What is \(\rm \int_0^{\pi/2}\frac{d(x)}{g(x)}\) equal to

Concept:

Integration using Substitution:

• This method involves replacing a complex expression with a single variable to simplify integration.
• Useful for trigonometric integrals where identities and substitutions simplify the expression.
• Here, the identity used is:
  • \( \sin x + \cos x = \frac{2\tan \frac{x}{2}}{1 + \tan^2 \frac{x}{2}} + \frac{1 - \tan^2 \frac{x}{2}}{1 + \tan^2 \frac{x}{2}} \)
  • \( 1 + \tan \frac{x}{2} = t \Rightarrow \frac{1}{2} \sec^2 \frac{x}{2} dx = dt \)
• The integral is solved using limits transformation and properties of logarithmic integrals.

Calculation:

Given: \( g(x) = \sin x + \cos x + 1 \)

We evaluate \( \int_{0}^{\frac{\pi}{2}} \frac{dx}{g(x)} = \int_{0}^{\frac{\pi}{2}} \frac{1}{\sin x + \cos x + 1} dx \)

⇒ Apply Weierstrass substitution: \( \tan \frac{x}{2} = t \)

⇒ Use identity: \( \sin x + \cos x = \frac{2\tan \frac{x}{2}}{1 + \tan^2 \frac{x}{2}} + \frac{1 - \tan^2 \frac{x}{2}}{1 + \tan^2 \frac{x}{2}} \)

⇒ Simplify denominator:

\( \sin x + \cos x + 1 = \frac{2\tan \frac{x}{2} + 1 - \tan^2 \frac{x}{2} + 1 + \tan^2 \frac{x}{2}}{1 + \tan^2 \frac{x}{2}} = \frac{2(1 + \tan \frac{x}{2})}{1 + \tan^2 \frac{x}{2}} \)

⇒ \( \int_{0}^{\frac{\pi}{2}} \frac{1}{2(1 + \tan \frac{x}{2})} \cdot \sec^2 \frac{x}{2} dx \)

Let \( 1 + \tan \frac{x}{2} = t \Rightarrow \frac{1}{2} \sec^2 \frac{x}{2} dx = dt \)

⇒ Limits change:

• \( x = 0 \Rightarrow t = 1 \)
• \( x = \frac{\pi}{2} \Rightarrow t = 2 \)

⇒ \( \int_{1}^{2} \frac{1}{t} dt = [\ln t]_{1}^{2} = \ln 2 - \ln 1 = \ln 2 \)

∴ The final answer is \( \ln 2 \)