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Example 35 - Find integral pi/6 to pi/3  1/ 1 + root(tan x) - Teachoo

Example 35 - Chapter 7 Class 12 Integrals - Part 2
Example 35 - Chapter 7 Class 12 Integrals - Part 3 Example 35 - Chapter 7 Class 12 Integrals - Part 4 Example 35 - Chapter 7 Class 12 Integrals - Part 5 Example 35 - Chapter 7 Class 12 Integrals - Part 6

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Example 33 (Method 1) Evaluate ∫_(πœ‹/6)^(πœ‹/3 )▒𝑑π‘₯/(1 +√(tan⁑π‘₯ )) Let I =∫_(πœ‹/6)^(πœ‹/3 )β–’1/(1 + √(tan⁑π‘₯ )).𝑑π‘₯ I = ∫_(πœ‹/6)^(πœ‹/3 )β–’1/(1 + √(sin⁑π‘₯/cos⁑π‘₯ )).𝑑π‘₯ I = ∫_(πœ‹/6)^(πœ‹/3 )β–’1/(1 + √(sin⁑π‘₯ )/√(cos⁑π‘₯ )).𝑑π‘₯ I = ∫_(πœ‹/6)^(πœ‹/3 )β–’1/( (√(cos⁑π‘₯ ) + √(sin⁑π‘₯ ))/√(cos⁑π‘₯ )) 𝑑π‘₯ ("Using t" π‘Žπ‘› π‘₯=sin⁑π‘₯/cos⁑π‘₯ ) I = ∫_(πœ‹/6)^(πœ‹/3 )β–’(√(cos⁑π‘₯ ) )/(√(cos⁑π‘₯ ) + √(sin⁑π‘₯ )) 𝑑π‘₯ ∴ I =∫_(πœ‹/6)^(πœ‹/3 )β–’(√(cos⁑〖 (πœ‹/6 + πœ‹/3 βˆ’ π‘₯)γ€— ) )/(√(γ€–cos 〗⁑(πœ‹/6 + πœ‹/3 βˆ’ π‘₯) ) +√(γ€–sin 〗⁑(πœ‹/6 + πœ‹/3 βˆ’ π‘₯) )) 𝑑π‘₯ I =∫_(πœ‹/6)^(πœ‹/3 )β–’(√(cos⁑〖 (πœ‹/2 βˆ’ π‘₯)γ€— ) )/(√(cos⁑〖 (πœ‹/2 βˆ’ π‘₯)γ€— ) +√(sin⁑(πœ‹/2 βˆ’ π‘₯) )) 𝑑π‘₯ I =∫_(πœ‹/6)^(πœ‹/3 )β–’(√(sin⁑π‘₯ ) )/(√(sin⁑π‘₯ ) + √(cos⁑π‘₯ )) 𝑑π‘₯ Adding (1) and (2) i.e. (1) + (2) I + I = ∫_(πœ‹/6)^(πœ‹/3 )β–’(√(cos π‘₯) )/(√(sin⁑π‘₯ ) + √(cos⁑π‘₯ )). 𝑑π‘₯+∫_(πœ‹/6)^(πœ‹/3 )β–’(√(sin⁑π‘₯ ) )/(√(sin⁑π‘₯ ) + √(cos⁑π‘₯ )) 𝑑π‘₯ Using :- sin (πœ‹/2βˆ’πœƒ)=cosβ‘πœƒ & cos (πœ‹/2βˆ’πœƒ)=sinβ‘πœƒ Adding (1) and (2) i.e. (1) + (2) I + I = ∫_(πœ‹/6)^(πœ‹/3 )β–’(√(cos π‘₯) )/(√(sin⁑π‘₯ ) + √(cos⁑π‘₯ )). 