∫ dx/(sin^2⁑x cos^2⁑x) is equal to

(A) tan⁑x + cot⁑x + C 

(B) (tan⁑x + cot⁑x )^2 + C 

(C) tan⁑x - cot⁑x + C 

(D) (tan⁑x - cot⁑x )^2 + C

This question is similar to Ex 7.2, 39 (MCQ) - Chapter 7 Class 12 - Integrals

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  1. Chapter 7 Class 12 Integrals (Term 2)
  2. Serial order wise

Transcript

Question 2 ∫1▒𝑑π‘₯/(sin^2⁑π‘₯ cos^2⁑π‘₯ ) is equal to (A) tan⁑π‘₯+cot⁑π‘₯+𝐢 (B) (tan⁑π‘₯+cot⁑π‘₯ )^2+𝐢 (C) tan⁑π‘₯βˆ’cot⁑π‘₯+𝐢 (D) (tan⁑π‘₯βˆ’cot⁑π‘₯ )^2+𝐢 ∫1β–’γ€–" " 𝑑π‘₯/(sin^2 π‘₯ cos^2⁑π‘₯ )γ€— = ∫1β–’γ€–" " 𝟏/(sin^2 π‘₯ cos^2⁑π‘₯ ) . 𝑑π‘₯γ€— = ∫1β–’γ€–" " (〖𝐬𝐒𝐧〗^𝟐 𝒙 +γ€– γ€–πœπ¨π¬γ€—^πŸγ€—β‘π’™)/(sin^2 π‘₯ cos^2⁑π‘₯ ) . 𝑑π‘₯γ€— = ∫1β–’γ€–" " (sin^2 π‘₯)/(sin^2 π‘₯ cos^2⁑π‘₯ ) . 𝑑π‘₯γ€— + ∫1β–’γ€–" " γ€– cos^2〗⁑π‘₯/(sin^2 π‘₯ cos^2⁑π‘₯ ) . 𝑑π‘₯γ€— = ∫1β–’γ€–" " 1/cos^2⁑π‘₯ . 𝑑π‘₯γ€— + ∫1β–’γ€–1/(sin^2 π‘₯) . 𝑑π‘₯γ€— = ∫1β–’γ€–sec^2⁑π‘₯. 𝑑π‘₯γ€— + ∫1β–’γ€–π‘π‘œπ‘ π‘’π‘^2 π‘₯ . 𝑑π‘₯γ€— = π’•π’‚π’β‘π’™βˆ’π’„π’π’•β‘π’™+π‘ͺ So, the correct answer is (c)

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.