13.证明方程 sin x+x+1=0 在开区间 (-dfrac (pi )(2),dfrac (pi )(2)) 内至少有一个根.

(sin x)=dfrac (1)({cos )^2x} in (0,dfrac (pi )(2)),则(sin x)=dfrac (1)({cos )^2x}

+dfrac ({a)_(n)}(n+1)=0, 证明方程 _(0)+(a)_(1)x+-|||-_(2)(x)^2+... +(a)_(n)(x)^n=0 在

(int )_(-dfrac {pi )(2)}^dfrac (pi {2)}dfrac (|x|sin x)(1+{cos )^3x}dx=(int )_(-

1.证明:方程 ^3-4(x)^2+1=0 在区间(0,1)内至少有一个根.

设=(int )_(-dfrac {pi )(2)}^dfrac (pi {2)}dfrac (sin x)(1+{x)^2}(cos )^4xdx, =(in

[问答题]证明方程x2x-1=0在（0,1）内至少有一根

已知函数 (x)=dfrac (1)(2)sin (2x-dfrac (pi )(3)) x∈R,-|||-(1)求f(x)的最小正周期;-|||-(2)求f(

( (int )_(dfrac {pi )(4)}^dfrac (pi {3)}dfrac (x)({sin )^2x}dx ;

计算 lim _(xarrow dfrac {pi )(2)}dfrac (ln sin x)({(pi -2x))^2}

求lim _(xarrow dfrac {pi )(2)}dfrac (ln sin x)({(pi -2x))^2}求