Question

求证cosx/(1- sinx)^2=1

Answer 1

cosx/(1- sinx)^2=1
<==>cosx=(1- sinx)^2,
令x=π/4，得1/√2=（1-1/√2）^2=3/2-√2，
3√2=3，矛盾，命题不成立。

Answer 2

Let Im,n=∫(sinx)^m*(cosx)^ndx
then Im,n=(sinx)^(m+1)*(cosx)^(n-1)-
∫(sinx)[(sinx)^m*(cosx)^(n-1)]'dx
=(sinx)^(m+1)*(cosx)^(n-1)-
∫[m(sinx)^m*(cosx)^n-(n-1)(sinx)^(m+2)*(cosx)^(n-1)]dx
=(sinx)^(m+1)*(cosx)^(n-1)-mIm,n+(n-1)Im+2,n-2
so (m+1)Im,n=(sinx)^(m+1)*(cosx)^(n-1)+(n-1)Im+2,n-2
用此递推公式求解
sin(ax)*cos(bx)
=(1/2)*[sin(a+b)x+sin(a-b)x]
so ∫sin(ax)*cos(bx)dx
=-(1/2)*[cos(a+b)x/(a+b)+cos(a-b)x/(a-b)]+C