Ex 7.3, 16 - Chapter 7 Class 12 Integrals (Term 2)

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Ex 7.3, 16 ∫1▒〖tan^4 𝑥〗 𝑑𝑥 ∫1▒〖tan^4 𝑥〗 𝑑𝑥=∫1▒〖tan^2 𝑥 .tan^2 𝑥〗 𝑑𝑥 =∫1▒〖(sec^2⁡𝑥− 1) tan^2⁡𝑥 〗 𝑑𝑥 =∫1▒(sec^2⁡𝑥.tan^2⁡𝑥−tan^2⁡𝑥 ) 𝑑𝑥 =∫1▒〖tan^2⁡𝑥.sec^2⁡𝑥 〗 𝑑𝑥−∫1▒〖tan^2 𝑥〗 𝑑𝑥 Solving both these integrals separately We know that 〖𝑡𝑎𝑛〗^2 𝜃=〖𝑠𝑒𝑐〗^2⁡〖𝜃−1〗 ∫1▒〖〖𝒕𝒂𝒏〗^𝟐⁡𝒙.〖𝒔𝒆𝒄〗^𝟐⁡𝒙 〗 𝒅𝒙 Let tan 𝑥=𝑡 sec^2⁡𝑥=𝑑𝑡/𝑑𝑥 𝑑𝑥=1/sec^2⁡𝑥 . 𝑑𝑡 Now, ∫1▒tan^2⁡𝑥 .sec^2⁡𝑥.𝑑𝑥 =∫1▒𝑡^2 .sec^2⁡𝑥. 1/sec^2⁡𝑥 .𝑑𝑡 =∫1▒𝑡^2 . 𝑑𝑡 =𝑡^(2 + 1)/(2 + 1) + C =𝑡^3/3+𝐶 Putting value of 𝑡=𝑡𝑎𝑛⁡𝑥 =tan^3⁡𝑥/3+𝐶1 ∫1▒〖〖𝒕𝒂𝒏〗^𝟐 𝒙〗 𝒅𝒙 =∫1▒(sec^2⁡𝑥−1) 𝑑𝑥 =∫1▒sec^2⁡𝑥 𝑑𝑥−∫1▒1⁡〖.𝑑𝑥〗 =tan⁡𝑥−𝑥+𝐶2 "As" ∫1▒〖𝑥^𝑛 𝑑𝑥=𝑥^(𝑛+1)/(𝑛+1)+𝐶〗 & ∫1▒sec^2⁡𝑥 𝑑𝑥=tan⁡𝑥+𝐶 Now, ∫1▒〖tan^4 𝑥〗 𝑑𝑥=∫1▒〖tan^2 𝑥 .sec^2 𝑥〗 𝑑𝑥−∫1▒〖tan^2 𝑥〗 𝑑𝑥 =tan^3⁡𝑥/3+𝐶1−(tan⁡𝑥−𝑥+𝐶2) =tan^3⁡𝑥/3 −tan⁡𝑥+𝑥+𝐶1−𝐶2 =〖𝒕𝒂𝒏〗^𝟑⁡𝒙/𝟑 −𝒕𝒂𝒏⁡𝒙+𝒙+𝑪 (Where 𝐶=𝐶1−𝐶2)