Question

\(\int_{0}^{\pi / 2} \frac{x d x}{1+\sin x}=\) (1) \(\operatorname{n} 2+\pi / 4\) (2) \(\ell \mathrm{n} 2-\pi / 4\) (3) \(1+\pi / 2+\ell n 2\) (4) \(\ell \mathrm{n} 2\)

Short answer

\ln 2 - \frac{\pi}{4}

Step-by-step solution

- Identify the integral

We need to evaluate the integral \(\int_{0}^{\frac{\pi}{2}} \frac{x \, dx}{1 + \sin x}\).

- Consider substitution

Let \(I = \int_{0}^{\frac{\pi}{2}} \frac{x \, dx}{1 + \sin x}.\) Now employ the substitution \(u = \frac{\pi}{2} - x\). This gives \(du = -dx\), and when \(x = 0\), \(u = \frac{\pi}{2}\), and when \(x = \frac{\pi}{2}\), \(u = 0\).

- Transform the integral

Substitute and reverse the limits of integration: \(I = \int_{\frac{\pi}{2}}^{0} \frac{(\frac{\pi}{2} - u)(-du)}{1 + \sin(\frac{\pi}{2} - u)} = \int_{0}^{\frac{\pi}{2}} \frac{(\frac{\pi}{2} - u) \, du}{1 + \cos u}\) because \(\sin (\frac{\pi}{2} - u) = \cos u\).

- Split the integral

Rewrite and split the integral: \(I = \int_{0}^{\frac{\pi}{2}} \frac{\frac{\pi}{2} \, du}{1 + \cos u} - \int_{0}^{\frac{\pi}{2}} \frac{u \, du}{1 + \cos u}\). Observe that the second integral is \(I\) itself.

- Simplify and solve

Let’s focus on the first integral: \(\int_{0}^{\frac{\pi}{2}} \frac{\frac{\pi}{2} \, du}{1 + \cos u} = \frac{\pi}{2} \int_{0}^{\frac{\pi}{2}} \frac{du}{1 + \cos u}\). To evaluate \(\int_{0}^{\frac{\pi}{2}} \frac{du}{1 + \cos u}$, use the identity \(1 + \cos u = 2\cos^2(\frac{u}{2})\), which gives: $$\int_{0}^{\frac{\pi}{2}} \frac{du}{1 + \cos u} = \int_{0}^{\frac{\pi}{2}} \frac{du}{2\cos^2(\frac{u}{2})}\). This simplifies to \(\frac{1}{2} \int_{0}^{\frac{\pi}{2}} \sec^2(\frac{u}{2}) d (\frac{u}{2}) = \frac{1}{2} \int_{0}^{\frac{\pi}{4}} \sec^2 t \, dt\), which further simplifies to \(\frac{1}{2}[\tan t ]_{0}^{\frac{\pi}{4}} = \frac{1}{2}[1 - 0] = \frac{1}{2}\).

- Combine results

So, \(\frac{\pi}{2} \int_{0}^{\frac{\pi}{2}} \frac{du}{1 + \cos u} = \frac{\pi}{2} \cdot \frac{1}{2} = \frac{\pi}{4}\). Therefore, we have: \(I = \frac{\pi}{4} - I\). Solving for \(I\), get: \(2I = \frac{\pi}{4}\), hence \(I = \frac{\pi}{8}\).

Key concepts

Integration Techniques

To solve the problem, we need to use a variety of integration techniques. Initially, we encounter a definite integral with a function in the denominator that involves a trigonometric component. For such integrals, substitution often helps to simplify the integrand. In this case, substituting \(u = \frac{\pi}{2} - x\) changes the integration limits and trigonometric function, making the integral more manageable.Another technique is splitting the integral into more straightforward parts. This involves recognizing patterns or identities that let us separate the integral into parts that are easier to handle or even repeat the original integral form.

Trigonometric Substitution

Trigonometric substitution is a useful tool in integration, especially when the integral includes trigonometric functions. Substitution simplifies these functions by leveraging known identities. For instance, in our problem, the substitution \(u = \frac{\pi}{2} - x\) transforms \(\sin x\) into \(\cos u\), utilizing the identity \(\sin(\frac{\pi}{2} - u) = \cos u\).Breaking down the integral further using trigonometric identities like \(1 + \cos u = 2\cos^2(\frac{u}{2})\) simplifies the integrand even more. We then integrate using basic trigonometric integrals, such as \(\int \sec^2 t \, dt = \tan t\). This step-by-step simplification underscores how trigonometric identities and substitutions make complex integrals solvable.

Definite Integrals

Definite integrals have specific limits of integration, which provide a concrete value for the area under the curve. In our problem, we deal with the definite integral \(\int_{0}^{\frac{\pi}{2}} \frac{x \, dx}{1 + \sin x}\).After transforming the integral using substitution \(u = \frac{\pi}{2} - x\), and reversing the limits, we simplify and split the integral. Solving these parts separately—one using a simplified trigonometric identity—helps us to re-evaluate the complicated integral back to known simpler forms. Reassembling these parts and combining the results, we use algebra to solve for the unknown.This approach of breaking down the integral, using known identities, and reassembling reflects the systematic process essential for definite integrals, providing clarity and manageable steps to reach the solution.