CBSE Guess > Papers > Important Questions > Class XII > 2010 > Maths > Mathematics By Mr. M.P.Keshari

Integrals

Q.16. Evaluate : 0∫12tan ^-1 x² dx.

Solution :

Let I = ∫ 2tan –1 x² dx = ∫ tan –1 x²(2dx)
Integrating by parts, we get
I = tan –1 x².2x – ∫[2x/(1 + x⁴)].2x dx
= 2xtan –1 x² – 2∫[2x²/(1 + x4)].dx
= 2xtan –1 x² – 2I1.
Now, I1 = ∫[2x²/(1 + x⁴)]dx
= ∫[(x² + 1 + x² – 1)/(x⁴ + 1)]dx
= ∫[(x² + 1)/(x⁴ + 1)]dx + ∫[(x² – 1)/(x⁴ + 1)]dx
= ∫[(1 + 1/x²)/(x² + 1/x²)]dx + ∫[(1 – 1/x²)/(x² + 1/x²)]dx
= ∫[(1 + 1/x²)/{(x – 1/x)² + 2}]dx + ∫[(1 – 1/x²)/{(x + 1/x⁾² – 2}]dx
Put x – 1/x = u and x + 1/x = v
Therefore, (1 + 1/x²) dx = du and (1 – 1/x²)dx = dv
Then I1 = ∫du/[u² + (√2)²] + ∫dv/[v² – (√2)²]
= 1/√2tan –1 (u/√2) + 1/2√2log|(v – √2)/(v + √2)|
= 1/√2tan –1{(x – 1/x)/√2} + 1/2√2log|(x + 1/x – √2)/(x + 1/x + √2)|
= tan –1{(x² – 1)/x√2} + 1/(2√2)log[(x2 + 1 – x√2)/(x² + 1 + x√2)]
Therefore, I = 2xtan –1 x² – √2tan –1[(x² – 1)/x√2]
– 1/√2log[(x² – x√2 + 1)/(x² + x√2 + 1)]
Therefore, 0∫12tan –1 x2dx = [2xtan –1 x2]01 – √2[tan –1{(x² – 1)/x√2}]01
1/√2[log{(x² – x√2 + 1)/(x² + x√2 + 1)}]01
= 2.(π/4) – √2(0 – π/2) – 1/√2[log{(1 – √2 + 1)/(1 + √2 + 1)} – 0]
= π/2 + π/√2 – 1/√2log[(2 – √2)/(2 + √2)]. [Ans.]

Q.17. Evaluate : ₀∫^a sin ^-1√[x/(a + x)]dx.

Solution :

Let I = ₀∫^a sin -1√[x/(x + a)]dx
Put x = a tan2 t => dx = 2a tan t sec² t dt
When x = 0, t = 0 ; when x = a, t = π/4.
Therefore, I = 0∫π/4 sin –1√[a tan2 t/(a + a tan² t)]. 2 a tan t sec² t dt
= ₀∫^π/4 sin ^-1√[(a tan² t)/(a sec² t)]. 2 a tan t sec² t dt
=₀∫^π/4 sin ^-1[tan t/sec t] . 2 a tan t sec2 t dt
= ₀∫^π/4 sin -1[(sin t/cos t).(1/cos t)]. 2 a tan t sec² t dt
= 2a₀∫^π/4 t tan t sec² t dt
= 2a{[t. tan² t/2]0π/4 – 0∫π/4[1.tan2 t/2.dt}
= 2a[π/4.(tan2 π/4)/2 – 0 – 1/20∫π/4 tan²t dt]
= 2a[π/8 – 1/20∫π/4(sec² t – 1)dt]
= πa/4 – a[tan t – t]0π/4
= πa/4 – a(tan π/4 – π/4 – 0)
= πa/4 – a (1 – π/4)
= πa/4 – a + πa/4
= πa/2 – a
= a(π – 2)/2. [Ans.]

Paper By Mr. M.P.Keshari
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