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

Integrals

Q.9. Using properties of definite integrals, evaluate the following :
₀∫^π/2 sin²x logtanx dx.

Solution :

Let I = ₀∫^π/2 sin2x logtanx dx. [₀∫^af(x)dx = ₀∫^af(a – x)dx]
= ₀∫^π/2 sin2(π/2 – x) logtan (π/2 – x) dx
= ₀∫^π/2 sin(π – 2x) logcotx dx
= – ₀∫^π/2 sin2x logtanx dx = – I
Therefore, 2I = 0 => I = 0. [Ans.]

Q.10. Evaluate : _π/6∫^π/3[1/{1 + √(tan x)}]dx.

Solution :

Let I = _π/6∫^π/3[1/{1 +√(tan x)} ]dx ------------------------- (1)
[Using a∫bf(x)dx = a∫bf(a + b – x)dx]
I = _π/6∫^π/3 dx/[1 + √{tan(π/3 + π/6 – x)}] = _π/6∫^π/3dx/[1 + √cot x] = _π/6∫^π/3[(√tan x)/(√tan x + 1)]dx ---------------------- (2)
Adding (1) and (2) we get
2I = _π/6∫^π/3 1. dx = [x]π/6π/3 = π/3 – π/6 = π/6. [Ans.]

Q.11. Evaluate : ₀∫^π [(x tan x)/(sec x cosec x)]dx.

Solution :

Let I = ₀∫^π[(x tan x)/(sec x cosec x)]dx
= ₀∫^π[{(x sin x/cos x)}/{(1/cos x)(1/sin x)}]dx
= ₀∫^πx sin2 x dx --------------- (1)
= ₀∫^π(π – x) sin² (π – x)dx
= ₀∫^π(π – x) sin² x dx -------------(2)
Adding (1) and (2), we get
2I = 0∫π(x + π – x) sin2 x dx
= π₀∫^π sin² x dx
= π₀∫^π[(1 – cos²x)/2]dx
= π/2[x – sin ²x/2]0π
= π/2[π – sin 2π/0 – 0]
= π/2.π = π2/2
Therefore, I = π2/4. [Ans.]

Q.12. Evaluate : ₀∫¹cot ^-1[1 – x + x²] dx.

Solution :

Let I = 0∫1cot ^-1[1 – x + x²]dx
= ₀∫¹tan ^-1[1/(1 – x + x²)]
= ₀∫¹tan ^-1[{x + (1 - x)}/{1 – x(1 – x)}]dx
= ₀∫¹[tan ^-1x + tan –1(1 – x)]dx
= ₀∫¹tan ^-1x dx + 0∫1tan –1{1 – (1 – x)} dx
= ₀∫¹tan ^-1 x dx + 0∫1tan –1x dx
= 2 ₀∫¹tan ^-1x dx
= 2 ₀∫¹tan ^-1. 1 dx
= 2{[x tan ^-1x]01 – 1/2 0∫12x/(1 – x²) dx }
= 2{[1.tan ^-1(1) – 0] – 1/2[log(1 + x²)]01}
= 2[π/4 ^-1/2(log 2 – log 1]
= 2[π/4 ^-1/2log2]
= π/2 – log2. [Ans.]

Paper By Mr. M.P.Keshari
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