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Example 44 - Chapter 7 Class 12 Integrals (Term 2)
Last updated at Dec. 23, 2019 by Teachoo
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Example 25 Important Deleted for CBSE Board 2023 Exams
Example 26 Deleted for CBSE Board 2023 Exams
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Example 44 Important You are here
Example 25 (Supplementary NCERT) Important Deleted for CBSE Board 2023 Exams
Chapter 7 Class 12 Integrals
Serial order wise
Transcript
Example 44 Evaluate β«_0^πβ(π₯ ππ₯)/(π^2 cos^2β‘γπ₯ + π^2 γ sin^2β‘π₯ ) Let I= β«_0^πβγπ₯/(π^2 πππ ^2 π₯ + π^2 π ππ^2 π₯) ππ₯γ β΄ I=β«_0^πβγ((π β π₯))/(π^2 πππ ^2 (π β π₯) + π^2 π ππ^2 (π β π₯) ) ππ₯γ I=β«_0^πβγ(π β π₯)/(π^2 [πππ (π β π₯)]^2 + π^2 [π ππ(π β π₯)]^2 ) ππ₯γ Using The Property, P4 β«_0^πβπ(π₯)ππ₯=β«_0^πβπ(πβπ₯)ππ₯ [β("Using " πππ (πβπ)=βcosβ‘π@ π ππ(πβπ)=sinβ‘π@" " )] I=β«_0^πβγ(π β π₯)/(π^2 [β πππ π₯]^2 + π^2 [π ππ π₯]^2 ) ππ₯γ I=β«_0^πβγ(π β π₯)/(π^2 cos^2β‘π₯ + π^2 sin^2β‘π₯ ) ππ₯γ Adding (1) and (2) i.e. (1) + (2) I+I=β«_0^πβγπ₯/(π^2 cos^2β‘π₯ + π^2 sin^2β‘π₯ ) ππ₯γ+β«1β(π β π₯)/(π^2 cos^2β‘π₯ + π^2 sin^2β‘π₯ ) ππ₯ 2I=β«_0^πβ(π₯ + π β π₯)/(π^2 cos^2β‘π₯ + π^2 sin^2β‘π₯ ) ππ₯ 2I=β«_0^πβ(π )/(π^2 cos^2β‘π₯ + π^2 sin^2β‘π₯ ) ππ₯ I=π/2 β«_0^πβγ1/(π^2 cos^2β‘π₯ + π^2 sin^2β‘π₯ ) ππ₯γ Dividing numerator and denominator by πππ ^2 π₯, we get I=π/2 β«_0^πβγ(1/cos^2β‘π₯ )/((π^2 cos^2β‘γπ₯ + π^2 sin^2β‘π₯ γ)/cos^2β‘π₯ ) ππ₯γ I=π/2 β«_0^πβγ(π ππ^2 π₯)/((π^2 cos^2β‘π₯)/cos^2β‘π₯ + (π^2 sin^2β‘π₯)/cos^2β‘π₯ ) ππ₯γ I=π/2 β«_0^πβγ(π ππ^2 π₯)/(π^2 + π^2 tan^2β‘π₯ ) ππ₯γ Let π(π₯)=sec^2β‘π₯/(π^2 + π^2 tan^2β‘π₯ ) and a = Ο Now, π(2πβπ₯)=sec^2β‘(π β π₯)/(π^2 + π^2 tan^2β‘(π β π₯) ) π(2πβπ₯)=[βπ ππ π₯]^2/(π^2 + π^2 [βtanβ‘π₯ ]^2 ) π(2πβπ₯)=(π ππ^2 π₯)/(π^2 + π^2 tan^2β‘π₯ ) Therefore, π(π₯)=π(2πβπ₯) Using the Property , P6 P6 :- β«_0^2πβγπ (π₯)ππ₯=2β«_0^πβγπ(π₯)ππ₯,γ γ if π(2πβπ₯)=π(π₯) Therefore, I=π/2 β«_0^πβγ(π ππ^2 π₯)/(π^2 + π^2 tan^2β‘π₯ ) ππ₯γ =π/2 Γ 2 β«_0^(π/2)βγ(π ππ^2 π₯)/(π^2 + π^2 tan^2β‘π₯ ) ππ₯γ =πβ«_0^(π/2)βγ(π ππ^2 π₯)/(π^2 + π^2 tan^2β‘π₯ ) ππ₯γ Let π tanβ‘γπ₯=π‘γ Differentiating both sides w.r.t. π₯ π π ππ^2 π₯ ππ₯=ππ‘ ππ‘=ππ‘/(π^2 π ππ^2 π₯) Putting the values of tan π₯ and ππ₯ , we get πΌ=πβ«1_0^(π/2)βγ(π ππ^2 π₯)/(π^2 + π‘^2 ) . ππ₯γ πΌ=π β«1_0^ββγ(π ππ^2 π₯)/(π^2 + π‘^2 ) .ππ‘/(π π ππ^2 π₯)γ πΌ=π/π β«1_0^ββππ‘/(π^2 + π‘^2 ) It is of form β«1βγππ₯/(π^2+π₯^2 )=1/π tan^(β1)β‘γπ₯/π+πΆγ γ Replacing π₯ by π‘ , we get πΌ= γπ/π [1/π tan^(β1)β‘(π‘/π) ]γ_0^β Putting limits, I=π/π [1/π γπ‘ππγ^(β1) (β/π)β1/π γπ‘ππγ^(β1) (0/π)] I =π/π [γ1/π tan^(β1)γβ‘γ(β)β1/π tan^(β1)β‘(0) γ ] I =π/π (1/π (π/2)β0) I =π ^π/πππ
