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

Integrals

Q.5. Evaluate : –a ∫a √[(a – x)/(a + x)]dx.

Solution :

We have,
–_a∫^a √[(a – x)/(x + a)]dx = –a∫a [(a – x)/√(a² – x2)]dx
= –a∫a[a/√(a² – x²)]dx – –a∫a[x/√(a² – x²)]dx
= 20∫a [a/√(a² – x²)]dx [even function]
– 0 [An odd function]
= 2a[ sin –1 (x/a)]0a
= 2a[sin –1(a/a) – sin –1(0/a)]
= 2a[π/2 – 0] = aπ [Ans.]

Q.6. Evaluate : ₀∫^π/4 sin 2x sin 3x dx.

Solution :

We have
₀∫^π/4sin 2x sin 3x dx = 1/2 ₀∫^π/4(2 sin 2x sin 3x) dx
= 1/2 0∫π/4(cos x – cos 5x) dx
= 1/2 [sin x – (sin 5x)/5]0π/4
= 1/2 [sin π/4 – 1/5 sin 5π/4 – 0]
= 1/2 [sin π/4 + 1/5 sin π/4]
= 1/2 (1 + 1/5)1/√2
= 3/(5√2). [Ans.]

Q.7. Evaluate : ₀∫^π/4 log(1 + tan x) dx.

Solution :

We have
I = ₀∫^π/4log (1 + tan x) dx (say)
= ₀∫^π/4 log[1 + tan (π/4 – x)] dx [0∫a f(x) dx = 0∫af(a – x )dx]
= ₀∫^π/4 log[1 + (1 – tan x)/(1 + tan x)] dx
= ₀∫^π/4 log [2/(1 + tan x)] dx
= ₀∫^π/4[log 2 – log(1 + tan x)] dx
= log 2₀∫^π/4 dx – 0∫π/4 log (1 + tan x) dx
= log 2₀∫^π/4 dx – I
Or, 2I = log 2₀∫^π/4dx = log2[x]0π/4 = log 2(π/4 – 0)
Or, I = π/8 log 2. [Ans.]

Q.8. Evaluate : ₀∫^π/2 x² cos²x dx.

Solution :

We have, I = ₀∫^π/2 x² cos²x dx.
Integrating by parts, we get
I = ₀∫^π/2 x² cos²x dx
= x². sin2x/2 – ∫2x. sin²x/2 dx
= 1/2 x² sin²x – ∫x. sin²x dx
= 1/2 x² sin²x – [x .(– cos²x/2) – ∫1 (– cos²x/2) dx]
= [1/2 x² sin²x + 1/2 x cos²x – 1/4 sin²x]0π/2
= – π/4. [Ans.]

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