Find ∫ sec⁑ x /(1 + cosec x)  dx

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

You can download the question paper here  https://www.teachoo.com/cbse/sample-papers/


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Question 18 Find ∫1β–’sec⁑π‘₯/(1 + π‘π‘œπ‘ π‘’π‘ π‘₯) dx ∫1β–’(𝑠𝑒𝑐 π‘₯)/(1 + π‘π‘œπ‘ π‘’π‘ π‘₯) 𝑑π‘₯ = ∫1β–’(1/π‘π‘œπ‘ β‘π‘₯ )/(1 + 1/𝑠𝑖𝑛⁑π‘₯ ) 𝑑π‘₯ = ∫1β–’(1/π‘π‘œπ‘ β‘π‘₯ )/((𝑠𝑖𝑛⁑π‘₯ + 1)/𝑠𝑖𝑛⁑π‘₯ ) 𝑑π‘₯ = ∫1β–’γ€–1/π‘π‘œπ‘ β‘π‘₯ Γ— sin⁑π‘₯/(𝑠𝑖𝑛⁑π‘₯ + 1)γ€— 𝑑π‘₯ = ∫1β–’(𝑠𝑖𝑛 π‘₯)/(π‘π‘œπ‘  π‘₯(1 + 𝑠𝑖𝑛 π‘₯)) 𝑑π‘₯ Multiplying and dividing by cos x = ∫1β–’(𝑠𝑖𝑛 π‘₯)/(π‘π‘œπ‘  π‘₯(1 + 𝑠𝑖𝑛 π‘₯)) Γ—cos⁑π‘₯/cos⁑π‘₯ 𝑑π‘₯ = ∫1β–’(𝑠𝑖𝑛 π‘₯ cos⁑π‘₯)/(cos^2⁑π‘₯ (1 + 𝑠𝑖𝑛 π‘₯)) 𝑑π‘₯ Using cos2 x = 1 – sin2 x = ∫1β–’(𝑠𝑖𝑛 π‘₯ cos⁑π‘₯)/((1 βˆ’ sin^2⁑π‘₯ )(1 + sin⁑〖π‘₯)γ€— ) 𝑑π‘₯ = ∫1β–’(𝑠𝑖𝑛 π‘₯ cos⁑π‘₯)/( (1 βˆ’ sin⁑〖π‘₯)γ€— (1 + sin⁑〖π‘₯)γ€— (1 + sin⁑〖π‘₯)γ€— ) 𝑑π‘₯ = ∫1β–’(𝑠𝑖𝑛 π‘₯ cos⁑π‘₯)/((1 + γ€–sin⁑〖π‘₯)γ€—γ€—^2 (1 βˆ’ sin⁑〖π‘₯)γ€— ) 𝑑π‘₯ Let sin x = t ∴ cos x dx = dt Putting values in equation = ∫1β–’(𝑑 )/((1 + 𝑑)^2 (1 βˆ’ 𝑑)) dt We solve this by partial fractions We can write the integrand as 𝑑/((1 + 𝑑)^2 (1 βˆ’ 𝑑)) = 𝐴/(1 + 𝑑) + 𝐡/(1 + 𝑑)^2 + 𝐢/(1 βˆ’ 𝑑) 𝑑/((1 + 𝑑)^2 (1 βˆ’ 𝑑)) = (𝐴(1 + 𝑑)(1 βˆ’ 𝑑) + 𝐡(1 βˆ’ 𝑑) + 𝐢(1 + 𝑑)^2)/((1 + 𝑑)^2 (1 βˆ’ 𝑑)) By cancelling denominator 𝑑 = 𝐴(1 + 𝑑)(1 βˆ’ 𝑑) + 𝐡(1 βˆ’ 𝑑) + 𝐢(1 + 𝑑)^2 Putting t = –1 in (1) –1 = 𝐴(1+(βˆ’1))(1 βˆ’(βˆ’1)) + 𝐡(1 βˆ’(βˆ’1)) + 𝐢(1 +(βˆ’1))^2 βˆ’1 = AΓ—0+BΓ—2+C(0)^2 βˆ’1 = 2B 2B = –1 B = (βˆ’1)/2 Putting t = 1 in (1) 1 = 𝐴(1+1)(1 βˆ’1) + 𝐡(1 βˆ’1) + 𝐢(1 +1)^2 1 = AΓ—0+BΓ—0+C(2)^2 1 = 4C 4C = 1 C = 1/4 Putting t = 0 in (1) 0 = 𝐴(1+0)(1 βˆ’0) + 𝐡(1 βˆ’0) + 𝐢(1 +0)^2 0 = 𝐴(1)(1) + 𝐡(1) + 𝐢(1)^2 0 = A+B+C 0 = A + ((βˆ’1)/2) + 1/4 1/2βˆ’1/4 = A 1/4 = A A = 1/4 Hence we can write ∫1β–’(𝑑 )/((1 + 𝑑)^2 (1 βˆ’ 𝑑)) dt = ∫1β–’(1/4)/((1 + 𝑑) ) dt + ∫1β–’((βˆ’1)/2)/(1 + 𝑑)^2 dt + ∫1β–’(1/4)/( (1 βˆ’ 𝑑)) dt

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.