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

Integrals

Q.13. Using properties of definite integrals, evaluate the following : ₀∫^πx tan x/(sec x + tan x) dx .

Solution :

Let I = 0∫πx tan x/(sec x + tan x)dx
= ₀∫^π[{x(sin x/cos x)}/{1/cos x + sin x/cos x}] dx
= ₀∫^π[(x sin x)/(1 + sin x)] dx
= ₀∫^π[(π – x) sin (π – x)}/{1 + sin (π – x )}]dx
= ₀∫^π[(π – x) sin x /(1 + sin x)] dx
Adding (1) and (2), we get
2I = ₀∫^π[(x + π – x)/(1 + sin x)]dx
= π0∫π [sin x/(1 + sin x)]dx
= π₀∫^π[(1 + sin x – 1)/(1 + sin x)]dx
= π₀∫^π[1 – 1/(1 + sin x)]dx
= π{0∫πdx – 0∫π[{1/(1 + sin x)}×{(1 – sin x)/(1 + sin x)}]dx }
= π{[x]0π0∫π[(1 – sin x)/cos2 x]dx
= π{(π – 0 ) – 0∫π(sec2 x – sec x tan x)dx }
= π{π – [tan x – sec x]0π}
= π[π – {(tan π – sec π) – (tan 0 – sec 0)}]
= π[π – (0 + 1) – (0 – 1)]
= π[π – (1 + 1)]
= π(π – 2)
Therefore, I = π/2(π – 2). [Ans.]

Q.14. Evaluate : 0∫1[2x/(1 + x²)] dx.

Solution :

Let I = 0∫1[2x/(1 + x²)] dx,
Put 1 + x² = t => 2x dx = dt
When x = 0 => t = 1 and when x = 1 => t = 2.
Therefore, I = 1∫2dt/t
= [log|t|]12
= log2 – log1 [As, log1 = 0]
= log2.[Ans.]

Q.15. Evaluate : ₀∫^π/2 logsin x dx.

Solution :

Let I = ₀∫^π/2 logsin x dx ---------------- (1)
I = ₀∫^π/2 logsin(π/2 – x) dx [ As, 0∫af(x) dx = 0∫af(a – x)dx]
= ₀∫^π/2log cos x dx ---------------- (2)
Adding (1) and (2), we get
2I = ₀∫^π/2[log sin x + log cos x] dx
= ₀∫^π/2log sinx cos x dx
= ₀∫^π/2 log[sin 2x/2]
= ₀∫^π/2 [log sin 2x – log 2]dx
= ₀∫^π/2 logsin 2x dx – 0∫π/2 log 2 dx
= ₀∫^π/2 logsin2x dx – π/2 log2.
Consider, 0∫π/2 logsin 2x dx,
Put 2x = t => 2dx = dt,
when x = 0 , t = 0; when x = π/2, t = 2.π/2 = π.
Therefore, 0∫π/2 logsin 2x dx = 0∫πlogsin t dt/2 = (1/2).2 ₀∫^π/2logsint.dt
[ 0∫2af(x)dx = 20∫af(x)dx if f(2a – x ) = f(x)]
= ₀∫^π/2 log sin x dx = I.
Therefore, 2I = I – π/2 log 2
Therefore, I = – π/2 log 2. [Ans.]

Paper By Mr. M.P.Keshari
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