KCET · Maths · Indefinite Integration
\(\int \frac{\sin x}{3+4 \cos ^2 x} d x\)
- A \(-\frac{1}{2 \sqrt{3}} \tan ^{-1}\left(\frac{2 \cos x}{\sqrt{3}}\right)+C\)
- B \(\frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{\cos x}{3}\right)+C\)
- C \(\frac{1}{2 \sqrt{3}} \tan ^{-1}\left(\frac{\cos x}{3}\right)+C\)
- D \(-\frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{2 \cos x}{3}\right)+C\)
Answer & Solution
Correct Answer
(A) \(-\frac{1}{2 \sqrt{3}} \tan ^{-1}\left(\frac{2 \cos x}{\sqrt{3}}\right)+C\)
Step-by-step Solution
Detailed explanation
\(\because I=\int \frac{\sin x}{3+4 \cos ^2 x} d x\)
\(=\frac{\sin x}{3\left[1+\left(\frac{2 \cos x}{\sqrt{3}}\right)^2\right]} d x\)
put \(\quad t=\frac{2 \cos x}{\sqrt{3}}\)
\(d t=\frac{-2 \sin x}{\sqrt{3}} d x\)
\(I=\int-\frac{\sqrt{3}}{2\left(3 t^2+3\right)} d t\)
\(=-\frac{1}{2 \sqrt{3}} \int \frac{1}{1+t^2} d t\)
\(=-\frac{1}{2 \sqrt{3}} \tan ^{-1}(t)+C\)
\(=\frac{-1}{2 \sqrt{3}} \tan ^{-1}\left(\frac{2 \cos x}{\sqrt{3}}\right)+C\)
\(=\frac{\sin x}{3\left[1+\left(\frac{2 \cos x}{\sqrt{3}}\right)^2\right]} d x\)
put \(\quad t=\frac{2 \cos x}{\sqrt{3}}\)
\(d t=\frac{-2 \sin x}{\sqrt{3}} d x\)
\(I=\int-\frac{\sqrt{3}}{2\left(3 t^2+3\right)} d t\)
\(=-\frac{1}{2 \sqrt{3}} \int \frac{1}{1+t^2} d t\)
\(=-\frac{1}{2 \sqrt{3}} \tan ^{-1}(t)+C\)
\(=\frac{-1}{2 \sqrt{3}} \tan ^{-1}\left(\frac{2 \cos x}{\sqrt{3}}\right)+C\)
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