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Chapter 7 Class 12 Integrals
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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 is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.