Hey there! We receieved your request
Stay Tuned as we are going to contact you within 1 Hour
One of our academic counsellors will contact you within 1 working day.
Click to Chat
1800-5470-145
+91 7353221155
Use Coupon: CART20 and get 20% off on all online Study Material
Complete Your Registration (Step 2 of 2 )
Sit and relax as our customer representative will contact you within 1 business day
OTP to be sent to Change
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
∫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
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
Get your questions answered by the expert for free
You will get reply from our expert in sometime.
We will notify you when Our expert answers your question. To View your Question
Win Gift vouchers upto Rs 500/-
Register Yourself for a FREE Demo Class by Top IITians & Medical Experts Today !