TS EAMCET · Maths · Indefinite Integration
\(\int \frac{1}{16-7 \sin ^2 x} d x=\)
- A \(\frac{1}{12} \operatorname{Tan}^{-1}\left(\frac{3 \tan x}{4}\right)+c\)
- B \(\frac{1}{3} \operatorname{Sin}^{-1}\left(\frac{3 \sin x}{4}\right)+c\)
- C \(\frac{1}{12} \log \left(\frac{4-\sqrt{7} \sin x}{4+\sqrt{7} \sin x}\right)+c\)
- D \(\frac{1}{12} \log \left(\frac{4+\sqrt{7} \sin x}{4-\sqrt{7} \sin x}\right)+c\)
Answer & Solution
Correct Answer
(A) \(\frac{1}{12} \operatorname{Tan}^{-1}\left(\frac{3 \tan x}{4}\right)+c\)
Step-by-step Solution
Detailed explanation
\[ \begin{aligned} & \int \frac{d x}{16-7 \sin ^2 x}=\int \frac{\operatorname{cosec}^2 x d x}{16 \operatorname{cosec}^2 x-7} \\ & =\int \frac{\operatorname{cosec}^2 x d x}{16+16 \cot ^2 x-7}=\int \frac{\operatorname{cosec}^2 x d x}{16 \cot ^2 x+9} \end{aligned} \] Let…
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