evaluate... the integral.. 1/5cosx-12sinx dx
Answers
Let I = ∫dx5 cos x − 12 sin x=∫dx5⎡⎣1−tan2(x/2)1+tan2(x/2)⎤⎦ − 12⎡⎣2 tan (x/2)1 + tan2(x/2)⎤⎦=∫sec2(x/2) dx5 − 5 tan2(x/2)− 24 tan(x/2)put tan (x/2) = t⇒sec2(x/2) dx = 2 dtSo, I = 2∫dt5 − 5t2−24t=2−5∫dtt2+245t−1=−25∫dtt2+24t5+14425−14425−1=−25∫dt[t+125]2 − (135)2=25∫dt(135)2 − (t+125)2=25 × 526 log ∣∣∣∣t+515−t∣∣∣∣=113 log ∣∣∣∣tan(x/2)+515−tan(x/2)∣∣∣∣ + C
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student-name Ricky answered this
7 helpful votes in Math, Class XII-Science
5cos x - 12sin x = 1+tan^2x/2 / ( 5 - 5tan^2x/2 + 24tan x/2)
= sec^2x/2 / ( 5 - 5tan^2x/2 + 24tan x/2)
Substitute tan x/2 = t
=> 1/2 sec^2x/2 dx = dt
=> dx = 2dt/sec^2x
You will get 1/5-5t^2 +24t
Solve it by property of partial fractions.
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student-name Jigar answered this
269 helpful votes in Math, Class X
Let tan(x/2) = t
》 sec^2 (x/2) × 1/2 dx = dt
》 dx = 2dt /[sec^2 (x/2)]
》 dx = 2dt /[1 + tan^2 (x/2)]
》dx = 2dt/[1+t^2]
Sin2x = 2tanx/[1+tan^2(x)]
》Sinx = 2t/(1+t^2)
similarly
Cos2x = [1-tan^2 (x)]/[1+tan^2(x)]
》Cosx = (1-t^2)/(1+t^2)
Now
I = dx /(5 cosx- 12sinx)
》I =
2dt/(1+t^2) × 1/[(5 (1-t^2)/(1+t^2) -
12 2t/(1+t^2)]
taking lcm & cancelling 1+t^2 from denominator of each term we get
I = 2dt /(5-5t^2 -24t)
I = -2dt/(5t^2 +24t-5)
=-2dt /(5t^2 +25t -t -5)
= -2dt /(t-5)(5t-1)
Solve using partial fractions !
Answer:
Step-by-step explanation:
Let I = ∫dx5 cos x − 12 sin x=∫dx5⎡⎣1−tan2(x/2)1+tan2(x/2)⎤⎦ − 12⎡⎣2 tan (x/2)1 + tan2(x/2)⎤⎦=∫sec2(x/2) dx5 − 5 tan2(x/2)− 24 tan(x/2)put tan (x/2) = t⇒sec2(x/2) dx = 2 dtSo, I = 2∫dt5 − 5t2−24t=2−5∫dtt2+245t−1=−25∫dtt2+24t5+14425−14425−1=−25∫dt[t+125]2 − (135)2=25∫dt(135)2 − (t+125)2=25 × 526 log ∣∣∣∣t+515−t∣∣∣∣=113 log ∣∣∣∣tan(x/2)+515−tan(x/2)∣∣∣∣ + C
Was this answer helpful875% users found this answer helpful.
student-name Ricky answered this
7 helpful votes in Math, Class XII-Science
5cos x - 12sin x = 1+tan^2x/2 / ( 5 - 5tan^2x/2 + 24tan x/2)
= sec^2x/2 / ( 5 - 5tan^2x/2 + 24tan x/2)
Substitute tan x/2 = t
=> 1/2 sec^2x/2 dx = dt
=> dx = 2dt/sec^2x
You will get 1/5-5t^2 +24t
Solve it by property of partial fractions.
Was this answer helpful483% users found this answer helpful.
student-name Jigar answered this
269 helpful votes in Math, Class X
Let tan(x/2) = t
》 sec^2 (x/2) × 1/2 dx = dt
》 dx = 2dt /[sec^2 (x/2)]
》 dx = 2dt /[1 + tan^2 (x/2)]
》dx = 2dt/[1+t^2]
Sin2x = 2tanx/[1+tan^2(x)]
》Sinx = 2t/(1+t^2)
similarly
Cos2x = [1-tan^2 (x)]/[1+tan^2(x)]
》Cosx = (1-t^2)/(1+t^2)
Now
I = dx /(5 cosx- 12sinx)
》I =
2dt/(1+t^2) × 1/[(5 (1-t^2)/(1+t^2) -
12 2t/(1+t^2)]
taking lcm & cancelling 1+t^2 from denominator of each term we get
I = 2dt /(5-5t^2 -24t)
I = -2dt/(5t^2 +24t-5)
=-2dt /(5t^2 +25t -t -5)
= -2dt /(t-5)(5t-1)