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The integral is computed by substituting the Maclaurin series for $\cos x$, using the DCT to interchange the sum and integral, integrating each monomial term using the power rule, simplifying the factorial denominator, recognizing the resulting series as the sine series, and evaluating it at the endpoints $0$ and $\pi/2$.

Conditions: The integration interval is $[0,\pi/2]$.; The cosine power series is $\cos x = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!} x^{2n}$.; The DCT is applicable because the series terms are dominated by $g(x)=1$ on $[0,\pi/2]$.

The integral is computed by substituting the Maclaurin series for $\cos x$, using the DCT to interchange the sum and integral, integrating each monomial term using the power rule, simplifying the factorial denominator, recognizing the resulting series as the sine series, and evaluating it at the endpoints $0$ and $\pi/2$.

Conditions: The integration interval is $[0,\pi/2]$.; The cosine power series is $\cos x = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!} x^{2n}$.; The DCT is applicable because the series terms are dominated by $g(x)=1$ on $[0,\pi/2]$.

The final value of the integral is $1$. This is obtained by evaluating the recognized sine series at the endpoints of the interval: $\sin(\pi/2) - \sin(0) = 1 - 0 = 1$.

Conditions: The integration interval is $[0,\pi/2]$.; The integrated series has been correctly identified as the Maclaurin series for $\sin x$.; The Fundamental Theorem of Calculus is applied to evaluate the definite integral.

The final value of the integral is $1$. This is obtained by evaluating the recognized sine series at the endpoints of the interval: $\sin(\pi/2) - \sin(0) = 1 - 0 = 1$.

Conditions: The integration interval is $[0,\pi/2]$.; The integrated series has been correctly identified as the Maclaurin series for $\sin x$.; The Fundamental Theorem of Calculus is applied to evaluate the definite integral.

The $(2n+1)!$ denominator comes from combining the original factorial $(2n)!$ from the cosine series coefficient with the new factor $2n+1$ introduced by the power rule of integration. Algebraically, $(2n)! \times (2n+1) = (2n+1)!$.

Conditions: The integration is performed on the term $x^{2n}$.; The original coefficient contains $(2n)!$ in the denominator.; The power rule is applied, raising the exponent to $2n+1$ and dividing by $2n+1$.

The $(2n+1)!$ denominator comes from combining the original factorial $(2n)!$ from the cosine series coefficient with the new factor $2n+1$ introduced by the power rule of integration. Algebraically, $(2n)! \times (2n+1) = (2n+1)!$.

Conditions: The integration is performed on the term $x^{2n}$.; The original coefficient contains $(2n)!$ in the denominator.; The power rule is applied, raising the exponent to $2n+1$ and dividing by $2n+1$.

The sum and integral can be interchanged because the Dominated Convergence Theorem applies. The terms of the cosine power series are bounded in magnitude by $1$ on the interval $[0,\pi/2]$, making the constant function $g(x)=1$ a valid dominating function with a finite integral over the interval.

Conditions: The integration interval is $[0,\pi/2]$.; The series terms are $\frac{(-1)^n}{(2n)!}x^{2n}$.; The dominating function is $g(x)=1$.

The sum and integral can be interchanged because the Dominated Convergence Theorem applies. The terms of the cosine power series are bounded in magnitude by $1$ on the interval $[0,\pi/2]$, making the constant function $g(x)=1$ a valid dominating function with a finite integral over the interval.

Conditions: The integration interval is $[0,\pi/2]$.; The series terms are $\frac{(-1)^n}{(2n)!}x^{2n}$.; The dominating function is $g(x)=1$.

The example is included to demonstrate that the domination hypothesis alone is insufficient if the dominating function's integral diverges. It serves as a warning that if the finite-integral condition is not met, the desired interchange of integral and sum may fail.

Conditions: The example is presented as a warning illustration on the interval $[0,\infty)$.; The function $e^{2x}$ dominates $e^x$, but $\int_0^{\infty} e^{2x}\,dx$ diverges.

The example is included to demonstrate that the domination hypothesis alone is insufficient if the dominating function's integral diverges. It serves as a warning that if the finite-integral condition is not met, the desired interchange of integral and sum may fail.

Conditions: The example is presented as a warning illustration on the interval $[0,\infty)$.; The function $e^{2x}$ dominates $e^x$, but $\int_0^{\infty} e^{2x}\,dx$ diverges.