Evaluate the integral: ${\int }_0^{\frac{\pi }{2}}\frac{\cos \left(t\right)}{1+{\sin }^t^2}dt$

• A. $\frac{1}{2}$
• B. $1$
• C. $0$
• D. $\frac{\pi }{4}$

Answer: (D)

Verified step by step guidance

1. Step 1: Recognize that the integral involves a trigonometric function and its derivative. The numerator, cos(t), is the derivative of sin(t), which appears in the denominator. This suggests a substitution method might simplify the integral.
2. Step 2: Let u = sin(t). Then, du = cos(t) dt. This substitution transforms the integral into ∫₀^{π/2} rac{ ext{cos}(t)}{1 + ext{sin}^2(t)} ext{d}t → ∫₀^{1} rac{1}{1 + u^2} ext{d}u, where the limits of integration change because sin(0) = 0 and sin(π/2) = 1.
3. Step 3: Recognize that the integral ∫ rac{1}{1 + u^2} ext{d}u is a standard form that evaluates to arctan(u). This is a result from the derivative of arctan(u), which is rac{1}{1 + u^2}.
4. Step 4: Apply the antiderivative formula to the integral: ∫₀^{1} rac{1}{1 + u^2} ext{d}u = [ ext{arctan}(u)]₀^{1}.
5. Step 5: Evaluate the definite integral by substituting the limits of integration into the antiderivative: [ ext{arctan}(1) - ext{arctan}(0)]. Recall that arctan(1) = π/4 and arctan(0) = 0, so the result simplifies to π/4.