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

Integrals

Q.18. Evaluate : ₀∫^π/2 xdx/(sin x + cos x).

Solution :

Let I = ₀∫^π/2 xdx/(sin x + cos x). ----------------------- (1)
I = ₀∫^π/2(π/2 – x)dx/[sin (π/2 – x) + cos(π/2 – x)] [As, ₀∫^af(x)dx = ₀∫^af(a – x)dx]
= ₀∫^π/2(π/2 – x)dx/(cos x + sin x) ----------------------- (2)
Adding (1) and (2), we get
2I = ₀∫^π/2(x + π/2 – x)dx/(sin x + cos x)
= π/2₀∫^π/2 dx/(sin x + cos x)
= π/2.₀∫^π/2(1/√2)dx/[1/√2 sin x + 1/√2cos x]
= π/2.(1/√2)₀∫^π/2dx/[cos π/4 sin x + sin π/4 cos x]
= π/(2√2)₀∫^π/2dx/sin (x + π/4)
= π/(2√2)₀∫^π/2 cosec (x + π/4) dx
= π/(2√2)[log|cosec (x + π/4) – cot(x + π/4)|]0^π/2
= π/(2√2)[log|cose (π/2 + π/4 – cot (π/2 + π/4)| – log|cosec π/4 – cot π/4|]
= π/(2√2)[log| sec π/4 + tan π/4| – log|cosec π/4 – cot π/4|]
= π/(2√2)[log|√2 + 1| – log|√2 – 1|
= π/(2√2)log|(√2 + 1)/(√2 – 1)|
= π/(2√2)log|{(√2 + 1)/(√2 – 1)}×{(√2 + 1)/(√2 + 1)}|
= π/(2√2)log|(√2 + 1)2/(2 – 1)|
= π/(2√2).2 log|√2 + 1|
= π/√2log|√2 + 1|
Therefore, I = π(2√2)log(√2 + 1). [Ans.]

Q.19. Using properties of definite integrals, evaluate the following :
₀∫^a[√x/{√x + √(a – x)}]dx.

Solution :

Let I = ₀∫^a[√x/{√x + √(a – x)}]dx ------------------------------ (1)
[Using ₀∫^af(x)dx = 0∫af(a – x)dx]
I = ₀∫^a[√(a – x)/{√(a – x) + √(a – |a – x|)}]dx
= ₀∫^a[√(a – x)/{√(a – x) + √x}]dx --------------------- (2)
Adding (1) and (2) we get
2I = ₀∫^a[{√x + √(a – x)}/{√x + √(a – x)}]dx
= ₀∫^adx
= [x]0a
= a – 0
= a
Therefore, I = a/2. [Ans.]

Q.20. Evaluate : –5∫5|x + 2|dx.

Solution :

We have, ∫x + 2 if x ≥ – 2
|x + 2| = |
∫– (x + 2) if x < – 2
Let I = –₅∫⁵|x + 2| dx
= _-5∫^-2|x + 2|dx + –2∫5|x + 2|dx
= _-5∫^-2 [– (x + 2)]dx + –2∫5|x + 2|dx
= – {[x²/2 + 2x]–5–2} + [x²/2 + 2x]–25
= – [(4/2 – 4) – (25/2 – 10)] + [(25/2 + 10) – (4/2 – 4)]
= – [– 2 – 5/2] + [45/2 + 2]
= 2 + 5/2 + 45/2 + 2
= 29. [Ans.]

Paper By Mr. M.P.Keshari
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