Integration Full Chapter Explained -



  1. Chapter 7 Class 12 Integrals
  2. Serial order wise


Ex 7.3, 19 Integrate the function 1/(sin⁑π‘₯ . cos^3⁑π‘₯ ) ∫1β–’1/(sin⁑π‘₯ . cos^3⁑π‘₯ ) 𝑑π‘₯ =∫1β–’(sin^2⁑π‘₯ + cos^2⁑π‘₯)/(sin⁑π‘₯ . cos^3⁑π‘₯ ) 𝑑π‘₯ =∫1β–’(sin^2⁑π‘₯/(sin⁑π‘₯ . cos^3⁑π‘₯ )+cos^2⁑π‘₯/(sin⁑π‘₯ . cos^3⁑π‘₯ )) 𝑑π‘₯ =∫1β–’(𝑠𝑖𝑛⁑π‘₯/cos^3⁑π‘₯ +π‘π‘œπ‘ β‘π‘₯/(sin⁑π‘₯ . cos^2⁑π‘₯ )) 𝑑π‘₯ =∫1β–’(𝑠𝑖𝑛⁑π‘₯/π‘π‘œπ‘ β‘π‘₯ Γ—1/cos^2⁑π‘₯ +π‘π‘œπ‘ β‘π‘₯/sin⁑π‘₯ Γ—1/cos^2⁑π‘₯ ) 𝑑π‘₯ =∫1β–’γ€–1/cos^2⁑π‘₯ (𝑠𝑖𝑛⁑π‘₯/π‘π‘œπ‘ β‘π‘₯ +π‘π‘œπ‘ β‘π‘₯/sin⁑π‘₯ ) γ€— 𝑑π‘₯ (As 〖𝑠𝑖𝑛〗^2β‘πœƒ+γ€–π‘π‘œπ‘ γ€—^2β‘πœƒ=1) =∫1β–’γ€–sec^2⁑π‘₯ (tan⁑π‘₯+cot⁑π‘₯ ) γ€— 𝑑π‘₯ =∫1β–’γ€–sec^2⁑π‘₯ (tan⁑π‘₯+1/tan⁑π‘₯ ) γ€— 𝑑π‘₯ =∫1β–’γ€–(tan⁑π‘₯+1/tan⁑π‘₯ ). sec^2⁑π‘₯ γ€— 𝑑π‘₯ Putting π‘‘π‘Žπ‘›β‘π‘₯=𝑑 Differentiating w.r.t.x sec^2⁑π‘₯=𝑑𝑑/𝑑π‘₯ 𝑑π‘₯=1/sec^2⁑π‘₯ 𝑑𝑑 Thus, our equation becomes =∫1β–’γ€–(𝑑+1/𝑑) sec^2⁑π‘₯Γ—1/sec^2⁑π‘₯ γ€— 𝑑𝑑 =∫1β–’(𝑑+1/𝑑) 𝑑𝑑 =∫1▒𝑑 𝑑𝑑+∫1β–’1/𝑑 𝑑𝑑 =𝑑^2/2 +log⁑|𝑑|+𝐢 Putting value of 𝑑=π‘‘π‘Žπ‘›β‘π‘₯ =(γ€–π‘‘π‘Žπ‘›γ€—^2 π‘₯)/2+π‘™π‘œπ‘”β‘|π‘‘π‘Žπ‘›β‘π‘₯ |+𝐢 =π’π’π’ˆβ‘|𝒕𝒂𝒏⁑𝒙 |+(〖𝒕𝒂𝒏〗^𝟐 𝒙)/𝟐 +π‘ͺ

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.