Question

`int(sin^6x + cos^6x)/(sin^2xcos^2x)` dx = ______ 

Options

• cotx + cot²x + c 

• tanx - cotx - 3x + c

• 2sin⁵x + 2cos⁵x + c

• sin³x + 2cos³x + c 

Answer 1

`int(sin^6x + cos^6x)/(sin^2xcos^2x)` dx = tanx - cotx - 3x + c

Explanation:

`int(sin^6x + cos^6x)/(sin^2xcos^2x) dx`

= `int((sin^2x + cos^2x)^3 - 3sin^2xcos^2x(sin^2x + cos^2x))/(sin^2xcos^2x)`dx ...................`[∵ a^3 + b^3 = (a + b)^3 - 3ab(a + b)]`

= `int(1 - 3sin^2x cos^2x(1))/(sin^2x cos^2x)`dx

= `(1/(sin^2x cos^2x) - 3)`dx

= `int((sin^2x + cos^2x)/(sin^2xcos^2x) - 3)`dx

= `int(1/cos^2x + 1/sin^2x - 3)dx`

= `intsec^2xdx + intcosec^2xdx - 3intdx`

= tanx - cotx - 3x + c