∫sin(2x+1)dx=?· ∫1/(1+x^2)dx=

∫sin(2x+1)dx=?· ∫1/(1+x^2)dx= ∫sin(1/x)dx ∫sin^2x/(1+sin^2x )dx求解, ∫(1-sin^2( x/2))dx ∫(cosx／1+sin^2x)dx ∫sin^2x(1+tanx)dx ∫1/sin(x/2)dx ∫1+sin^2x/1-cos2x dx∫1+sin^2x/1-cos2x dx 不定积分∫sin(2x+1)dx=? ∫ x sin(x+1) dx 求定积分：d/dx*[∫ (1到2)sin(x^2)dx]= 定积分d/dx*[∫ （1到2）sin x^2dx]= ∫（1/x)sin(1/x)dx=? ∫1/x^2 * sin 1/x dx ∫(1+sin^2x)/(cos^2x)dx ∫e^(-2x)sin(1/2x)dx 求∫1／tan^2x+sin^2x dx ∫[1/(sin^2(x)cos^4(x)]dx