Question ID: #746
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 $\text{gcd}(p_{i},q_{i})=1$ for $i=1,2,3,4$ and $C$ is the constant of integration, then $\frac{15p_{1}p_{2}p_{3}p_{4}}{q_{1}q_{2}q_{3}q_{4}}$ is equal to
Solution:
Let $I = \int (\sin x)^{-11/2} (\cos x)^{-5/2} dx$.
$$ I = \int \frac{1}{(\sin x)^{11/2} (\cos x)^{5/2}} dx $$
Divide numerator and denominator by $\cos^{11/2} x$ to convert $\sin x$ to $\tan x$.
$$ I = \int \frac{1}{\frac{(\sin x)^{11/2}}{(\cos x)^{11/2}} \cdot (\cos x)^{5/2} \cdot (\cos x)^{11/2}} dx $$
$$ I = \int \frac{1}{(\tan x)^{11/2} \cdot (\cos x)^{8}} dx $$
$$ I = \int (\tan x)^{-11/2} \sec^8 x dx $$
$$ I = \int (\tan x)^{-11/2} (1 + \tan^2 x)^3 \sec^2 x dx $$
Let $\tan x = t \Rightarrow \sec^2 x dx = dt$.
$$ I = \int t^{-11/2} (1 + t^2)^3 dt $$
$$ I = \int t^{-11/2} (1 + 3t^2 + 3t^4 + t^6) dt $$
$$ I = \int (t^{-11/2} + 3t^{-7/2} + 3t^{-3/2} + t^{1/2}) dt $$
Integrate using power rule $\int x^n dx = \frac{x^{n+1}}{n+1}$:
$$ I = \frac{t^{-9/2}}{-9/2} + 3\frac{t^{-5/2}}{-5/2} + 3\frac{t^{-1/2}}{-1/2} + \frac{t^{3/2}}{3/2} + C $$
$$ I = -\frac{2}{9}t^{-9/2} – \frac{6}{5}t^{-5/2} – 6t^{-1/2} + \frac{2}{3}t^{3/2} + C $$
Substitute $t = \tan x = \frac{1}{\cot x}$:
$$ t^{-9/2} = (\cot x)^{9/2} $$
$$ t^{-5/2} = (\cot x)^{5/2} $$
$$ t^{-1/2} = (\cot x)^{1/2} $$
$$ t^{3/2} = (\tan x)^{3/2} = (\cot x)^{-3/2} $$
$$ I = -\frac{2}{9}(\cot x)^{9/2} – \frac{6}{5}(\cot x)^{5/2} – \frac{6}{1}(\cot x)^{1/2} + \frac{2}{3}(\cot x)^{-3/2} + C $$
Comparing with the given expression:
$p_1 = 2, q_1 = 9$
$p_2 = 6, q_2 = 5$
$p_3 = 6, q_3 = 1$
$p_4 = 2, q_4 = 3$
Calculate the required value:
$$ \text{Value} = \frac{15 p_1 p_2 p_3 p_4}{q_1 q_2 q_3 q_4} $$
$$ = \frac{15 (2)(6)(6)(2)}{(9)(5)(1)(3)} $$
$$ = \frac{15 \cdot 144}{135} $$
$$ = \frac{15}{135} \cdot 144 = \frac{1}{9} \cdot 144 = 16 $$
Ans. (16)
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