Question

sin2 x/2 (0;п/2) как решить этот интеграл

Answer 1

$\int\limits^{ \pi /2}_0 {sin^{2} \frac{x}{2} } \, dx$
Для взятие этого интеграла понизим степень синуса по формуле:
$sin^{2} \frac{x}{2}= \frac{1}{2} (1-cosx)$

$\int\limits^{ \pi /2}_0 {sin^{2} \frac{x}{2} } \, dx=\frac{1}{2} \int\limits^{ \pi /2}_0 {(1-cosx)} \, dx=\frac{1}{2}( \int\limits^{ \pi /2}_0 {1} \, dx - \int\limits^{ \pi /2}_0 {cosx} \, dx)=$

$=\frac{1}{2}( x - sinx)|^{ \pi /2}_0 =\frac{1}{2}( \frac{ \pi }{2} -0 - (1-0))=\frac{1}{2}( \frac{ \pi }{2} -1)= \frac{ \pi }{4} -\frac{1}{2}$

Answer 2

Solving the Integral sin^2(x/2) from 0 to π/2

To solve the integral of sin^2(x/2) from 0 to π/2, we can use the trigonometric identity that relates sin^2(x) to 1 - cos(2x)/2. This will allow us to express sin^2(x/2) in terms of cos(2x) and then integrate with respect to x.

Using the trigonometric identity: sin^2(x) = (1 - cos(2x))/2

We can rewrite the integral as: ∫(1 - cos(x)) / 2 dx

Now, we can proceed to integrate this expression.

Integration Process

The integration of (1 - cos(x))/2 with respect to x involves finding the antiderivative of the function and evaluating it at the limits of integration.

The antiderivative of (1 - cos(x))/2 is: 1/2 * x - 1/2 * sin(x)

Now, we can evaluate this antiderivative at the limits of integration, which are 0 and π/2.

Evaluating the Antiderivative at the Limits of Integration

At x = π/2: 1/2 * (π/2) - 1/2 * sin(π/2) = π/4 - 1/2

At x = 0: 1/2 * 0 - 1/2 * sin(0) = 0

Final Result

Subtracting the value at the lower limit from the value at the upper limit gives us the value of the definite integral: π/4 - 1/2

Therefore, the value of the integral of sin^2(x/2) from 0 to π/2 is π/4 - 1/2.