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प्रश्न
Integrate the following w.r.t.x : sec4x cosec2x
उत्तर
Let I = `int sec^4x "cosec"^2x*dx`
= `int sec^4x "cosec"^2x* sec^2x*dx`
Put tan x = t
∴ sec2x·dx = d
Also, sec2x cosec2x = (1 + tan2x)(1 + cot2x)
= `(1 + t^2)(1 + 1/t^2)`
= `(1 + t^2)((t^2 + 1)/t^2)`
= `(t^4 + 2t^2 + 1)/t^2`
= `t^2 + 2 + (1)/t^2`
∴ I = `int (t^2 + 2 + 1/t^2)*dt`
= `int t^2*dt + 2 int *dt + int 1/t^2*dt`
= `t^3/(3) + 2t + (t^-1)/((-1)) + c`
= `(1)/(3)tan^3x + 2tanx - (1)/tanx + c`
= `(1)/(3cot^3x) + (2)/(cotx) - cot x + c`.
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