Source: Extra Practice

Evaluate the definite integral: 0π/2sinθcosθdθ\int_{0}^{\pi/2} \sin\theta \cos\theta \, d\theta.

Options

Option A is correct

1/2

Option B

1

Option C

0

Option D

1/4

Explanation

To solve this definite integral, we can use the substitution method. Let u=sinθu = \sin\theta. Then, the differential is: du=cosθdθdu = \cos\theta \, d\theta

Now, we change the limits of integration according to our substitution:

  • When θ=0\theta = 0, u=sin(0)=0u = \sin(0) = 0
  • When θ=π/2\theta = \pi/2, u=sin(π/2)=1u = \sin(\pi/2) = 1

Substituting these values into the integral: 01udu=[u22]01\int_{0}^{1} u \, du = \left[ \frac{u^2}{2} \right]_{0}^{1} [122022]=12\left[ \frac{1^2}{2} - \frac{0^2}{2} \right] = \frac{1}{2}

Alternative Method: Using double-angle identity sinθcosθ=12sin(2θ)\sin\theta \cos\theta = \frac{1}{2}\sin(2\theta): 0π/212sin(2θ)dθ=12[cos(2θ)2]0π/2=14[cos(π)cos(0)]=14[11]=12\int_{0}^{\pi/2} \frac{1}{2}\sin(2\theta) \, d\theta = \frac{1}{2} \left[ -\frac{\cos(2\theta)}{2} \right]_{0}^{\pi/2} = -\frac{1}{4} [\cos(\pi) - \cos(0)] = -\frac{1}{4} [-1 - 1] = \frac{1}{2}

  • Option A is correct.
  • Option B is incorrect as it neglects the factor of 1/21/2 from the integration of uu.
  • Option C is incorrect and typically arises from a sign error in evaluating cosine values.
  • Option D is incorrect and comes from double-counting the division factor in the trigonometric identity method.