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Question

If $\int (\sin x)^{\frac{-11}{2}} (\cos x)^{\frac{-5}{2}} dx = - \frac{p_1}{q_1} (\cot x)^{\frac{9}{2}} - \frac{p_2}{q_2} (\cot x)^{\frac{5}{2}} - \frac{p_3}{q_3} (\cot x)^{\frac{1}{2}} + \frac{p_4}{q_4} (\cot x)^{\frac{-3}{2}} + C$, where $p_i$ and $q_i$ are positive integers with $\gcd(p_i, q_i) = 1$ for $i = 1, 2, 3, 4$ and C is the constant of integration, then $\frac{15 p_1 p_2 p_3 p_4}{q_1 q_2 q_3 q_4}$ is equal to _________

Integral Rewriting and Substitution

The integral to solve is $I = \int (\sin x)^{\frac{-11}{2}} (\cos x)^{\frac{-5}{2}} dx$. First, rewrite the integrand:

$ (\sin x)^{\frac{-11}{2}} (\cos x)^{\frac{-5}{2}} = \frac{\cos^{\frac{-5}{2}} x}{\sin^{\frac{11}{2}} x} $ $ = \frac{\cos^{\frac{-5}{2}} x}{\sin^{\frac{-5}{2}} x} \cdot \frac{1}{\sin^{\frac{11}{2}} x \sin^{\frac{5}{2}} x} = (\cot x)^{\frac{-5}{2}} \csc^6 x $ Using the identity $\csc^2 x = 1 + \cot^2 x$, we get: $ (\cot x)^{\frac{-5}{2}} (\csc^2 x)^3 = (\cot x)^{\frac{-5}{2}} (1 + \cot^2 x)^3 $ Let $u = \cot x$. Then the differential is $du = -\csc^2 x dx$. Substituting $u$ into the integral expression, we get the integrand in terms of $u$ and $du$: $ I = \int - u^{\frac{-5}{2}} (1 + u^2)^3 du $

Integration Calculation

Expand the term $(1 + u^2)^3 = 1 + 3u^2 + 3u^4 + u^6$. The integral becomes:

$ I = - \int u^{\frac{-5}{2}} (1 + 3u^2 + 3u^4 + u^6) du $ $ I = - \int (u^{\frac{-5}{2}} + 3u^{\frac{-1}{2}} + 3u^{\frac{3}{2}} + u^{\frac{7}{2}}) du $ Integrate term by term:

$ I = - \left[ \frac{u^{\frac{-5}{2}+1}}{\frac{-5}{2}+1} + 3 \frac{u^{\frac{-1}{2}+1}}{\frac{-1}{2}+1} + 3 \frac{u^{\frac{3}{2}+1}}{\frac{3}{2}+1} + \frac{u^{\frac{7}{2}+1}}{\frac{7}{2}+1} \right] + C $ $ I = - \left[ \frac{u^{\frac{-3}{2}}}{-\frac{3}{2}} + 3 \frac{u^{\frac{1}{2}}}{\frac{1}{2}} + 3 \frac{u^{\frac{5}{2}}}{\frac{5}{2}} + \frac{u^{\frac{9}{2}}}{\frac{9}{2}} \right] + C $ $ I = - \left[ -\frac{2}{3} u^{\frac{-3}{2}} + 6 u^{\frac{1}{2}} + \frac{6}{5} u^{\frac{5}{2}} + \frac{2}{9} u^{\frac{9}{2}} \right] + C $ $ I = \frac{2}{3} u^{\frac{-3}{2}} - 6 u^{\frac{1}{2}} - \frac{6}{5} u^{\frac{5}{2}} - \frac{2}{9} u^{\frac{9}{2}} + C $ Substitute back $u = \cot x$:

$ I = \frac{2}{3} (\cot x)^{\frac{-3}{2}} - 6 (\cot x)^{\frac{1}{2}} - \frac{6}{5} (\cot x)^{\frac{5}{2}} - \frac{2}{9} (\cot x)^{\frac{9}{2}} + C $

Coefficient Identification

The calculated integral is: $ - \frac{2}{9} (\cot x)^{\frac{9}{2}} - \frac{6}{5} (\cot x)^{\frac{5}{2}} - 6 (\cot x)^{\frac{1}{2}} + \frac{2}{3} (\cot x)^{\frac{-3}{2}} + C $ The given form is: $ - \frac{p_1}{q_1} (\cot x)^{\frac{9}{2}} - \frac{p_2}{q_2} (\cot x)^{\frac{5}{2}} - \frac{p_3}{q_3} (\cot x)^{\frac{1}{2}} + \frac{p_4}{q_4} (\cot x)^{\frac{-3}{2}} + C $ Comparing the coefficients for each power of $(\cot x)$, and ensuring $p_i, q_i$ are positive coprime integers:

  • For $(\cot x)^{\frac{9}{2}}$: $-\frac{p_1}{q_1} = -\frac{2}{9} \implies \frac{p_1}{q_1} = \frac{2}{9}$. So, $p_1=2, q_1=9$.
  • For $(\cot x)^{\frac{5}{2}}$: $-\frac{p_2}{q_2} = -\frac{6}{5} \implies \frac{p_2}{q_2} = \frac{6}{5}$. So, $p_2=6, q_2=5$.
  • For $(\cot x)^{\frac{1}{2}}$: $-\frac{p_3}{q_3} = -6 \implies \frac{p_3}{q_3} = 6$. So, $p_3=6, q_3=1$.
  • For $(\cot x)^{\frac{-3}{2}}$: $\frac{p_4}{q_4} = \frac{2}{3}$. So, $p_4=2, q_4=3$.

Final Expression Evaluation

We need to calculate $\frac{15 p_1 p_2 p_3 p_4}{q_1 q_2 q_3 q_4}$. Substitute the values:

$ p_1=2, q_1=9 $ $ p_2=6, q_2=5 $ $ p_3=6, q_3=1 $ $ p_4=2, q_4=3 $ $ \frac{15 p_1 p_2 p_3 p_4}{q_1 q_2 q_3 q_4} = \frac{15 \times (2 \times 6 \times 6 \times 2)}{9 \times 5 \times 1 \times 3} $ $ = \frac{15 \times 144}{135} $ $ = \frac{2160}{135} $ Simplify the fraction:

$ \frac{2160 \div 135}{135 \div 135} = 16 $

The value is 16.

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