3. 计算下列不定积分


(3)
\[
    \int  \sin^2 x \tan x \sec x dx
=  \int \sin^2 x d (\sec x)
   =\int (1-\frac{1}{\sec^2 x}) d(\sec x)
=\sec x +\frac{1}{\sec x}+C.
\]
注意到常用的凑微分  $\tan x \sec x dx =d(\sec x)$


(4)
令 $u=\tan \frac x2$


(5)
\[
\begin{aligned}
   \int \tan^5 x \sec^3 x dx& = \int \tan^4 x \sec^2 x \cdot \tan x \sec xdx
\\
  &=  \int \tan^4 x \sec^2 x d(\sec x)
\\
  &= (1-\sec^2 x)^2 \sec^2 x d(\sec x)
\\ &=
   \frac13 \sec^3 x  -\frac25 \sec^5 x + \frac17 \sec^7 x +C.
\end{aligned}
\]


(6)
\[
   \int  \frac{1}{1+\sin x}dx =\int \frac{1-\sin x}{1-\sin^2 x} dx = \int \frac{1}{\cos^2 x} dx - \int \frac{\sin x}{\cos^2 x} dx
=   \tan x - \sec x +C.
\]


(7)
\[
   \mbox{原函数}= -\frac{x}{x+\ln x}
\]


(8)
\[
  \int \frac{x^3}{\sqrt{1+x^2}} dx = \frac12 \int \frac{x^2}{\sqrt{1+x^2}} d(x^2) =
  \frac12 \int \sqrt{1+x^2} d(x^2) -\frac12 \int \frac{1}{\sqrt{1+x^2}}d(x^2)
=  \frac13 (1+x^2)^{3/2}-\sqrt{1+x^2} +C.
\]


(9)
\[
   \int  \frac{1}{e^x +e^{-x}} dx= \int \frac{e^x}{e^{2x}+1}dx
= \arctan (e^x) +C.
\]


(10)
\[
   \mbox{原函数}= -\frac{1}{x\ln x}
\]


4.
(1) 首先, 假设 $x>1$, 则
\[
    \int \frac{dx}{x\sqrt{x^2 -1}}
=  
   \int \frac{1}{x|x| \sqrt{1- (\frac{1}{x})^2 } } dx
=
  \int \frac{1}{x^2 \sqrt{1- (\frac{1}{x})^2 } } dx
=
-\int \frac{1}{ \sqrt{1- (\frac{1}{x})^2 } } d(\frac1x)
=  -\arctan (\frac1x)+C,
\]
当 $x<-1$ 时, 令 $t=-x$, 则 $t>1$, 即
\[
    \int \frac{dx}{x\sqrt{x^2 -1}}= \int \frac{dt}{t\sqrt{t^2 -1}}=-\arctan (\frac 1t) +C=
   -\arctan (-\frac1x)+C,
\]
因此,
\[
 \int \frac{dx}{x\sqrt{x^2 -1}}  = - \arctan |\frac1x|+C.
\]

作业 21 不定积分换元