Q.33. Integration

Question 33.
∫ dx/(secx + sinx).

Answer 33.
1/(secx + sinx) = cosx/(1 + sinx cosx)
= 2cosx / (2 + 2 sinx cosx)
= (cosx + sinx) / (2 + 2 sinx cosx) + (cosx - sinx) / (2 + 2 sinx cosx)
= (cosx + sinx) / [3 - ( sinx - cosx )^2] + (cos x - sinx) / [1 + (sinx + cosx)^2]

=> ∫ dx/(secx + sinx)
= ∫ (cosx + sinx) dx / [3 - (sinx - cosx)^2] + ∫ (cosx - sinx) dx / [1 + (sinx + cosx)^2]
Substituting sinx - cosx = u in the first integral and sinx + cosx = v in the second,
(cosx + sinx) dx = du in the first and (cosx - sinx) dx = dv in the second integral
=> Integral
= ∫ du / (3 - u^2) + ∫ dv / (1 + v^2)
= (1/2√3) ln l (√3+u)/(√3-u) l + arctan v + c
= (1/2√3) ln l [√3 + (sinx-cosx)] / (√3 - (sinx - cosx) l + arctan (sinx + cosx) + c.

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