If \(\int(\sin x)^{-11/2} (\cos x)^{-5/2} dx = \frac{p_1}{q_1}(\cot x)^{9/2} - \frac{p_2}{q_2}(\cot x)^{5/2} - \frac{p_3}{q_3}(\cot x)^{1/2} + \frac{p_4}{q_4}(\cot x)^{-3/2} + C\) where H.C.F. \(\{p_i, q_i\} = 1\) \& \(i \in \{1, 2, 3, 4\}\). Then value of \(\sum_{i=1}^{4}(p_i +q_i)\) is:
Show Hint
For integrals of the form \(\int\sin^m x \cos^n x \, dx\), if \(m+n\) is a negative even integer, the substitution \(t = \tan x\) or \(t = \cot x\) is a very effective strategy to simplify the integral.
Step 1: Understanding the Question:
We are asked to evaluate a complex trigonometric integral and then compare the resulting coefficients with a given form to find the sum of certain values.
Step 2: Key Formula or Approach:
The integral is \(I = \int(\sin x)^{-11/2} (\cos x)^{-5/2} dx\).
The sum of the powers is \(-11/2 - 5/2 = -16/2 = -8\), which is a negative even integer. In such cases, substitution with \(t = \tan x\) or \(t = \cot x\) is effective. Let's use \(t = \cot x\).
If \(t = \cot x\), then \(dt = -\csc^2 x \, dx\), which means \(dx = -\sin^2 x \, dt\).
We also need expressions for \(\sin x\) and \(\cos x\) in terms of \(t\):
From \(1+\cot^2 x = \csc^2 x\), we get \(\sin^2 x = \frac{1}{1+\cot^2 x} = \frac{1}{1+t^2}\).
From \(\tan^2 x = \frac{1}{\cot^2 x} = \frac{1}{t^2}\), we get \(\sec^2 x = 1+\tan^2 x = 1+\frac{1}{t^2} = \frac{t^2+1}{t^2}\), so \(\cos^2 x = \frac{t^2}{t^2+1}\).
Step 3: Detailed Explanation:
The integral is \(I = \int \sin^{-11/2}x \cos^{-5/2}x \, dx\).
Substitute \(dx = -\sin^2 x \, dt\):
\[ I = \int \sin^{-11/2}x \cos^{-5/2}x (-\sin^2 x) \, dt = -\int \sin^{-7/2}x \cos^{-5/2}x \, dt \]
Now substitute for \(\sin x\) and \(\cos x\) in terms of \(t\):
\(\sin x = (1+t^2)^{-1/2}\) and \(\cos x = t(1+t^2)^{-1/2}\).
\[ I = -\int ( (1+t^2)^{-1/2} )^{-7/2} ( t(1+t^2)^{-1/2} )^{-5/2} \, dt \]
\[ I = -\int (1+t^2)^{7/4} \cdot t^{-5/2} (1+t^2)^{5/4} \, dt \]
\[ I = -\int t^{-5/2} (1+t^2)^{7/4+5/4} \, dt = -\int t^{-5/2} (1+t^2)^{12/4} \, dt \]
\[ I = -\int t^{-5/2} (1+t^2)^3 \, dt \]
Expand \((1+t^2)^3 = 1 + 3t^2 + 3t^4 + t^6\).
\[ I = -\int t^{-5/2} (1 + 3t^2 + 3t^4 + t^6) \, dt \]
\[ I = -\int (t^{-5/2} + 3t^{-1/2} + 3t^{3/2} + t^{7/2}) \, dt \]
Now, integrate term by term using \(\int t^n \, dt = \frac{t^{n+1}}{n+1}\):
\[ I = - \left[ \frac{t^{-3/2}}{-3/2} + \frac{3t^{1/2}}{1/2} + \frac{3t^{5/2}}{5/2} + \frac{t^{9/2}}{9/2} \right] + C \]
\[ I = - \left[ -\frac{2}{3}t^{-3/2} + 6t^{1/2} + \frac{6}{5}t^{5/2} + \frac{2}{9}t^{9/2} \right] + C \]
\[ I = -\frac{2}{9}t^{9/2} - \frac{6}{5}t^{5/2} - 6t^{1/2} + \frac{2}{3}t^{-3/2} + C \]
Substitute back \(t = \cot x\):
\[ I = -\frac{2}{9}(\cot x)^{9/2} - \frac{6}{5}(\cot x)^{5/2} - 6(\cot x)^{1/2} + \frac{2}{3}(\cot x)^{-3/2} + C \]
Now, compare this with the given form: \(\frac{p_1}{q_1}(\cot x)^{9/2} - \frac{p_2}{q_2}(\cot x)^{5/2} - \frac{p_3}{q_3}(\cot x)^{1/2} + \frac{p_4}{q_4}(\cot x)^{-3/2} + C\).
Assuming the first term in the problem statement should have a negative sign, making it \(-\frac{p_1}{q_1}(\cot x)^{9/2}\), we compare coefficients:
\(-\frac{p_1}{q_1} = -\frac{2}{9} \Rightarrow p_1 = 2, q_1 = 9\) (HCF(2,9)=1)
\(-\frac{p_2}{q_2} = -\frac{6}{5} \Rightarrow p_2 = 6, q_2 = 5\) (HCF(6,5)=1)
\(-\frac{p_3}{q_3} = -6 \Rightarrow p_3 = 6, q_3 = 1\) (HCF(6,1)=1)
\(\frac{p_4}{q_4} = \frac{2}{3} \Rightarrow p_4 = 2, q_4 = 3\) (HCF(2,3)=1)
Now we calculate the required sum:
\[ \sum_{i=1}^{4}(p_i + q_i) = (p_1+q_1) + (p_2+q_2) + (p_3+q_3) + (p_4+q_4) \]
\[ = (2+9) + (6+5) + (6+1) + (2+3) \]
\[ = 11 + 11 + 7 + 5 = 34 \]
Step 4: Final Answer:
The value of the sum is 34.
