Question

निम्नलिखित के मान निकालिए-

`int (sin^6 x + cos^6 x)/(sin^2 x cos^2 x)"d"x`

Answer 1

मान लीजिए I = `int (sin^6x + cos^6x)/(sin^2x * cos^2x) "d"x`

= `int ((sin^2x)^3 + (cos^2x)^3)/(sin^2x * cos^2x) "d"x`

= `int ((sin^2x + cos^2x)^3 - 3sin^2x cos^2x(sin^2x + cos^2x))/(sin^2x * cos^2x) "d"x` ......[∵ a³ + b³ = (a + b)³ – 3ab(a + b)]

= `int ((1)^3 - 3sin^2x cos^2x * (1))/(sin^2x cos^2x) "d"x`

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

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

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

= `int ((sin^2x + cos^2x)/(sin^2x cos^2x) - 3) "d"x`

= `int [(1/(cos^2x) + 1/(sin^2x)) - 3]"d"x`

= `int (sec^2x + "cosec"^2x - 3) "d"x`

= `int sec^2x "d"x + int "cosec"^2x "d"x - 3 int 1"d"x`

= tan x – cot x – 3x + C

अत:, I = tan x – cot x – 3x + C.