AP 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 \(\log \left(\frac{1+\cos 2 x}{1+\cos ^2 2 x}\right)+\sec ^2 x+c\)
- B \(\log \frac{(1+\cos 2 x)^2}{\left(1+\cos ^2 x\right)}+\sec x+c\)
- C \(\log \frac{(1+\cos 2 x)^2}{\left(1+\cos ^2 2 x\right)}+\sec ^2 x+c\)
- D \(\log \frac{1+\cos ^2 2 x}{(1+\cos 2 x)^2}+\sec x+c\)
Answer & Solution
Correct Answer
(C) \(\log \frac{(1+\cos 2 x)^2}{\left(1+\cos ^2 2 x\right)}+\sec ^2 x+c\)
Step-by-step Solution
Detailed explanation
\(I=4 \int \frac{\cos 2 x \cdot(2 \sin 2 x \cos 2 x)}{(1+\cos 2 x)^2\left(1+\cos ^2 2 x\right)} d x\) Put \(\cos 2 x=t\) \(\Rightarrow -2 \sin 2 x d x=d t\) \(\therefore \quad I=-4 \int \frac{t^2}{(1+t)^2\left(1+t^2\right)} d t\) Split into partial fractions…
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