What is I equal to?

Question

Question

Direction : Consider the following for the items that follow :

Let \(\rm I=\int_0^{\pi/2}\frac{f(x)}{g(x)}dx\), where f(x) = sin x and g(x) = sin x + cos x + 1

This question was previously asked in

NDA-II 2024 (Maths) Official Paper (Held On: 01 Sept, 2024)

Answer 1

Concept:

Property of Definite Integrals (Symmetry):

If \( f(x) \) is continuous on \( [0, a] \), then \( \int_0^a f(x)\,dx = \int_0^a f(a - x)\,dx \).
This property helps in evaluating integrals involving trigonometric identities when limits are symmetric.
Standard result used: \( \int_0^{\frac{\pi}{2}} \frac{1}{\sin x + \cos x + 1} dx = \ln 2 \).
Combining two symmetrical integrals can simplify expressions and eliminate complex terms.

Calculation:

Let \( I = \int_0^{\frac{\pi}{2}} \frac{\sin x}{\sin x + \cos x + 1} dx \)

⇒ Using substitution \( x \rightarrow \frac{\pi}{2} - x \)

⇒ \( I = \int_0^{\frac{\pi}{2}} \frac{\cos x}{\sin x + \cos x + 1} dx \)

⇒ Adding both expressions, we get:

\( 2I = \int_0^{\frac{\pi}{2}} \frac{\sin x + \cos x}{\sin x + \cos x + 1} dx \)

⇒ \( 2I = \int_0^{\frac{\pi}{2}} \left(1 - \frac{1}{\sin x + \cos x + 1} \right) dx \)

⇒ \( 2I = \int_0^{\frac{\pi}{2}} 1 dx - \int_0^{\frac{\pi}{2}} \frac{1}{\sin x + \cos x + 1} dx \)

⇒ \( 2I = \frac{\pi}{2} - \ln 2 \)

⇒ \( I = \frac{\pi}{4} - \frac{\ln 2}{2} \)

∴ The value of the integral is \( \frac{\pi}{4} - \frac{\ln 2}{2} \).