i need the antiderivative of sinxcosx dx for intergral pie/2 to 0 thanks.

sin(x)cos(x)dx = sin(x)d[sin(x)] =

1/2d[sin^2(x)]

To evaluate the antiderivative of sin(x)cos(x)dx, you can use the substitution method. Let's follow these steps:

Step 1: Apply the substitution u = sin(x).

Step 2: Calculate du by taking the derivative of u with respect to x. du/dx = cos(x).

Step 3: Rearrange the equation to solve for dx. dx = du/cos(x).

Step 4: Substitute u and dx into the integral.

∫sin(x)cos(x)dx = ∫u du/cos(x).

Step 5: Simplify the integral.

Taking out the constant 1/2 and using a trigonometric identity (cos^2(x) = 1 - sin^2(x)), the integral becomes:

(1/2) ∫ (u du) / (√(1 - u^2)).

Step 6: Integrate the new expression.

The integral (u du) / (√(1 - u^2)) is known as the arcsine or inverse sine function.

So we get:

(1/2) arcsin(u) + C,

where C is the constant of integration.

Step 7: Substitute back u = sin(x) into the expression.

(1/2) arcsin(sin(x)) + C.

Step 8: Simplify the final result.

The arcsine function, when applied to the sine function, is equal to x for -π/2 ≤ x ≤ π/2.

Therefore, the antiderivative of sin(x)cos(x)dx from π/2 to 0 is:

(1/2) [arcsin(sin(0)) - arcsin(sin(π/2))] = (1/2) [arcsin(0) - arcsin(1)].

arcsin(0) = 0 and arcsin(1) = π/2, so the final result is:

(1/2) [0 - π/2] = -π/4.

Hence, the value of the integral is -π/4.