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प्रश्न
`int sec^2x sqrt(tan^2x + tanx - 7) "d"x`
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उत्तर
Let I = `int sec^2x sqrt(tan^2x + tanx - 7) "d"x`
Put tan x = t
∴ sec2x dx = dt
∴ I = `int sqrt("t"^2 + "t" - 7) "dt"`
= `int sqrt("t"^2 + "t" + 1/4- 1/4 - 7) "dt"`
= `intsqrt(("t" + 1/2)^2 - 29/4) "dt"`
= `int sqrt(("t" +1/2)^2 - ((sqrt(29))/2)^2) "dt"`
= `("t" + 1/2)/2 sqrt(("t" + 1/2)^2 - (sqrt(29)/2)^2`
= `- (sqrt(29)/2)^2/2 log|"t" + 1/2 + sqrt("t"^2 + "t" -7)| + "c"`
= `(2"t" + 1)/4 sqrt("t"^2 + "t" - 7) - 29/8 log|"t" + 1/2 + sqrt("t"^2 + "t" - 7)| + "c"`
∴ I = `((2tan x + 1))/4 sqrt(tan^2 x + tanx - 7) - 29/8 log|tanx + 1/2 + sqrt(tan^2x + tanx - 7)| + "c"`
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