$∫\frac{sin(tan^{-1}x)}{1+x^2}$dx=

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Home Questions 78GPUCTJ49

Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Indefinite Integration

Question:

$∫\frac{sin(tan^{-1}x)}{1+x^2}$dx=

Options:

-cos(tan^-1 x) + C

cos(tan^-1 x) + C

tan(cos^-1 x) + C

-tan(cos^-1 x) + C

Correct Answer:

-cos(tan^-1 x) + C

Explanation:

Let t = tan^-1 x ......(i)

$dt=\frac{dx}{1+x^2}$

it becomes ⇒ $∫sin\, t\, dt$

= - cot t + c ......(ii)

Substituting the value of t in equation (ii)

= -cos(tan^-1 x) + C

Option 1 is correct.