Question

Evaluate the following definite integrals. $$ \int_{0}^{\pi} \frac{\sin x}{\sqrt{1+\cos x}} d x $$

Short answer

Question: Evaluate the definite integral $$\int_{0}^{\pi} \frac{\sin x}{\sqrt{1+\cos x}} dx$$ Answer: $$2\sqrt{2}$$

Step-by-step solution

Identify a suitable substitution

Let's use the substitution $$u = \cos x$$. The derivative of u with respect to x is then $$du = -\sin x \, dx$$. This will help simplify the integral because $$-\sin x \,dx$$ is present in the given integral.

Compute the new limits of integration

Since we made the substitution $$u = \cos x$$, we need to change the limits of integration in terms of u: For the lower limit, when $$x = 0$$, we have $$u = \cos 0 = 1$$. For the upper limit, when $$x = \pi$$, we have $$u = \cos \pi = -1$$. The new limits of integration are from 1 to -1.

Rewrite the integral in terms of u

Applying the substitution $$u = \cos x$$ and $$du = -\sin x \, dx$$, we can rewrite the integral as: $$ \int_{0}^{\pi} \frac{\sin x}{\sqrt{1+\cos x}} d x = -\int_{1}^{-1} \frac{1}{\sqrt{1+u}} du $$

Evaluate the new integral

Now we need to evaluate the integral $$-\int_{1}^{-1} \frac{1}{\sqrt{1+u}} du$$. First, we compute the antiderivative of $$\frac{1}{\sqrt{1+u}}$$: Let $$v = 1 + u$$. Then, $$dv = du$$, and the integral becomes $$ -\int \frac{1}{\sqrt{v}} dv $$ The antiderivative of $$-\frac{1}{\sqrt{v}}$$ is $$-2\sqrt{v}$$. Now, replace v with the original expression, $$1 + u$$: $$ -2\sqrt{1+u} $$ Now, we need to apply the limits of integration: $$ -2\sqrt{1+(-1)} - (-2\sqrt{1+1}) = -2\sqrt{0} + 2\sqrt{2} = 2\sqrt{2} $$ Thus, the value of the definite integral is: $$ \int_{0}^{\pi} \frac{\sin x}{\sqrt{1+\cos x}} d x = 2\sqrt{2} $$

Key concepts

U-Substitution

The method of **U-Substitution** is a powerful technique to simplify the process of integration, primarily used in calculus to find the integral of complex functions. It's similar to the reverse of the chain rule used in differentiation. The main goal is to transform a difficult integral into a simpler one, making it easier to evaluate.

To use U-Substitution, we choose a new variable, often called "u," to substitute part of the original function. The chosen "u" should simplify the integrand (the function being integrated). In our example, we selected \( u = \cos x \). This choice was strategic because it pairs well with the derivative \( du = -\sin x \, dx \), which exists within the original integral.

• Identify a part of the integrand that can be substituted as "u."
• Differential of "u," denoted as "du," should match some part of the integrand to facilitate substitution.
• Rewrite the entire integral in terms of "u." The original limits also need to be converted.

By substituting and changing both the function and differential, we transform the integral into one that is usually simpler to solve.

Limits of Integration

When applying U-Substitution in definite integrals, the **Limits of Integration** must also be converted to reflect the new variable. This is a crucial step and ensures that the new integral evaluates properly.

In the given problem, after deciding on \( u = \cos x \), the next step was to translate the limits of integration from \( x \)-terms into \( u \)-terms.

• Original lower limit for \( x \) is 0. Substituting \( x = 0 \) into \( u = \cos x \), we get the new lower limit as \( u = 1 \).
• Original upper limit for \( x \) is \( \pi \). Substituting \( x = \pi \) gives the new upper limit as \( u = -1 \).

These new limits are critical, as they determine the interval over which we compute the integral in terms of the newly substituted variable "u." This re-evaluation of limits ensures that the final computation reflects the original problem's domain and range accurately.

Antiderivative Calculation

With the integral rewritten in terms of the new variable "u," the next step is to find the **Antiderivative**. The antiderivative, often referred to also as the indefinite integral, is the reverse of differentiation. It involves finding a function whose derivative matches the given integrand.

For the converted integral \(-\int \frac{1}{\sqrt{1+u}} du\), we seek the antiderivative of \( \frac{1}{\sqrt{1+u}} \).

• Make another substitution if needed, such as letting \( v = 1 + u \), transforming the function again to \( -\int \frac{1}{\sqrt{v}} dv \).
• The antiderivative of \( \frac{1}{\sqrt{v}} \) is \( 2\sqrt{v} \).

Hence, the antiderivative of \( \frac{1}{\sqrt{1+u}} \) now becomes \(-2\sqrt{1+u}\). Converting it back to our original terms ensures the solution aligns with the initial substitution.

Integral Evaluation Steps

The final stage in solving a definite integral is the **Evaluation Steps**, where we apply what we've resolved to calculate the definite value of the integral. These steps follow finding the antiderivative when applied to the specified interval of integration.

In our example, having derived the antiderivative as \(-2\sqrt{1+u}\), we fulfill the integral evaluation by substituting back using the revised limits of integration from the U-Substitution stage:

• First, compute with the upper limit. Substituting \( u = -1 \) gives \( -2\sqrt{0} = 0 \).
• Then, compute with the lower limit. Substituting \( u = 1 \) results in \( -2\sqrt{2} \).

Apply the definite integral formula: substituting the results and finding the difference with correct signs, we achieve \( 2\sqrt{2} \). Hence, the evaluated integral provides the exact area under the curve as described by the original problem statement.