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Question
Evaluate the following:
`int (sin^6x + cos^6x)/(sin^2x cos^2x) "d"x`
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Solution
Let 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
Hence, I = tan x – cot x – 3x + C.
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