TS EAMCET · Maths · Indefinite Integration
\(\int \frac{1+\cos 4 x}{\cot x-\tan x} d x\) is equal to
- A \(-\frac{1}{4} \cos 4 x+C\)
- B \(\frac{1}{8} \cos 4 x+C\)
- C \(\frac{1}{4} \sin 4 x+C\)
- D \(-\frac{1}{8} \cos 4 x+C\)
Answer & Solution
Correct Answer
(D) \(-\frac{1}{8} \cos 4 x+C\)
Step-by-step Solution
Detailed explanation
\begin{aligned} & \text { Let } I=\int \frac{1+\cos 4 x}{\cot x-\tan x} d x \\ & =\int \frac{\frac{2 \cos ^2 2 x}{\cos ^2 x-\sin ^2 x}}{\sin x \cos x} d x \\ & =\int \frac{\sin 2 x \cos ^2 2 x}{\cos 2 x} d x \\ & =\frac{1}{2} \int \sin 4 x d x=-\frac{\cos 4 x}{8}+C \\ &…
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