TS EAMCET · Maths · Indefinite Integration
\(\int \frac{\cos 2 x \cdot \sin 4 x}{\cos ^4 x\left(1+\cos ^2 2 x\right)} d x=\)
- A \(\log \left[\frac{1+\cos ^2 2 x}{1+\cos 2 x}\right]-\tan ^2 x+c\)
- B \(\log \left(\frac{1+\cos ^2 2 x}{1+\cos 2 x}\right)+\tan ^2 x+c\)
- C \(\log \left(\frac{1+\cos 2 x}{1+\cos ^2 2 x}\right)+\sec ^2 x+c\)
- D \(\log \left(\frac{(1+\cos 2 x)^2}{1+\cos ^2 2 x}\right)+\sec ^2 x+c\)
Answer & Solution
Correct Answer
(D) \(\log \left(\frac{(1+\cos 2 x)^2}{1+\cos ^2 2 x}\right)+\sec ^2 x+c\)
Step-by-step Solution
Detailed explanation
\[ \begin{aligned} & \text { Let } I=\int \frac{\cos 2 x \sin 4 x}{\cos ^4 x\left(1+\cos ^2 2 x\right)} d x \\ & =4 \int \frac{\cos 2 x(2 \sin 2 x \cos 2 x)}{(1+\cos 2 x)^2\left(1+\cos ^2 2 x\right)} d x \end{aligned} \] Put \(\cos 2 x=t \Rightarrow-2 \sin 2 x d x=d t\)…
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