𝑑π‘₯+∫_(πœ‹/6)^(πœ‹/3 )β–’(√(sin⁑π‘₯ ) )/(√(sin⁑π‘₯ ) + √(cos⁑π‘₯ )) 𝑑π‘₯ 2I =∫_(πœ‹/6)^(πœ‹/3 )β–’(√(cos π‘₯) + √(sin⁑π‘₯ ))/(√(cos π‘₯) + √(sin⁑π‘₯ )) 𝑑π‘₯ 2I=∫_(πœ‹/6)^(πœ‹/3 )β–’γ€–1.γ€— 𝑑π‘₯ 2I= [π‘₯]_(πœ‹/6)^(πœ‹/3) I= 1/2 [πœ‹/3βˆ’πœ‹/6] I= 1/2 [(2πœ‹ βˆ’ πœ‹)/6] 𝐈= 𝝅/𝟏𝟐 Example 33 (Method 2) Evaluate ∫_(πœ‹/6)^(πœ‹/3 )▒𝑑π‘₯/(1 +√(tan⁑π‘₯ )) Let I =∫_(πœ‹/6)^(πœ‹/3 )β–’1/(1 + √(tan⁑π‘₯ )).𝑑π‘₯ I = ∫_(πœ‹/6)^(πœ‹/3 )β–’1/(1 + √(tan⁑(πœ‹/3 + πœ‹/6 βˆ’ π‘₯) )) 𝑑π‘₯ I = ∫_(πœ‹/6)^(πœ‹/3 )β–’1/(1 +√(tan⁑(πœ‹/2 βˆ’ π‘₯) )) 𝑑π‘₯ Using the property P_4 ∫1_π‘Ž^𝑏▒〖𝑓(π‘₯) 𝑑π‘₯= ∫1_π‘Ž^𝑏▒𝑓(π‘Ž+π‘βˆ’π‘₯)𝑑π‘₯γ€— …(1) I = ∫_(πœ‹/6)^(πœ‹/3 )β–’(1 )/(1 + √(co𝑑⁑π‘₯ ) ) 𝑑π‘₯ Adding (1) and (2) I+I=∫_(πœ‹/6)^(πœ‹/3 )β–’(( 1 )/( 1 + √(tan⁑π‘₯ ))+1/(1 + √(cot⁑π‘₯ ))) 𝑑π‘₯ 2I=∫_(πœ‹/6)^(πœ‹/3 )β–’(( 1 + √(cot⁑π‘₯ ) + 1 + √(tan⁑π‘₯ ) ))/( 1 + √(tan π‘₯) )(1 + √(cot⁑π‘₯ ) ) 𝑑π‘₯ 2I=∫_(πœ‹/6)^(πœ‹/3 )β–’( 2 + √(cot⁑π‘₯ ) + √(tan⁑π‘₯ ))/( 1 + √(cot⁑π‘₯ ) + √(tan⁑π‘₯ ) + √(π­πšπ§β‘γ€–π’™ γ€— ) . √(πœπ¨π­β‘π’™ )) 𝑑π‘₯ 2I=∫_(πœ‹/6)^(πœ‹/3 )β–’( 2 + √(cot⁑π‘₯ ) + √(tan⁑π‘₯ ))/( 1 + √(cot⁑π‘₯ ) + √(tan⁑π‘₯ ) + 𝟏) 𝑑π‘₯ (As tan⁑〖(πœ‹/2 βˆ’πœƒ)=cotβ‘πœƒ γ€—) 2I=∫_(πœ‹/6)^(πœ‹/3 )β–’(2 + √(cot⁑π‘₯ ) + √(tan⁑π‘₯ ))/(2 + √(cot⁑π‘₯ ) + √(tan⁑π‘₯ )) 𝑑π‘₯ 2I=∫_(πœ‹/6)^(πœ‹/3 )▒𝑑π‘₯ 2I= [I]_(πœ‹/6)^(πœ‹/3) I= 1/2 [πœ‹/3βˆ’πœ‹/6] I= 1/2 [πœ‹/6] 𝐈= 𝝅/𝟏𝟐

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.