TS EAMCET · Maths · Indefinite Integration
\(\int \frac{d x}{\cos (x+4) \cos (x+2)}\) is equal to
- A \(\frac{1}{\sin 2} \log \left|\cos (x+4)^2\right|+C\)
- B \(\frac{1}{2} \log \left|\frac{\sec (x+2)}{\sec (x+4)}\right|+C\)
- C \(\frac{1}{\sin 2} \log \left|\frac{\sec (x+4)}{\sec (x+2)}\right|+C\)
- D \(\log \left|\frac{\sec (x+4)}{\sec (x+2)}\right|+C\)
Answer & Solution
Correct Answer
(C) \(\frac{1}{\sin 2} \log \left|\frac{\sec (x+4)}{\sec (x+2)}\right|+C\)
Step-by-step Solution
Detailed explanation
\begin{aligned} & \text { 73. Let } I=\int \frac{d x}{\cos (x+4) \cos (x+2)} \\ & =\frac{1}{\sin 2} \int \frac{\sin 2}{\cos (x+4) \cos (x+2)} d x \\ & =\frac{1}{\sin 2} \int \frac{\sin [(x+4)-(x+2)]}{\cos (x+4) \cos (x+2)} d x=\frac{1}{\sin 2} \\ & \int \frac{[\sin (x+4) \cos…
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