∫sin^2(x)cos^2(x)dx对于三角函数来说,形如∫sin^n(x)cos^m(x)有没有通解?

∫sin(x) cos^2(x)dx ∫dx/[sin^2(x/2)cos^2(x/2)] ∫(1+sin^2x)/(cos^2x)dx ∫dx/cos^2X+4sin^2X ∫ dx/[sin^(2) x·cos^(2) x] ∫(cos^3x/sin^2x)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 ∫sin^3(x)cos^2(x)dx= 求积分：∫sin^2 (x) /cos^3 (x) dx ∫[1/(sin^2(x)cos^4(x)]dx 求解∫1/(cos^4(x)sin^2(x))dx 求不定积分∫(cos2x)/(sin^2x)(cos^2x)dx∫(cos2x)/(sin^2x)(cos^2x)dx 求不定积分∫[1/(sin^2 cos^2(x)]dx 求不定积分∫[1/sin^2cos^2 (x)]dx 求不定积分∫(cos2x)/(sin^2x)(cos^2x)dx求不定积分,∫(cos2x)/(sin^2x)(cos^2x)dx,∫cos二倍x除以sin平方x 乘cos平方x dx ∫3/(x*x*x+1)dx ∫1/(2+sin x)dx 和 ∫1/(2sin x-cos x+5)dx的解法