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Example 44 - Evaluate x dx / a2 cos2 x + b2 sin2 x - Examples

Example 44 - Chapter 7 Class 12 Integrals - Part 2
Example 44 - Chapter 7 Class 12 Integrals - Part 3
Example 44 - Chapter 7 Class 12 Integrals - Part 4
Example 44 - Chapter 7 Class 12 Integrals - Part 5
Example 44 - Chapter 7 Class 12 Integrals - Part 6
Example 44 - Chapter 7 Class 12 Integrals - Part 7

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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 has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.