Question

Find \(\displaystyle\int e^x\left(\dfrac{1+\sin x}{1+\cos x}\right)dx\).

Hint:

Use \(1+\cos x=2\cos^2(x/2)\) and \(\sin x=2\sin(x/2)\cos(x/2)\), then apply \(\int e^x[f(x)+f'(x)]dx=e^xf(x)+C\).

Answer 1

Step 1: Understanding the Concept:
Many integrals of the form \(\int e^x[f(x)+f'(x)]dx\) simplify directly to \(e^xf(x)+C\).
The plan is to rewrite \(\dfrac{1+\sin x}{1+\cos x}\) in the form \(f(x)+f'(x)\) for some function f.

Step 2: Use half-angle identities:
Use \(1+\cos x=2\cos^2(x/2)\) and \(\sin x=2\sin(x/2)\cos(x/2)\).
\[ \frac{1+\sin x}{1+\cos x}=\frac{1+2\sin(x/2)\cos(x/2)}{2\cos^2(x/2)}=\frac{1}{2\cos^2(x/2)}+\frac{\sin(x/2)}{\cos(x/2)} \]
\[ =\frac12\sec^2\frac x2+\tan\frac x2 \]

Step 3: Identify f(x) and f'(x):
Let \(f(x)=\tan\dfrac x2\).
Then \(f'(x)=\dfrac12\sec^2\dfrac x2\), which is exactly the other term found above.
So the integrand is \(e^x[f(x)+f'(x)]\) with \(f(x)=\tan(x/2)\).

Step 4: Apply the standard formula:
\[ \int e^x[f(x)+f'(x)]dx=e^xf(x)+C \]

Final Answer:
Substitute \(f(x)=\tan(x/2)\) into the formula. \[ \boxed{\int e^x\left(\frac{1+\sin x}{1+\cos x}\right)dx=e^x\tan\frac x2+C} \]