
+$$
+1 + \cos^2 t = 1 + (1 - u^2) = 2 - u^2。
+$$
   
 由 $x = \sec t$,得 $\cos t = \frac{1}{x}$，所以
 
 $$\begin{align*}\sin t = \sqrt{1 - \frac{1}{x^2}} = \frac{\sqrt{x^2-1}}{x} \quad \\(\text{当 }x>1\text{ 时取正})\\
+\\
+原式= \int \frac{du}{2 - u^2}=  \frac{1}{2\sqrt{2}} \ln\left| \frac{\sqrt{x^2-1} + x\sqrt{2}}{\sqrt{x^2-1} - x\sqrt{2}} \right| + C\\
 
-原式等于  \frac{1}{2\sqrt{2}} \ln\left| \frac{\sqrt{x^2-1} + x\sqrt{2}}{\sqrt{x^2-1} - x\sqrt{2}} \right| + C\\
 \end{align*}$$
-$$
+
 【2.2】
 
 $$\int \frac{1}{x\sqrt{x^2-1}} dx$$