## β« dx/(sin^2β‘x cos^2β‘x) is equal to

## (A) tanβ‘x + cotβ‘x + CΒ

## (B) (tanβ‘x + cotβ‘x )^2 + CΒ

## (C) tanβ‘x - cotβ‘x + CΒ

## (D) (tanβ‘x - cotβ‘x )^2 + C

This question is

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

Last updated at Jan. 11, 2022 by

This question is

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

Transcript

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)

Chapter 7 Class 12 Integrals (Term 2)

Serial order wise

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