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Integrals

7.3. Integration using Trigonometric Identities.

Q.1. Evaluate : ∫√tan θ dθ.

Solution :

We have, I = ∫√tan θ dθ (say).
Put √tan θ = t, then tan θ = t² and sec² θ dθ = 2t dt => dθ = 2t/(1 + t4) dt. Therefore, I = ∫t.²t/(1 + t4) dt
= ∫2t²/(1 + t⁴) dt
= ∫{(t² + 1) + (t² – 1)}/(1 + t⁴) dt
= ∫(t² + 1)/(t⁴ + 1) dt + ∫(t² – 1)/(t⁴ + 1) dt
= I₁ + I₂ + c ------------------- (1)
Where, I1 = ∫(t² + 1)/(t⁴ + 1) dt
= ∫(1 + 1/t²)/(t² + 1/t²) dt
Put t – 1/t = x, then (1 + 1/t²) dt = dx and t² + 1/t² = (t – 1/t)² + 2 = x² + 2.
Therefore, I1 = ∫dx/[x² + (√2)²]
= 1/√2. tan –1 (x/√2)
= 1/√2. tan –1 [(t2 – 1)/√2 t] ---------------- (2)
and I₂ = ∫(t² – 1)/(t⁴ + 1) dt
= ∫(1 – 1/t²)/(t² + 1/t²) dt
Put t + 1/t = y, then (1 – 1/t²) dt = dy and t² + 1/t² = (t + 1/t)2 – 2 = y² – 2.
Therefore, I2 = ∫dy/[y² – (√2)²]
= 1/(2√2) log[|y – √2|/|y + √2|]
= 1/(2√2) log[|t + 1/t – √2|/|t + 1/t +√2|]
= 1/(2√2) log [|t2 – √2 t + 1|/|t2 + √2 t + 1|] ------------- (3)
Using (1), (2) and (3), we get
I = 1/√2 tan –1 [(t2 – 1)/√2.t] + 1/(2√2) log [|t2 – √2.t + 1|/|t2 + √2 t + 1|] + c. where, t = √tan θ.

Q.2. Evaluate : ∫[sin (x – a)/sin (x + a)]dx.

Solution :

We have,
∫[sin (x – a)/sin (x + a)] dx
= ∫[sin {(x + a) – 2a}/sin(x + a)] dx
= ∫[{sin (x + a) cos 2a – cos (x + a) sin 2a}/sin (x + a)] dx
= cos 2a ∫dx – sin 2a ∫cot (x + a) dx
= x cos 2a – sin 2a log|sin (x + a)| + c. [Ans.]

Q.3. Evaluate : ∫[(2sin x + 3cos x)/(3sin x + 4cos x)] dx.

Solution :

Let I = ∫[(2sin x + 3cos x)/(3sin x + 4cos x)] dx.
Now let 2sin x + 3cos x ≡ Ad/dx(3sin x + 4cos x) + B (3sin x + 4cos x)
≡ A(3cos x – 4sin x) + B(3sin x + 4cos x)
=> 2 = – 4A + 3 B and 3 = 3 A + 4 B.
Solving for A and B, we get
A = 1/25, B = 18/25 ---------- (1)
Hence, I = ∫[{A(3cos x – 4sin x) +B (3sin x + 4cos x)}/(3sin x + 4cos x)] dx
= A ∫[(3cos x – 4sin x)/(3sin x + 4cos x)] dx + B ∫dx + c
= 1/25 ∫dt/t + 18/25 x + c [Putting 3sin x + 4cos x = t.]
= 1/25 log|t| + 18/25 x + c
= 1/25 log|3sin x + 4cos x| + 18/25 x + c. [Ans.]

Paper By Mr. M.P.Keshari
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