CBSE Class 12 Sample Paper for 2018 Boards

Class 12
Solutions of Sample Papers and Past Year Papers - for Class 12 Boards

Find β« secβ‘ x /(1 + cosec x)Β  dx

This is a question of CBSE Sample Paper - Class 12 - 2017/18.

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Question 18 Find β«1βsecβ‘π₯/(1 + πππ ππ π₯) dx β«1β(π ππ π₯)/(1 + πππ ππ π₯) ππ₯ = β«1β(1/πππ β‘π₯ )/(1 + 1/π ππβ‘π₯ ) ππ₯ = β«1β(1/πππ β‘π₯ )/((π ππβ‘π₯ + 1)/π ππβ‘π₯ ) ππ₯ = β«1βγ1/πππ β‘π₯ Γ sinβ‘π₯/(π ππβ‘π₯ + 1)γ ππ₯ = β«1β(π ππ π₯)/(πππ  π₯(1 + π ππ π₯)) ππ₯ Multiplying and dividing by cos x = β«1β(π ππ π₯)/(πππ  π₯(1 + π ππ π₯)) Γcosβ‘π₯/cosβ‘π₯ ππ₯ = β«1β(π ππ π₯ cosβ‘π₯)/(cos^2β‘π₯ (1 + π ππ π₯)) ππ₯ Using cos2 x = 1 β sin2 x = β«1β(π ππ π₯ cosβ‘π₯)/((1 β sin^2β‘π₯ )(1 + sinβ‘γπ₯)γ ) ππ₯ = β«1β(π ππ π₯ cosβ‘π₯)/( (1 β sinβ‘γπ₯)γ (1 + sinβ‘γπ₯)γ (1 + sinβ‘γπ₯)γ ) ππ₯ = β«1β(π ππ π₯ cosβ‘π₯)/((1 + γsinβ‘γπ₯)γγ^2 (1 β sinβ‘γπ₯)γ ) ππ₯ 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 + πΆ/(1 β π‘) π‘/((1 + π‘)^2 (1 β π‘)) = (π΄(1 + π‘)(1 β π‘) + π΅(1 β π‘) + πΆ(1 + π‘)^2)/((1 + π‘)^2 (1 β π‘)) By cancelling denominator π‘ = π΄(1 + π‘)(1 β π‘) + π΅(1 β π‘) + πΆ(1 + π‘)^2 Putting t = β1 in (1) β1 = π΄(1+(β1))(1 β(β1)) + π΅(1 β(β1)) + πΆ(1 +(β1))^2 β1 = AΓ0+BΓ2+C(0)^2 β1 = 2B 2B = β1 B = (β1)/2 Putting t = 1 in (1) 1 = π΄(1+1)(1 β1) + π΅(1 β1) + πΆ(1 +1)^2 1 = AΓ0+BΓ0+C(2)^2 1 = 4C 4C = 1 C = 1/4 Putting t = 0 in (1) 0 = π΄(1+0)(1 β0) + π΅(1 β0) + πΆ(1 +0)^2 0 = π΄(1)(1) + π΅(1) + πΆ(1)^2 0 = A+B+C 0 = A + ((β1)/2) + 1/4 1/2β1/4 = 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((β1)/2)/(1 + π‘)^2 dt + β«1β(1/4)/( (1 β π‘)) dt