∫ √sin 𝜃 𝑑𝜃

To solve the integral ∫ √sin 𝜃 𝑑𝜃, we can use a trigonometric substitution. Let's substitute sin 𝜃 with u^2, where u = sin(𝜃).

First, let's find the derivative of u with respect to 𝜃. We have du/d𝜃 = cos(𝜃).

Next, we need to express 𝑑𝜃 in terms of du. Rearranging the equation, we have 𝑑𝜃 = du/cos(𝜃).

Now, let's substitute sin 𝜃 with u^2 and 𝑑𝜃 with du/cos(𝜃) in the integral. The integral becomes:

∫ √sin 𝜃 𝑑𝜃 = ∫ √(u^2) (du/cos(𝜃))

Simplifying, we have:

∫ u du/cos(𝜃) = ∫ u du/cos(𝜃)

Since u = sin(𝜃), we can rewrite cos(𝜃) as √(1 - u^2) using the Pythagorean identity.

∫ u du/√(1 - u^2)

Now, let's make a substitution by letting v = 1 - u^2. Taking the derivative of v with respect to u, we have dv/du = -2u.

Rearranging the equation, we have du = -dv/(2u).

Substituting du and u in the integral, we get:

∫ u du/√(1 - u^2) = ∫ (-v/(2u)) du/√v

Simplifying, we have:

∫ (-v/(2u)) du/√v = -1/2 ∫ (v/u) du/√v

Now, let's integrate with respect to u:

-1/2 ∫ (v/u) du/√v = -1/2 ∫ (v/u) (1/√v) du

Simplifying further, we have:

-1/2 ∫ (v/u) (1/√v) du = -1/2 ∫ (1/√u) du

Integrating, we get:

-1/2 ∫ (1/√u) du = -1/2 * 2√u + C

Finally, substituting back u = sin(𝜃), we have:

-1/2 * 2√(sin(𝜃)) + C

Simplifying, we get:

-√(sin(𝜃)) + C

Therefore, the solution to the integral ∫ √sin 𝜃 𝑑𝜃 is -√(sin(𝜃)) + C, where C is the constant of integration.