`int(sin2x-cos2x)/(sin2x*cos2x)dx=?`

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int (sin^2x-cos^2x)/(sin^2x cos^2 x) dx is equal to:

int(sin2x)/((1-cos2x)(2-cos2x))dx=

int_(0)^( pi/4)(x^(2)(sin2x-cos2x))/((1+sin2x)cos^(2)x)dx

int(cos2x)/(sin^(2)xcos^(2)x)dx=

int(cos2x)/((sin x+cos x)^(2))dx=

I=int(sin^(2)x-cos^(2)x)/(sin^(2)x*cos^(2)x)dx

Evaluate: int(cos4x-cos2x)/(sin4x-sin2x)dx

Solve int(cos4x-cos2x)/(sin4x-sin2x)dx

int(sin2x)/((sin x+cos x)^(2))dx

Evaluate: (i) int(sin x+2cos x)/(2sin x+cos x)dx (ii) int(cos4x-cos2x)/(sin4x-sin2x)dx