Example 33 - Chapter 7 Class 12 Integrals
Last updated at Dec. 16, 2024 by Teachoo
Last updated at Dec. 16, 2024 by Teachoo
Example 33 (Method 1) Evaluate β«_(π/6)^(π/3 )βππ₯/(1 +β(tanβ‘π₯ )) Let I =β«_(π/6)^(π/3 )β1/(1 + β(tanβ‘π₯ )).ππ₯ I = β«_(π/6)^(π/3 )β1/(1 + β(sinβ‘π₯/cosβ‘π₯ )).ππ₯ I = β«_(π/6)^(π/3 )β1/(1 + β(sinβ‘π₯ )/β(cosβ‘π₯ )).ππ₯ I = β«_(π/6)^(π/3 )β1/( (β(cosβ‘π₯ ) + β(sinβ‘π₯ ))/β(cosβ‘π₯ )) ππ₯ I = β«_(π/6)^(π/3 )β(β(cosβ‘π₯ ) )/(β(cosβ‘π₯ ) + β(sinβ‘π₯ )) ππ₯ β΄ I =β«_(π/6)^(π/3 )β(β(cosβ‘γ (π/6 + π/3 β π₯)γ ) )/(β(γcos γβ‘(π/6 + π/3 β π₯) ) +β(γsin γβ‘(π/6 + π/3 β π₯) )) ππ₯ I =β«_(π/6)^(π/3 )β(β(cosβ‘γ (π/2 β π₯)γ ) )/(β(cosβ‘γ (π/2 β π₯)γ ) +β(sinβ‘(π/2 β π₯) )) ππ₯ I =β«_(π/6)^(π/3 )β(β(sinβ‘π₯ ) )/(β(sinβ‘π₯ ) + β(cosβ‘π₯ )) ππ₯ Adding (1) and (2) i.e. (1) + (2) I + I = β«_(π/6)^(π/3 )β(β(cos π₯) )/(β(sinβ‘π₯ ) + β(cosβ‘π₯ )). ππ₯+β«_(π/6)^(π/3 )β(β(sinβ‘π₯ ) )/(β(sinβ‘π₯ ) + β(cosβ‘π₯ )) ππ₯ Adding (1) and (2) i.e. (1) + (2) I + I = β«_(π/6)^(π/3 )β(β(cos π₯) )/(β(sinβ‘π₯ ) + β(cosβ‘π₯ )). ππ₯+β«_(π/6)^(π/3 )β(β(sinβ‘π₯ ) )/(β(sinβ‘π₯ ) + β(cosβ‘π₯ )) ππ₯ 2I =β«_(π/6)^(π/3 )β(β(cos π₯) + β(sinβ‘π₯ ))/(β(cos π₯) + β(sinβ‘π₯ )) ππ₯ 2I=β«_(π/6)^(π/3 )βγ1.γ ππ₯ 2I= [π₯]_(π/6)^(π/3) I= 1/2 [π/3βπ/6] I= 1/2 [(2π β π)/6] π= π /ππ Example 33 (Method 2) Evaluate β«_(π/6)^(π/3 )βππ₯/(1 +β(tanβ‘π₯ )) Let I =β«_(π/6)^(π/3 )β1/(1 + β(tanβ‘π₯ )).ππ₯ I = β«_(π/6)^(π/3 )β1/(1 + β(tanβ‘(π/3 + π/6 β π₯) )) ππ₯ I = β«_(π/6)^(π/3 )β1/(1 +β(tanβ‘(π/2 β π₯) )) ππ₯ I = β«_(π/6)^(π/3 )β(1 )/(1 + β(coπ‘β‘π₯ ) ) ππ₯ Adding (1) and (2) I+I=β«_(π/6)^(π/3 )β(( 1 )/( 1 + β(tanβ‘π₯ ))+1/(1 + β(cotβ‘π₯ ))) ππ₯ 2I=β«_(π/6)^(π/3 )β(( 1 + β(cotβ‘π₯ ) + 1 + β(tanβ‘π₯ ) ))/( 1 + β(tan π₯) )(1 + β(cotβ‘π₯ ) ) ππ₯ 2I=β«_(π/6)^(π/3 )β( 2 + β(cotβ‘π₯ ) + β(tanβ‘π₯ ))/( 1 + β(cotβ‘π₯ ) + β(tanβ‘π₯ ) + β(πππ§β‘γπ γ ) . β(ππ¨πβ‘π )) ππ₯ 2I=β«_(π/6)^(π/3 )β( 2 + β(cotβ‘π₯ ) + β(tanβ‘π₯ ))/( 1 + β(cotβ‘π₯ ) + β(tanβ‘π₯ ) + π) ππ₯ 2I=β«_(π/6)^(π/3 )β(2 + β(cotβ‘π₯ ) + β(tanβ‘π₯ ))/(2 + β(cotβ‘π₯ ) + β(tanβ‘π₯ )) ππ₯ 2I=β«_(π/6)^(π/3 )βππ₯ 2I= [I]_(π/6)^(π/3) I= 1/2 [π/3βπ/6] I= 1/2 [π/6] π= π /ππ
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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