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Ex 7.3, 16 - Chapter 7 Class 12 Integrals - Part 2
Ex 7.3, 16 - Chapter 7 Class 12 Integrals - Part 3

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Ex 7.3, 16 ∫1β–’γ€–tan^4 π‘₯γ€— 𝑑π‘₯ ∫1β–’γ€–tan^4 π‘₯γ€— 𝑑π‘₯=∫1β–’γ€–tan^2 π‘₯ .tan^2 π‘₯γ€— 𝑑π‘₯ =∫1β–’γ€–(sec^2⁑π‘₯βˆ’ 1) tan^2⁑π‘₯ γ€— 𝑑π‘₯ =∫1β–’(sec^2⁑π‘₯.tan^2⁑π‘₯βˆ’tan^2⁑π‘₯ ) 𝑑π‘₯ =∫1β–’γ€–tan^2⁑π‘₯.sec^2⁑π‘₯ γ€— 𝑑π‘₯βˆ’βˆ«1β–’γ€–tan^2 π‘₯γ€— 𝑑π‘₯ Solving both these integrals separately We know that γ€–π‘‘π‘Žπ‘›γ€—^2 πœƒ=〖𝑠𝑒𝑐〗^2β‘γ€–πœƒβˆ’1γ€— ∫1▒〖〖𝒕𝒂𝒏〗^πŸβ‘π’™.〖𝒔𝒆𝒄〗^πŸβ‘π’™ γ€— 𝒅𝒙 Let tan π‘₯=𝑑 sec^2⁑π‘₯=𝑑𝑑/𝑑π‘₯ 𝑑π‘₯=1/sec^2⁑π‘₯ . 𝑑𝑑 Now, ∫1β–’tan^2⁑π‘₯ .sec^2⁑π‘₯.𝑑π‘₯ =∫1▒𝑑^2 .sec^2⁑π‘₯. 1/sec^2⁑π‘₯ .𝑑𝑑 =∫1▒𝑑^2 . 𝑑𝑑 =𝑑^(2 + 1)/(2 + 1) + C =𝑑^3/3+𝐢 Putting value of 𝑑=π‘‘π‘Žπ‘›β‘π‘₯ =tan^3⁑π‘₯/3+𝐢1 ∫1▒〖〖𝒕𝒂𝒏〗^𝟐 𝒙〗 𝒅𝒙 =∫1β–’(sec^2⁑π‘₯βˆ’1) 𝑑π‘₯ =∫1β–’sec^2⁑π‘₯ 𝑑π‘₯βˆ’βˆ«1β–’1⁑〖.𝑑π‘₯γ€— =tan⁑π‘₯βˆ’π‘₯+𝐢2 "As" ∫1β–’γ€–π‘₯^𝑛 𝑑π‘₯=π‘₯^(𝑛+1)/(𝑛+1)+𝐢〗 & ∫1β–’sec^2⁑π‘₯ 𝑑π‘₯=tan⁑π‘₯+𝐢 Now, ∫1β–’γ€–tan^4 π‘₯γ€— 𝑑π‘₯=∫1β–’γ€–tan^2 π‘₯ .sec^2 π‘₯γ€— 𝑑π‘₯βˆ’βˆ«1β–’γ€–tan^2 π‘₯γ€— 𝑑π‘₯ =tan^3⁑π‘₯/3+𝐢1βˆ’(tan⁑π‘₯βˆ’π‘₯+𝐢2) =tan^3⁑π‘₯/3 βˆ’tan⁑π‘₯+π‘₯+𝐢1βˆ’πΆ2 =〖𝒕𝒂𝒏〗^πŸ‘β‘π’™/πŸ‘ βˆ’π’•π’‚π’β‘π’™+𝒙+π‘ͺ (Where 𝐢=𝐢1βˆ’πΆ2)

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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.