Question
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Answer
التكامل يساوي \(\frac{1}{2} \tan x + C\).
Solution
1. نبدأ بالتعرف على هوية مثلثية:
\[
1+\cos 2x = 2\cos^2 x
\]
2. نستبدل ذلك في التكامل:
\[
\int \frac{1}{1+\cos 2x} \, dx = \int \frac{1}{2\cos^2 x} \, dx = \frac{1}{2}\int \sec^2 x \, dx
\]
3. نعلم أن:
\[
\int \sec^2 x \, dx = \tan x + C
\]
4. بالتعويض نجد:
\[
\frac{1}{2}\int \sec^2 x \, dx = \frac{1}{2}\tan x + C
\]
لذا، الحل النهائي هو:
\[
\frac{1}{2}\tan x + C
\]
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Simplify this solution Mind Expander
To solve the integral \( \int \frac{1}{1 + \cos 2x} \, dx \), we can use the identity \( \cos 2x = 2\cos^2 x - 1 \). This simplifies our denominator to \( 1 + \cos 2x = 2\cos^2 x \), leading to: \[ \int \frac{1}{1 + \cos 2x} \, dx = \int \frac{1}{2\cos^2 x} \, dx = \frac{1}{2} \int \sec^2 x \, dx \] The antiderivative of \( \sec^2 x \) is \( \tan x \). Therefore, we have: \[ \frac{1}{2} \int \sec^2 x \, dx = \frac{1}{2} \tan x + C \] So the final result is: \[ \int \frac{1}{1 + \cos 2x} \, dx = \frac{1}{2} \tan x + C \] where \( C \) is the constant of integration.
