∫ 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
NCERT Exemplar MCQ
Last updated at Dec. 14, 2024 by Teachoo
This question is similar to Ex 7.2, 39 (MCQ) - Chapter 7 Class 12 - Integrals
Question 2 β«1βππ₯/(sin^2β‘π₯ cos^2β‘π₯ ) is equal to (A) tanβ‘π₯+cotβ‘π₯+πΆ (B) (tanβ‘π₯+cotβ‘π₯ )^2+πΆ (C) tanβ‘π₯βcotβ‘π₯+πΆ (D) (tanβ‘π₯βcotβ‘π₯ )^2+πΆ β«1βγ" " ππ₯/(sin^2 π₯ cos^2β‘π₯ )γ = β«1βγ" " π/(sin^2 π₯ cos^2β‘π₯ ) . ππ₯γ = β«1βγ" " (γπ¬π’π§γ^π π +γ γππ¨π¬γ^πγβ‘π)/(sin^2 π₯ cos^2β‘π₯ ) . ππ₯γ = β«1βγ" " (sin^2 π₯)/(sin^2 π₯ cos^2β‘π₯ ) . ππ₯γ + β«1βγ" " γ cos^2γβ‘π₯/(sin^2 π₯ cos^2β‘π₯ ) . ππ₯γ = β«1βγ" " 1/cos^2β‘π₯ . ππ₯γ + β«1βγ1/(sin^2 π₯) . ππ₯γ = β«1βγsec^2β‘π₯. ππ₯γ + β«1βγπππ ππ^2 π₯ . ππ₯γ = πππβ‘πβπππβ‘π+πͺ So, the correct answer is (c)