Question 18 - CBSE Class 12 Sample Paper for 2018 Boards
Last updated at Sept. 14, 2018 by Teachoo

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Find ∫ secβ‘ x /(1 + cosec x) dx
This is a question of CBSE Sample Paper - Class 12 - 2017/18.

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Transcript

Question 18
Find β«1βsecβ‘π₯/(1 + πππ ππ π₯) dx
β«1β(π ππ π₯)/(1 + πππ ππ π₯) ππ₯
= β«1β(1/πππ β‘π₯ )/(1 + 1/π ππβ‘π₯ ) ππ₯
= β«1β(1/πππ β‘π₯ )/((π ππβ‘π₯ + 1)/π ππβ‘π₯ ) ππ₯
= β«1βγ1/πππ β‘π₯ Γ sinβ‘π₯/(π ππβ‘π₯ + 1)γ ππ₯
= β«1β(π ππ π₯)/(πππ π₯(1 + π ππ π₯)) ππ₯
Multiplying and dividing by cos x
= β«1β(π ππ π₯)/(πππ π₯(1 + π ππ π₯)) Γcosβ‘π₯/cosβ‘π₯ ππ₯
= β«1β(π ππ π₯ cosβ‘π₯)/(cos^2β‘π₯ (1 + π ππ π₯)) ππ₯
Using cos2 x = 1 β sin2 x
= β«1β(π ππ π₯ cosβ‘π₯)/((1 β sin^2β‘π₯ )(1 + sinβ‘γπ₯)γ ) ππ₯
= β«1β(π ππ π₯ cosβ‘π₯)/( (1 β sinβ‘γπ₯)γ (1 + sinβ‘γπ₯)γ (1 + sinβ‘γπ₯)γ ) ππ₯
= β«1β(π ππ π₯ cosβ‘π₯)/((1 + γsinβ‘γπ₯)γγ^2 (1 β sinβ‘γπ₯)γ ) ππ₯
Let sin x = t
β΄ cos x dx = dt
Putting values in equation
= β«1β(π‘ )/((1 + π‘)^2 (1 β π‘)) dt
We solve this by partial fractions
We can write the integrand as
π‘/((1 + π‘)^2 (1 β π‘)) = π΄/(1 + π‘) + π΅/(1 + π‘)^2 + πΆ/(1 β π‘)
π‘/((1 + π‘)^2 (1 β π‘)) = (π΄(1 + π‘)(1 β π‘) + π΅(1 β π‘) + πΆ(1 + π‘)^2)/((1 + π‘)^2 (1 β π‘))
By cancelling denominator
π‘ = π΄(1 + π‘)(1 β π‘) + π΅(1 β π‘) + πΆ(1 + π‘)^2
Putting t = β1 in (1)
β1 = π΄(1+(β1))(1 β(β1)) + π΅(1 β(β1)) + πΆ(1 +(β1))^2
β1 = AΓ0+BΓ2+C(0)^2
β1 = 2B
2B = β1
B = (β1)/2
Putting t = 1 in (1)
1 = π΄(1+1)(1 β1) + π΅(1 β1) + πΆ(1 +1)^2
1 = AΓ0+BΓ0+C(2)^2
1 = 4C
4C = 1
C = 1/4
Putting t = 0 in (1)
0 = π΄(1+0)(1 β0) + π΅(1 β0) + πΆ(1 +0)^2
0 = π΄(1)(1) + π΅(1) + πΆ(1)^2
0 = A+B+C
0 = A + ((β1)/2) + 1/4
1/2β1/4 = A
1/4 = A
A = 1/4
Hence we can write
β«1β(π‘ )/((1 + π‘)^2 (1 β π‘)) dt = β«1β(1/4)/((1 + π‘) ) dt + β«1β((β1)/2)/(1 + π‘)^2 dt + β«1β(1/4)/( (1 β π‘)) dt

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