Question

积分计算题

积分计算题如图

Answer 1

解答令t=tan(x/2)
则sinx= (2t)/(1+t²),
cosx=(1-t²)(1+t²),
dx= (2dt)/(1+t²)
于是 1+sinx+cosx=1+[(2t)/(1+t²)＋(1-t²)/(1+t²)]=(2+2t)/(1+t²),
即 1/(1+sinx+cosx)=(1+t²)/(2+2t)
故
∫1/(1+sinx+cOsx)dx =f(1+t²)/(2+2t)]* (2dt)(1+t²)]=f［1/(1+t)]dt=ln|1+t|+C
又
t=tan (x/2),
所以 J1/(1+sinx+cosx)dx=In|+tan(x /2)|+C

Answer 2

很简单。