Evaluate Definite Integrals

We calculate the value of each definite integral.

Integral A: $\int_0^\pi \frac{1}{1+\sin x} dx$

To evaluate this integral, we can multiply the numerator and denominator by $(1-\sin x)$:

$ \int_0^\pi \frac{1-\sin x}{1-\sin^2 x} dx = \int_0^\pi \frac{1-\sin x}{\cos^2 x} dx $

$ = \int_0^\pi (\sec^2 x - \sec x \tan x) dx $

$ = [\tan x - \sec x]_0^\pi $

This integral is improper at $x = \frac{\pi}{2}$. Using the substitution $t = \tan(x/2)$, we find the value:

$ \int_0^\infty \frac{2}{(1+t)^2} dt = 2 \left[ \frac{-1}{1+t} \right]_0^\infty = 2 (0 - (-1)) = 2 $

Value of A = 2

Integral B: $\int_1^2 x^2 dx$

Using the power rule for integration:

$ \int_1^2 x^2 dx = \left[ \frac{x^3}{3} \right]_1^2 $

$ = \frac{2^3}{3} - \frac{1^3}{3} = \frac{8}{3} - \frac{1}{3} = \frac{7}{3} $

Value of B = $\frac{7}{3} \approx 2.333$

Integral C: $\int_0^{\frac{\pi}{2}} \sin x dx$

The integral of $\sin x$ is $-\cos x$:

$ \int_0^{\frac{\pi}{2}} \sin x dx = [-\cos x]_0^{\frac{\pi}{2}} $

$ = -\cos\left(\frac{\pi}{2}\right) - (-\cos(0)) = -0 - (-1) = 1 $

Value of C = 1

Integral D: $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin^3 x dx$

The integrand $f(x) = \sin^3 x$ is an odd function because $f(-x) = \sin^3(-x) = (-\sin x)^3 = -\sin^3 x = -f(x)$.

The integral of an odd function over a symmetric interval $[-a, a]$ is zero.

Value of D = 0

Order Integral Values

We list the calculated values:

• D = 0
• C = 1
• A = 2
• B = $\frac{7}{3} \approx 2.333$

Arranging these values in ascending order:

0 < 1 < 2 < $\frac{7}{3}$

This corresponds to the order: D, C, A, B.