1/(2+sin^2x) dx cos根号下x dx 求高数大神指导

∫sin(x) cos^2(x)dx ∫(1+sin^2x)/(cos^2x)dx ∫[1/(sin^2(x)cos^4(x)]dx 求解∫1/(cos^4(x)sin^2(x))dx 求不定积分∫[1/(sin^2 cos^2(x)]dx 求不定积分∫[1/sin^2cos^2 (x)]dx 1/(2+sin^2x) dx cos根号下x dx 求高数大神指导 求不定积分：$(sin x +cos x)/(1+cos^2x) dx ∫3/(x*x*x+1)dx ∫1/(2+sin x)dx 和 ∫1/(2sin x-cos x+5)dx的解法 ∫dx/[sin^2(x/2)cos^2(x/2)] ∫dx/cos^2X+4sin^2X ∫ dx/[sin^(2) x·cos^(2) x] ∫(cos^3x/sin^2x)dx 求不定积分 ：| ( sin^2 x / cos^3x ) dx 求微分sin^2x*cos^5x*dx 求微积分 ∫sin^2(x)cos^4(x) dx 求不定积分 ∫ (ln sin x) / (cos^2 x) dx ∫ [cos^3(x)]/[sin^2 (x)]dx