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Chapter 7 Class 12 Integrals
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Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


Example 42 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€— 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‘₯) 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 ) 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 has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.