- ∫(0->π) sin^2(x)(1+cosx)dxsin(x)的平方 乘以(1+cos(x))

已知-π/2＜x＜0,sin x+cos x=1/5求sin 2x+2 sin 定积分0到π x*((sin(x))^2-(sin(x))^4)^(1/2) ∫sin^2x/(1+sin^2x )dx求解, f(x)=(√2 sin x + cos x)/(sin x +√(1- sin x)) (x∈ [0,π] 的最大值? - ∫(0->π) sin^2(x)(1+cosx)dxsin(x)的平方 乘以(1+cos(x)) 求极限 lim sin(x^2 * sin (1/x))/x x->0求极限 lim sin(x^2 * sin (1/x))/x x->0 求证:(sin 2x /(1-cos 2x) )·(sin x /(1+sin x))=tan (π/4-x/2). 证明∫( 0,π／2 ) (f sin x/(f sin x+f cos x) dx=π /4 ∫e^sin x/(e^sin x+e^cos x)dx在0~π／2上的积分 若x∈(0,π) ,且sin(π/2+x)+sin(π-x)=1/5,则tanx的值 已知函数f(x)=(1+1 anx)sin^2x+m sin(x+π/4)sin(x-π/4) f(x)=（1+cotx)sin^2x-2sin(x+π/4)sin(x-π/4)还有一题, sin(sinx)/x,x 趋于0的极限sin(sin(x))/x=[sin(sin(x))/sin(x)]*sin(x)/x=1*1=1sin(sin(x))/sin(x)这一部我不明白阿 求定积分 ∫[0,π]sin 2x dx ∫（0→2π） |sin x | dx 已知sin(x+π/6)=1/3,求sin(5π/6-x)+sin^2(π/3-x) 已知sin(x+π/6)=1/4,求sin(5π/6-x)+sin^2(π/3-x) ∫(1-sin^2( x/2))dx