Question

$\displaystyle \int \frac{dx}{1+\cos x}$ is equal to

• A. $\dfrac{1}{2}\tan\dfrac{x}{2}+C$
• B. $-\dfrac{1}{2}\cot\dfrac{x}{2}+C$
• C. $\cot\dfrac{x}{2}+C$
• D. $\tan\dfrac{x}{2}+C$

Hint:

Whenever integrals contain \(1+\cos x\) or \(1-\cos x\), convert them using half-angle identities such as \(1+\cos x=2\cos^2(x/2)\). This simplifies the integral immediately.

Answer 1

The Correct Option is D

Step 1: Simplify the denominator using trigonometric identity.
We use the identity:
\[ 1+\cos x = 2\cos^2\left(\frac{x}{2}\right) \]
Substitute into the integral:
\[ \int \frac{dx}{1+\cos x} = \int \frac{dx}{2\cos^2(x/2)} \]
Step 2: Rewrite the expression.
\[ \frac{1}{2}\int \sec^2\left(\frac{x}{2}\right)dx \]
Step 3: Use substitution.
Let \[ u=\frac{x}{2} \] then \[ dx=2\,du \]
Substitute:
\[ \frac{1}{2}\int \sec^2(u)\,(2du) = \int \sec^2(u)du \]
Step 4: Integrate.
\[ \int \sec^2(u)du=\tan(u) \]
Substitute back \(u=\frac{x}{2}\):
\[ \tan\frac{x}{2}+C \]
Conclusion:
\[ \int \frac{dx}{1+\cos x} = \tan\frac{x}{2}+C \]
Final Answer: $\boxed{\tan\frac{x}{2}+C}$