KCET · Maths · Continuity and Differentiability
\( \int \frac{\sin 2 x}{\left(\sin ^{2} x+2 \cos ^{2} x\right.} d x= \)
- A \( \log \left(1+\cos ^{2} x\right)+C \)
- B \( \log \left(1+\tan ^{2} x\right)+C \)
- C \( -\log \left(1+\sin ^{2} x\right)+C \)
- D \( -\log \left(1+\cos ^{2} x\right)+C \)
Answer & Solution
Correct Answer
(D) \( -\log \left(1+\cos ^{2} x\right)+C \)
Step-by-step Solution
Detailed explanation
Given that,\(I=\int \frac{\sin 2 x}{\sin ^{2} x+2 \cos ^{2} x} d x\)
Since, \(\sin ^{2} x=1-\cos ^{2} x .\) So,
\(=\int \frac{\sin 2 x d x}{1+\cos ^{2} x}\)
Let \(1+\cos ^{2} x=t .\) So,
\(-2 \cos x \sin x d x=d t\)
\(\Rightarrow \sin 2 x d x=-d t\)
\(\Rightarrow-\int \frac{d t}{t}=-\log t+c=-\log \left(1+\cos ^{2} x\right)+c\)
Since, \(\sin ^{2} x=1-\cos ^{2} x .\) So,
\(=\int \frac{\sin 2 x d x}{1+\cos ^{2} x}\)
Let \(1+\cos ^{2} x=t .\) So,
\(-2 \cos x \sin x d x=d t\)
\(\Rightarrow \sin 2 x d x=-d t\)
\(\Rightarrow-\int \frac{d t}{t}=-\log t+c=-\log \left(1+\cos ^{2} x\right)+c\)
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