askiitianexpert arulmani
Last Activity: 15 Years ago
∫dx/(sin x+sec x)=∫dx/(sin x + (1/cos x))=∫cos x dx /(1+sin x cos x)
Multiplying by 2, we get = ∫2 cos x dx / (2 + 2 sin x cos x)
2 cos x dx can be substituted with (cos x + sin x) + (cos x - sin x) and
2 + 2 sin x cos x can be substituted with either (3 - (sin x - cos x)2) or (1 + (sin x + cos x)2)
Hence we get,
∫2 cos x dx / (2 + 2 sin x cos x) = ∫ (cos x + sin x) dx / (3 - (sin x - cos x)2) + ∫(cos x - sin x) dx / (1 + (sin x + cos x)2)
FIRST PART OF THE INTEGRAL
∫ (cos x + sin x) dx / (3 - (sin x - cos x)2)
Integrate by substitution, put y = sin x - cos x
then dy = (cos x +sin x) dx,
and we get ∫ dy / (3 - y2) = (1/(2*√3)) [ ∫ dy / (√3 + y) + ∫ dy / (√3 - y) ]
=> (1/(2*√3)) [ log(√3 + y) - log(√3 - y) ] + C
replacing y with its original value, we get
=> (1/(2*√3)) [ log(√3 + (sin x - cos x)) - log(√3 - (sin x - cos x)) ] + C
SECOND PART OF THE INTEGRAL
∫(cos x - sin x) dx / (1 + (sin x + cos x)2)
Integrate by substitution, put z = sin x + cos x
then dz = (cos x - sin x) dx,
and we get ∫dz / (1 + (z)2)
Let z = tan Θ, then dz = sec2Θ dΘ
∫dz / (1 + (z)2) = ∫sec2Θ dΘ / (1 + (tan Θ)2) = ∫sec2Θ dΘ / sec2Θ = ∫dΘ = Θ = tan-1z + C
replacing z with its original value, we get,
=> tan-1(sin x + cos x)+ C
FINAL ANSWER
Now adding the first and second parts, we get the result as
=> (1/(2*√3)) [ log(√3 + (sin x - cos x)) - log(√3 - (sin x - cos x)) ] + tan-1(sin x + cos x)+ C