# Hello there:I have been trying to do the following integral but have not been successful. Could someone please lend a hand?1 / (sin(x) + sec(x))Thanks a ton!Manish

6 Points
14 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

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

Anurag Kishore
37 Points
14 years ago

Hi, the integral is

 ∫ cosx dx / (sinx + cosx) = 1/2 [2 cosx dx /(sinx + cosx) = 1/2 [(cosx + sinx) + (cosx - sinx) dx] / (sinx + cosx) Now divide individually put sinx + cosx = t and integrate   Final answer   x/2  +  1/2  log I sinx + cosx I  + c