∫ 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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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 has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo