Integration Full Chapter Explained - https://you.tube/Integration-Class-12

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

Transcript

Example 44 Evaluate ∫_0^πœ‹β–’(π‘₯ 𝑑π‘₯)/(π‘Ž^2 cos^2⁑〖π‘₯ + 𝑏^2 γ€— sin^2⁑π‘₯ ) Let I= ∫_0^πœ‹β–’γ€–π‘₯/(π‘Ž^2 π‘π‘œπ‘ ^2 π‘₯ + 𝑏^2 𝑠𝑖𝑛^2 π‘₯) 𝑑π‘₯γ€— ∴ I=∫_0^πœ‹β–’γ€–((πœ‹ βˆ’ π‘₯))/(π‘Ž^2 π‘π‘œπ‘ ^2 (πœ‹ βˆ’ π‘₯) + 𝑏^2 𝑠𝑖𝑛^2 (πœ‹ βˆ’ π‘₯) ) 𝑑π‘₯γ€— I=∫_0^πœ‹β–’γ€–(πœ‹ βˆ’ π‘₯)/(π‘Ž^2 [π‘π‘œπ‘ (πœ‹ βˆ’ π‘₯)]^2 + 𝑏^2 [𝑠𝑖𝑛(πœ‹ βˆ’ π‘₯)]^2 ) 𝑑π‘₯γ€— Using The Property, P4 ∫_0^π‘Žβ–’π‘“(π‘₯)𝑑π‘₯=∫_0^π‘Žβ–’π‘“(π‘Žβˆ’π‘₯)𝑑π‘₯ [β–ˆ("Using " π‘π‘œπ‘ (πœ‹βˆ’πœƒ)=βˆ’cosβ‘πœƒ@ 𝑠𝑖𝑛(πœ‹βˆ’πœƒ)=sinβ‘πœƒ@" " )] I=∫_0^πœ‹β–’γ€–(πœ‹ βˆ’ π‘₯)/(π‘Ž^2 [βˆ’ π‘π‘œπ‘  π‘₯]^2 + 𝑏^2 [𝑠𝑖𝑛 π‘₯]^2 ) 𝑑π‘₯γ€— I=∫_0^πœ‹β–’γ€–(πœ‹ βˆ’ π‘₯)/(π‘Ž^2 cos^2⁑π‘₯ + 𝑏^2 sin^2⁑π‘₯ ) 𝑑π‘₯γ€— Adding (1) and (2) i.e. (1) + (2) I+I=∫_0^πœ‹β–’γ€–π‘₯/(π‘Ž^2 cos^2⁑π‘₯ + 𝑏^2 sin^2⁑π‘₯ ) 𝑑π‘₯γ€—+∫1β–’(πœ‹ βˆ’ π‘₯)/(π‘Ž^2 cos^2⁑π‘₯ + 𝑏^2 sin^2⁑π‘₯ ) 𝑑π‘₯ 2I=∫_0^πœ‹β–’(π‘₯ + πœ‹ βˆ’ π‘₯)/(π‘Ž^2 cos^2⁑π‘₯ + 𝑏^2 sin^2⁑π‘₯ ) 𝑑π‘₯ 2I=∫_0^πœ‹β–’(πœ‹ )/(π‘Ž^2 cos^2⁑π‘₯ + 𝑏^2 sin^2⁑π‘₯ ) 𝑑π‘₯ I=πœ‹/2 ∫_0^πœ‹β–’γ€–1/(π‘Ž^2 cos^2⁑π‘₯ + 𝑏^2 sin^2⁑π‘₯ ) 𝑑π‘₯γ€— Dividing numerator and denominator by π‘π‘œπ‘ ^2 π‘₯, we get I=πœ‹/2 ∫_0^πœ‹β–’γ€–(1/cos^2⁑π‘₯ )/((π‘Ž^2 cos^2⁑〖π‘₯ + 𝑏^2 sin^2⁑π‘₯ γ€—)/cos^2⁑π‘₯ ) 𝑑π‘₯γ€— I=πœ‹/2 ∫_0^πœ‹β–’γ€–(𝑠𝑒𝑐^2 π‘₯)/((π‘Ž^2 cos^2⁑π‘₯)/cos^2⁑π‘₯ + (𝑏^2 sin^2⁑π‘₯)/cos^2⁑π‘₯ ) 𝑑π‘₯γ€— I=πœ‹/2 ∫_0^πœ‹β–’γ€–(𝑠𝑒𝑐^2 