Β  Β  Example 33 - Find integral pi/6 to pi/3  1/ 1 + root(tan x) - Teachoo - Examples

part 2 - Example 33 - Examples - Serial order wise - Chapter 7 Class 12 Integrals
part 3 - Example 33 - Examples - Serial order wise - Chapter 7 Class 12 Integrals
part 4 - Example 33 - Examples - Serial order wise - Chapter 7 Class 12 Integrals
part 5 - Example 33 - Examples - Serial order wise - Chapter 7 Class 12 Integrals part 6 - Example 33 - Examples - Serial order wise - Chapter 7 Class 12 Integrals

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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‘₯ 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‘₯ 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‘₯ 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‘₯ 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 14 years. He provides courses for Maths, Science and Computer Science at Teachoo