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प्रश्न
उत्तर
Given I = `int 1/(cos x - sin x)dx`
We know that sin x = `(2 tan (x/2))/(1 + tan^2 (x/2)) and cos x = (1 - tan^2 (x/2))/(1 + tan^2 (x/2))`
⇒ `int 1/(-sin x + cos x)dx = int 1/(- (2 tan (x/2))/(1 + tan^2 (x/2)) + (1 - tan^2 (x/2))/(1 + tan^2 (x/2)))`
= `int (1 + tan^2 (x/2))/(-2 tan (x/2)+1 - tan^2 (x/2))dx`
Replacing 1 + tan2 x/2 in numerator by sec2 x/2 and putting tan x/2 = t and sec2 x/2 dx =
⇒ `int (1 + tan^2 (x/2))/(-2 tan (x/2) + 1 - tan^2 (x/2))dx`
= `int (sec^2 (x/2))/(- tan^2 (x/2) - 2 tan (x/2) + 1) dx`
= `- int (2dt)/(t^2 + 2t - 1)`
= `-2 int 1/((t + 1)^2 - (sqrt2)^2)dt`
= `2 int 1/((sqrt2)^2 - (t + 1)^2)dt`
We know that `int 1/(a^2 - x^2)dx = 1/(2a) log |(a + x)/(a - x)| + c`
= `2 int 1/((sqrt2)^2 - (t + 1)^2)dt`
= `2/(2sqrt2)log|(sqrt2 + t + 1)/(sqrt2 - t - 1)|+c`
= `1/sqrt2 log|(sqrt2 + tan (x/2)+1)/(sqrt2 - tan (x/2)-1)| + c`
= `1/sqrt2 log |(sqrt2 + tan (x/2) +1)/(sqrt2 - tan (x/2)-1)| + c`
∴ I = `int 1/(cos x - sin x)dx = 1/sqrt2 log |(sqrt2 + tan (x/2)+ 1)/(sqrt2 - tan (x/2) - 1)|+x`
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