Question Statement

Evaluate the definite integral:

${\int }_0^{\frac{\pi }{2}}\frac{\sin x}{(2+\cos x)(5+\cos x)}dx$

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Background and Explanation

This problem combines integration by substitution with partial fraction decomposition. The presence of $\sin x$ in the numerator and $\cos x$ in the denominator suggests substituting $t=\cos x$, while the product of linear factors in the denominator calls for breaking the rational function into simpler fractions.

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Solution

Let $I$ denote the integral:

$I={\int }_0^{\frac{\pi }{2}}\frac{\sin x}{(2+\cos x)(5+\cos x)}dx$

Step 1: Substitution

Put $\cos x=t$. Differentiating gives $-\sin xdx=dt$, or equivalently $\sin xdx=-dt$.

Change the limits of integration:

• When $x=0$, $t=\cos (0)=1$
• When $x=\frac{\pi }{2}$, $t=\cos \left(\frac{\pi }{2}\right)=0$

Substituting into the integral:

$I={\int }_1^0\frac{-dt}{(2+t)(5+t)}$

Using the property ${\int }_a^{b}f(x)dx=-{\int }_b^{a}f(x)dx$ to reverse the limits and cancel the negative signs:

$I={\int }_0^1\frac{1}{(2+t)(5+t)}dt$

Step 2: Partial Fraction Decomposition

We express the integrand as partial fractions:

$\frac{1}{(2+t)(5+t)}=\frac{A}{2+t}+\frac{B}{5+t}$

Multiplying both sides by $(2+t)(5+t)$:

$1=A(5+t)+B(2+t)$

Finding $A$: Put $t=-2$: $1=A(5-2)+B(0)\Rightarrow 1=3A\Rightarrow A=\frac{1}{3}$

Finding $B$: Put $t=-5$: $1=A(0)+B(2-5)\Rightarrow 1=-3B\Rightarrow B=-\frac{1}{3}$

Substituting back:

$\frac{1}{(2+t)(5+t)}=\frac{1}{3(2+t)}-\frac{1}{3(5+t)}$

Step 3: Integration

Now the integral becomes:

$I={\int }_0^1\left(\frac{1}{3(2+t)}-\frac{1}{3(5+t)}\right)dt$

Split and factor out constants:

$I=\frac{1}{3}{\int }_0^1\frac{1}{2+t}dt-\frac{1}{3}{\int }_0^1\frac{1}{5+t}dt$

Integrate using $\int \frac{1}{u}du=\ln |u|$:

$I=\frac{1}{3}{\left[\ln |2+t|\right]}_0^1-\frac{1}{3}{\left[\ln |5+t|\right]}_0^1$

Step 4: Evaluation and Simplification

Apply the limits $t=1$ and $t=0$:

$I=\frac{1}{3}[\ln (3)-\ln (2)]-\frac{1}{3}[\ln (6)-\ln (5)]$

Factor out $\frac{1}{3}$ and rearrange terms:

$I=\frac{1}{3}\left[\ln 3-\ln 2-\ln 6+\ln 5\right]$

Combine using logarithm properties $\ln a+\ln b=\ln (ab)$ and $\ln a-\ln b=\ln \left(\frac{a}{b}\right)$:

$I=\frac{1}{3}\left[\ln \left(\frac{3\times 5}{2\times 6}\right)\right]=\frac{1}{3}\left[\ln \left(\frac{15}{12}\right)\right]=\frac{1}{3}\ln \left(\frac{5}{4}\right)$

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Key Formulas or Methods Used

• Substitution Rule: $\int f(g(x))g^{\prime }(x)dx=\int f(u)du$ where $u=g(x)$
• Partial Fractions (distinct linear factors): $\frac{1}{(x+a)(x+b)}=\frac{1}{b-a}\left(\frac{1}{x+a}-\frac{1}{x+b}\right)$
• Logarithmic Integration: $\int \frac{1}{x}dx=\ln |x|+C$
• Definite Integral Property: ${\int }_a^{b}f(x)dx=-{\int }_b^{a}f(x)dx$
• Logarithm Properties: $\ln a-\ln b=\ln \left(\frac{a}{b}\right)$ and $\ln a+\ln b=\ln (ab)$

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Summary of Steps

1. Substitute $\cos x=t$ with $\sin xdx=-dt$, changing limits from $[0,\frac{\pi }{2}]$ to $[1,0]$
2. Reverse limits to eliminate the negative sign, obtaining ${\int }_0^1\frac{dt}{(2+t)(5+t)}$
3. Decompose using partial fractions: $\frac{1}{(2+t)(5+t)}=\frac{1}{3(2+t)}-\frac{1}{3(5+t)}$
4. Integrate term by term to get $\frac{1}{3}[\ln (2+t){]}_0^1-\frac{1}{3}[\ln (5+t){]}_0^1$
5. Evaluate at bounds: $\frac{1}{3}[(\ln 3-\ln 2)-(\ln 6-\ln 5)]$
6. Simplify logarithms to final answer: $\frac{1}{3}\ln \left(\frac{5}{4}\right)$