Integration Full Chapter Explained - Integration Class 12 - Everything you need

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  1. Chapter 7 Class 12 Integrals
  2. Serial order wise

Transcript

Example 35 (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 35 (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 is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.