If ∫ d x 16 - 9 x 2 = A sin - 1 ⁡ B x + C then A + B =, 9 4, 19 4, 3 4, 13 12

I = ∫ d x 4 2 − ( 3 x ) 2 I = 1 3 sin − 1 ( 3 x 4 ) + C A + B = 1 3 + 3 4 = 4 + 9 12 = 13 12

If ∫ d x 16 - 9 x 2 = A sin - 1 ⁡ B x + C then A + B =

Question: Answered & Verified by Expert

\int \frac{d x}{\sqrt{16 - 9 x^{2}}} = A \sin^{- 1} (B x) + C

then

A + B =

MathematicsIndefinite IntegrationJEE Main

Options:

Solution:

2594 Upvotes Verified Answer

The correct answer is:

\frac{13}{12}

I = \int \frac{d x}{\sqrt{4^{2} - {(3 x)}^{2}}}

I = \frac{1}{3} \sin^{- 1} (\frac{3 x}{4}) + C

A + B = \frac{1}{3} + \frac{3}{4} = \frac{4 + 9}{12} = \frac{13}{12}

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