∫ 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

Slide2.JPG

Slide3.JPG

Go Ad-free

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)

Davneet Singh's photo - Co-founder, Teachoo

Made by

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, Social Science, Physics, Chemistry, Computer Science at Teachoo.