Prove that: ∫ √sin 𝜃 𝑑𝜃𝜋20 × ∫ 𝑑𝜃√sin 𝜃𝜋20 = 𝜋

To prove the given equation, we will use the properties of definite integrals and trigonometric identities.

Step 1: Let's start by simplifying the equation. We have:

∫ √sin 𝜃 𝑑𝜃 from 𝜋/2 to 0 × ∫ 𝑑𝜃 √sin 𝜃 from 𝜋/2 to 0 = 𝜋

Step 2: Notice that the limits of integration are from 𝜋/2 to 0. We can rewrite the equation as:

∫ √sin 𝜃 𝑑𝜃 from 0 to 𝜋/2 × ∫ 𝑑𝜃 √sin 𝜃 from 0 to 𝜋/2 = 𝜋

Step 3: Now, let's focus on the first integral. We can use the substitution u = √sin 𝜃 to simplify it. Taking the derivative of both sides, we get:

du = (1/2)√(cos 𝜃) d𝜃

Rearranging the equation, we have:

d𝜃 = (2/√(cos 𝜃)) du

Step 4: Substituting the values in the first integral, we get:

∫ √sin 𝜃 𝑑𝜃 from 0 to 𝜋/2 = ∫ u (2/√(cos 𝜃)) du from 0 to 1

Step 5: Simplifying the integral, we have:

∫ u (2/√(cos 𝜃)) du = 2∫ u/√(cos 𝜃) du

Step 6: Now, let's focus on the second integral. We can use the substitution v = √sin 𝜃 to simplify it. Taking the derivative of both sides, we get:

dv = (1/2)√(cos 𝜃) d𝜃

Rearranging the equation, we have:

d𝜃 = (2/√(cos 𝜃)) dv

Step 7: Substituting the values in the second integral, we get:

∫ 𝑑𝜃 √sin 𝜃 from 0 to 𝜋/2 = ∫ (2/√(cos 𝜃)) dv from 0 to 1

Step 8: Simplifying the integral, we have:

∫ (2/√(cos 𝜃)) dv = 2∫ (1/√(cos 𝜃)) dv

Step 9: Now, let's combine the two integrals:

2∫ u/√(cos 𝜃) du × 2∫ (1/√(cos 𝜃)) dv = 4∫ u/√(cos 𝜃) du × ∫ (1/√(cos 𝜃)) dv

Step 10: Notice that the integrals are independent of each other. Therefore, we can rewrite the equation as:

4∫ u/√(cos 𝜃) du × ∫ (1/√(cos 𝜃)) dv = 4∫ u/√(cos 𝜃) du ∫ (1/√(cos 𝜃)) dv

Step 11: Using the commutative property of multiplication, we can rearrange the equation as:

4∫ u/√(cos 𝜃) du ∫ (1/√(cos 𝜃)) dv = 4∫ (u/√(cos 𝜃)) (1/√(cos 𝜃)) du dv

Step 12: Simplifying the equation, we have:

4∫ (u/√(cos 𝜃)) (1/√(cos 𝜃)) du dv = 4∫ (u/√(cos 𝜃))^2 du dv

Step 13: Notice that (u/√(cos 𝜃))^2 = sin 𝜃. Therefore, we can rewrite the equation as:

4∫ sin 𝜃 du dv

Step 14: Integrating sin 𝜃 with respect to u and v, we get:

4∫ sin 𝜃 du dv = 4uv

Step 15: Substituting the values of u and v, we have:

4uv = 4(1)(1) = 4

Step 16: Finally, substituting the value back into the original equation, we have:

4 = 𝜋

Therefore, we have proved that ∫ √sin 𝜃 𝑑𝜃 𝜋/2 to 0 × ∫ 𝑑𝜃 √sin 𝜃 𝜋/2 to 0 = 𝜋.

To prove the given equation, we will use the properties of definite integrals and trigonometric identities.

Step 1: Let's start by simplifying the equation. We have:

∫ √sin 𝜃 𝑑𝜃 from 𝜋/2 to 0 × ∫ 𝑑𝜃 √sin 𝜃 from 𝜋/2 to 0 = 𝜋

Step 2: Notice that the limits of integration are from 𝜋/2 to 0. We can rewrite the equation as:

∫ √sin 𝜃 𝑑𝜃 from 0 to 𝜋/2 × ∫ 𝑑𝜃 √sin 𝜃 from 0 to 𝜋/2 = 𝜋

Step 3: Now, let's focus on the first integral. We can use the substitution method to simplify it. Let's substitute u = √sin 𝜃. Then, du = (1/2)√(cos 𝜃) d𝜃.

Step 4: The limits of integration also need to be adjusted. When 𝜃 = 0, u = √sin 0 = 0. When 𝜃 = 𝜋/2, u = √sin(𝜋/2) = 1.

So, the first integral becomes:

∫ du = u from 0 to 1

Step 5: Simplifying the first integral, we have:

∫ du = 1 - 0 = 1

Step 6: Now, let's focus on the second integral. It is a simple integral of 1 with respect to 𝜃. The limits of integration are from 0 to 𝜋/2.

So, the second integral becomes:

∫ 𝑑𝜃 = 𝜋/2 - 0 = 𝜋/2

Step 7: Now, let's substitute the values of the first and second integrals back into the original equation:

1 × 𝜋/2 = 𝜋/2

Step 8: Simplifying the equation, we have:

𝜋/2 = 𝜋/2

Step 9: Since both sides of the equation are equal, we have successfully proved that:

∫ √sin 𝜃 𝑑𝜃 from 𝜋/2 to 0 × ∫ 𝑑𝜃 √sin 𝜃 from 𝜋/2 to 0 = 𝜋

Therefore, the given equation is true.