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

Integrals

Q.2. Prove that 0∫a f(x)dx = 0∫a f(a – x)dx.
Hence, evaluate ₀∫^π/2 dx/(1 + tan x).

Solution :

R. H. S. = 0∫a f(a – x)dx
Put a – x = t then, – dx = dt => dx = – dt.
When x = 0, t = a and when x = a, t = 0.
Therefore, R. H. S. = – a∫0 f(t)dt = 0∫a f(t)dt = 0∫a f(x)dx = L. H. S.
Now let I = 0∫π/2 dx/(1 + tan x) --------------- (1)
Using the above result, I = ₀∫^π/2 dx/[1 + tan (π/2 – x)]
= ₀∫^π/2 dx/(1 + cot x)
= ₀∫^π/2 [tan x/(1 + tan x)]dx. -------------- (2)
Adding (1) and (2) we get,
2I = 0∫π/2 [1/(1 + tan x)]dx + ₀∫^π/2 [tan x/(1 + tan x)]dx
= ₀∫^π/2[(1 + tan x)/(1 + tan x)]dx
= ₀∫^π/2 1.dx
= [x]0π/2
= π/2 – 0
= π/2.
Therefore, I = π/4. [Ans.]

Q.3. Using properties of definite integrals, evaluate the following :
₀∫^π/4 log(1 + tan x)dx.

Solution :

Let I = ₀∫^π/4 log(1 + tan x)dx ------------------------ (1)
= ₀∫^π/4 log{1 + tan(π/4 – x)}dx
= ₀∫^π/4 log[1 + {(tan π/4 – tan x)/(1 + tan π/4.tan x)}]dx
= ₀∫^π/4 log[1 + (1 – tan x)/(1 + tan x)]dx
= ₀∫^π/4 log[(1 + tan x + 1 – tan x)/(1 + tan x)]dx
= ₀∫^π/4 log[2/(1 + tan x)]dx
= ₀∫^π/4 log(2)dx – ₀∫^π/4 log(1 + tan x)dx
= log 2[x]0π/4 – I
Or, I + I = log2[π/4 – 0] = π/4 log²
Therefore, I = π/8 log². [Ans.]

Q.4. Evaluate : ₀∫^π (x sin x)/(1 + cos² x)dx.

Solution :

We have, I = ₀∫^π(x sin x)/(1 + cos² x)dx
Therefore, I = ₀∫^π [(π – x) sin (π – x)]/[1 + cos² (π – x)]dx,
[As, ₀∫^a f(x)dx = ₀∫^a f(a – x)dx ]
= ₀∫^π[(π – x) sin x]/[1 + cos² x]dx
= π ₀∫^πsin x/(1 + cos²x)dx – I
Or, 2 I = π 0 ₀∫^π sin x/(1 + cos² x) dx
Or, I = π/2 ₀∫^π sin x/(1 + cos² x)dx
Put cos x = t so that – sin x dx = dt. When x = 0, t = 1 and when x = π, t = –1. Therefore, I = – π/2 1∫–1 dt/(1 + t2)
= π/2 –1 ∫1 dt/(1 + t²)
= π 0∫1 dt/(1 + t2) [As, 1/(1 + t²) is an even function]
= π[tan –1 t]01
= π[tan –1 1 – tan –1 0]
= π[π/4 – 0] = π2/4. [Ans.]

Paper By Mr. M.P.Keshari
Email Id : [email protected]
Ph No. : 09434150289