Examples

Chapter 7 Class 12 Integrals
Serial order wise

### Transcript

Example 33 (Method 1) Evaluate β«_(π/6)^(π/3 )βππ₯/(1 +β(tanβ‘π₯ )) Let I =β«_(π/6)^(π/3 )β1/(1 + β(tanβ‘π₯ )).ππ₯ I = β«_(π/6)^(π/3 )β1/(1 + β(sinβ‘π₯/cosβ‘π₯ )).ππ₯ I = β«_(π/6)^(π/3 )β1/(1 + β(sinβ‘π₯ )/β(cosβ‘π₯ )).ππ₯ I = β«_(π/6)^(π/3 )β1/( (β(cosβ‘π₯ ) + β(sinβ‘π₯ ))/β(cosβ‘π₯ )) ππ₯ I = β«_(π/6)^(π/3 )β(β(cosβ‘π₯ ) )/(β(cosβ‘π₯ ) + β(sinβ‘π₯ )) ππ₯ β΄ I =β«_(π/6)^(π/3 )β(β(cosβ‘γ (π/6 + π/3 β π₯)γ ) )/(β(γcos γβ‘(π/6 + π/3 β π₯) ) +β(γsin γβ‘(π/6 + π/3 β π₯) )) ππ₯ 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₯) )) ππ₯ I =β«_(π/6)^(π/3 )β(β(sinβ‘π₯ ) )/(β(sinβ‘π₯ ) + β(cosβ‘π₯ )) ππ₯ 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₯ )). ππ₯+β«_(π/6)^(π/3 )β(β(sinβ‘π₯ ) )/(β(sinβ‘π₯ ) + β(cosβ‘π₯ )) ππ₯ 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₯ )). ππ₯+β«_(π/6)^(π/3 )β(β(sinβ‘π₯ ) )/(β(sinβ‘π₯ ) + β(cosβ‘π₯ )) ππ₯ 2I =β«_(π/6)^(π/3 )β(β(cos π₯) + β(sinβ‘π₯ ))/(β(cos π₯) + β(sinβ‘π₯ )) ππ₯ 2I=β«_(π/6)^(π/3 )βγ1.γ ππ₯ 2I= [π₯]_(π/6)^(π/3) I= 1/2 [π/3βπ/6] I= 1/2 [(2π β π)/6] π= π/ππ Example 33 (Method 2) Evaluate β«_(π/6)^(π/3 )βππ₯/(1 +β(tanβ‘π₯ )) Let I =β«_(π/6)^(π/3 )β1/(1 + β(tanβ‘π₯ )).ππ₯ I = β«_(π/6)^(π/3 )β1/(1 + β(tanβ‘(π/3 + π/6 β π₯) )) ππ₯ I = β«_(π/6)^(π/3 )β1/(1 +β(tanβ‘(π/2 β π₯) )) ππ₯ I = β«_(π/6)^(π/3 )β(1 )/(1 + β(coπ‘β‘π₯ ) ) ππ₯ Adding (1) and (2) I+I=β«_(π/6)^(π/3 )β(( 1 )/( 1 + β(tanβ‘π₯ ))+1/(1 + β(cotβ‘π₯ ))) ππ₯ 2I=β«_(π/6)^(π/3 )β(( 1 + β(cotβ‘π₯ ) + 1 + β(tanβ‘π₯ ) ))/( 1 + β(tan π₯) )(1 + β(cotβ‘π₯ ) ) ππ₯ 2I=β«_(π/6)^(π/3 )β( 2 + β(cotβ‘π₯ ) + β(tanβ‘π₯ ))/( 1 + β(cotβ‘π₯ ) + β(tanβ‘π₯ ) + β(π­ππ§β‘γπ γ ) . β(ππ¨π­β‘π )) ππ₯ 2I=β«_(π/6)^(π/3 )β( 2 + β(cotβ‘π₯ ) + β(tanβ‘π₯ ))/( 1 + β(cotβ‘π₯ ) + β(tanβ‘π₯ ) + π) ππ₯ 2I=β«_(π/6)^(π/3 )β(2 + β(cotβ‘π₯ ) + β(tanβ‘π₯ ))/(2 + β(cotβ‘π₯ ) + β(tanβ‘π₯ )) ππ₯ 2I=β«_(π/6)^(π/3 )βππ₯ 2I= [I]_(π/6)^(π/3) I= 1/2 [π/3βπ/6] I= 1/2 [π/6] π= π/ππ