1.∫ (1/x^2)*cos(1/x) dx 2.设∫xf(x)dx =arcsinx+c ,则 ∫[1/f(x)] dx=?3.∫ lnx/x dx

cos x>-1/2 证明cos 2x cos x=1/2(cos x+cos 3x) 2cos x (sin x -cos x)+1 ∫1/(1+cos^2(x)) dx ∫(1-cos^(2)2x)dx ∫(1+sinx) / cos^2 x dx 2（cos x）^2=1+cos 2x, y=(cos x-2)/(cos x-1)的值域 Cos x=-1/2 y=cos^2 x-cos x +1的值域y=cos^2 x+cos x +1的值域, cos(2x)/(1-x)的导数 ∫cos²x d(cos x)为什么等于1/3cos³x 化简(sin x+cos x-1)(sin x-cos x+1)/sin 2x ∫(1+sin^2x)/(cos^2x)dx 计算不定积分 ∫cos^2x/1+e^(-x) ∫[1/(sin^2(x)cos^4(x)]dx 求解∫1/(cos^4(x)sin^2(x))dx ∫ x cos(x^2 + 1)dx 请问怎决解?