π‘₯)/(π‘Ž^2 + 𝑏^2 tan^2⁑π‘₯ ) 𝑑π‘₯γ€— Let 𝑓(π‘₯)=sec^2⁑π‘₯/(π‘Ž^2 + 𝑏^2 tan^2⁑π‘₯ ) and a = Ο€ Now, 𝑓(2π‘Žβˆ’π‘₯)=sec^2⁑(πœ‹ βˆ’ π‘₯)/(π‘Ž^2 + 𝑏^2 tan^2⁑(πœ‹ βˆ’ π‘₯) ) 𝑓(2π‘Žβˆ’π‘₯)=[βˆ’π‘ π‘’π‘ π‘₯]^2/(π‘Ž^2 + 𝑏^2 [βˆ’tan⁑π‘₯ ]^2 ) 𝑓(2π‘Žβˆ’π‘₯)=(𝑠𝑒𝑐^2 π‘₯)/(π‘Ž^2 + 𝑏^2 tan^2⁑π‘₯ ) Therefore, 𝑓(π‘₯)=𝑓(2π‘Žβˆ’π‘₯) Using the Property , P6 P6 :- ∫_0^2π‘Žβ–’γ€–π‘“ (π‘₯)𝑑π‘₯=2∫_0^π‘Žβ–’γ€–π‘“(π‘₯)𝑑π‘₯,γ€— γ€— if 𝑓(2π‘Žβˆ’π‘₯)=𝑓(π‘₯) Therefore, I=πœ‹/2 ∫_0^πœ‹β–’γ€–(𝑠𝑒𝑐^2 π‘₯)/(π‘Ž^2 + 𝑏^2 tan^2⁑π‘₯ ) 𝑑π‘₯γ€— =πœ‹/2 Γ— 2 ∫_0^(πœ‹/2)β–’γ€–(𝑠𝑒𝑐^2 π‘₯)/(π‘Ž^2 + 𝑏^2 tan^2⁑π‘₯ ) 𝑑π‘₯γ€— =πœ‹βˆ«_0^(πœ‹/2)β–’γ€–(𝑠𝑒𝑐^2 π‘₯)/(π‘Ž^2 + 𝑏^2 tan^2⁑π‘₯ ) 𝑑π‘₯γ€— Let 𝑏 tan⁑〖π‘₯=𝑑〗 Differentiating both sides w.r.t. π‘₯ 𝑏 𝑠𝑒𝑐^2 π‘₯ 𝑑π‘₯=𝑑𝑑 𝑑𝑑=𝑑𝑑/(𝑏^2 𝑠𝑒𝑐^2 π‘₯) Putting the values of tan π‘₯ and 𝑑π‘₯ , we get 𝐼=πœ‹βˆ«1_0^(πœ‹/2)β–’γ€–(𝑠𝑒𝑐^2 π‘₯)/(π‘Ž^2 + 𝑑^2 ) . 𝑑π‘₯γ€— 𝐼=πœ‹ ∫1_0^βˆžβ–’γ€–(𝑠𝑒𝑐^2 π‘₯)/(π‘Ž^2 + 𝑑^2 ) .𝑑𝑑/(𝑏 𝑠𝑒𝑐^2 π‘₯)γ€— 𝐼=πœ‹/𝑏 ∫1_0^βˆžβ–’π‘‘π‘‘/(π‘Ž^2 + 𝑑^2 ) It is of form ∫1▒〖𝑑π‘₯/(π‘Ž^2+π‘₯^2 )=1/π‘Ž tan^(βˆ’1)⁑〖π‘₯/π‘Ž+𝐢〗 γ€— Replacing π‘₯ by 𝑑 , we get 𝐼= γ€–πœ‹/𝑏 [1/π‘Ž tan^(βˆ’1)⁑(𝑑/π‘Ž) ]γ€—_0^∞ Putting limits, I=πœ‹/𝑏 [1/π‘Ž γ€–π‘‘π‘Žπ‘›γ€—^(βˆ’1) (∞/π‘Ž)βˆ’1/π‘Ž γ€–π‘‘π‘Žπ‘›γ€—^(βˆ’1) (0/π‘Ž)] I =πœ‹/𝑏 [γ€–1/π‘Ž tan^(βˆ’1)〗⁑〖(∞)βˆ’1/π‘Ž tan^(βˆ’1)⁑(0) γ€— ] I =πœ‹/𝑏 (1/π‘Ž (πœ‹/2)βˆ’0) I =𝝅^𝟐/πŸπ’‚π’ƒ

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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.