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Evaluate $\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \frac{\mathrm{~d} x}{\sin ^{2} x...

Oct 17, 2024

Solution by Steps

step 1

We start with the integral to evaluate:

I = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \frac{\mathrm{d} x}{\sin^2 x (\sin^2 x + 1)(\sin^2 x + 2)}

step 2

To simplify the integrand, we can use the substitution

u = \sin^2 x

, \mathrm{d}u = 2\sin x \cos x \mathrm{d}x

. The limits change accordingly: when

x = \frac{\pi}{4}

,

u = \frac{1}{2}

; when

x = \frac{\pi}{2}

,

u = 1

. Thus, the integral becomes:

I = \frac{1}{2} \int_{\frac{1}{2}}^{1} \frac{\mathrm{d}u}{u(u + 1)(u + 2)}$

step 3

We can perform partial fraction decomposition on the integrand:

\frac{1}{u(u + 1)(u + 2)} = \frac{A}{u} + \frac{B}{u + 1} + \frac{C}{u + 2}

. Solving for

A

B

, and

C

gives us:

A = 1

B = -2

C = 1

. Thus, we rewrite the integral as:

I = \frac{1}{2} \int_{\frac{1}{2}}^{1} \left( \frac{1}{u} - \frac{2}{u + 1} + \frac{1}{u + 2} \right) \mathrm{d}u

step 4

Evaluating the integral, we have:

I = \frac{1}{2} \left[ \ln |u| - 2\ln |u + 1| + \ln |u + 2| \right]_{\frac{1}{2}}^{1}

. Plugging in the limits gives us:

I = \frac{1}{2} \left[ \ln(1) - 2\ln(2) + \ln(3) - \left( \ln(\frac{1}{2}) - 2\ln(\frac{3}{2}) + \ln(2) \right) \right]

step 5

Simplifying the expression results in:

I = \frac{1}{2} \left[ 0 - 2\ln(2) + \ln(3) + \ln(2) - 2\ln(\frac{3}{2}) \right]

. Further simplification leads to:

I = \frac{1}{2} \left[ -\ln(2) + \ln(3) - 2(\ln(3) - \ln(2)) \right]

step 6

Finally, we combine the logarithmic terms to find the value of the integral:

I = \frac{1}{2} \left[ -\ln(2) + \ln(3) - 2\ln(3) + 2\ln(2) \right] = \frac{1}{2} \left[ \ln(2) - \ln(3) \right] = \frac{1}{2} \ln\left(\frac{2}{3}\right)

Answer

\frac{1}{2} \ln\left(\frac{2}{3}\right)

Key Concept

Evaluating definite integrals using substitution and partial fractions

Explanation

The integral was simplified using a substitution and then evaluated using partial fraction decomposition, leading to the final result.