TS EAMCET · Maths · Indefinite Integration
If \(\int \frac{d x}{\left(x^2+9\right) \sqrt{x^2+16}}=\frac{1}{3 \sqrt{7}} \operatorname{Tan}^{-1}\left(\mathrm{~K} \frac{x}{\sqrt{16+x^2}}\right)+\mathrm{c}\), then \(\mathrm{K}=\)
- A \(\frac{\sqrt{7}}{3}\)
- B \(3 \sqrt{7}\)
- C \(\frac{3}{\sqrt{7}}\)
- D \(\frac{3}{7}\)
Answer & Solution
Correct Answer
(A) \(\frac{\sqrt{7}}{3}\)
Step-by-step Solution
Detailed explanation
Let \( x = 4 \tan \theta \). Then \( dx = 4 \sec^2 \theta d\theta \). \( \sqrt{x^2+16} = \sqrt{16 \tan^2 \theta + 16} = 4 \sec \theta \). \( \int \frac{4 \sec^2 \theta d\theta}{(16 \tan^2 \theta + 9)(4 \sec \theta)} = \int \frac{\sec \theta d\theta}{16 \tan^2 \theta + 9} \).…
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