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.
Similar Questions
- Let the solution curve of the differential equation $x dy - y dx = \sqrt{x^2 + y^2} dx, x > 0$, $y(1) = 0$, be $y = y(x)$. Then $y(3)$ is equal to
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$16(\sqrt{x+ 9\sqrt{x}})(4 + \sqrt{9 + \sqrt{x}}) \cos y \, dy = (1 + 2 \sin y) dx, x > 0$ and $y(256) = \frac{\pi}{2}, y(49) = \alpha$, then $2 \sin \alpha$ is equal to :- Let a differentiable function $f$ satisfy the equation $\int_{0}^{36} f\left(\frac{tx}{36}\right) dt = 4\alpha f(x)$. If $y = f(x)$ is a standard parabola passing through the points (2, 1) and (– 4, $\beta$), then $\beta^\alpha$ is equal to ______.
- Let $f(\alpha)$ denote the area of the region in the first quadrant bounded by $x = 0, x = 1, y^2 = x$ and $y = |\alpha x - 5| - |1 - \alpha x| + \alpha x^2$. Then $(f(0) + f(1))$ is equal to
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If $A = \begin{bmatrix} 0 & 0 & 1 \\ \frac{1}{4} & f'(1) & 1 \\ \alpha^2 & 4 & 1 \end{bmatrix}$ and $B = \text{adj}(\text{adj } A)$ be such that $|B| = 81$, then $\alpha^2$ is equal to
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Let $f$ be a differentiable function satisfying $f(x) = 1 - 2x + \int_{0}^{x} e^{(x - t)} f(t) dt$, $x \in \mathbf{R}$ and let $g(x) = \int_{0}^{x} (f(t) + 2)^{15} (t - 4)^{6} (t + 12)^{17} dt$, $x \in \mathbf{R}$. If $\text{p}$ and $\text{q}$ are respectively the points of local minima and local maxima of $g$, then the value of $|\text{p} + \text{q}|$ is equal to _________.
Important Questions from Integral Calculus
- Let the solution curve of the differential equation $x dy - y dx = \sqrt{x^2 + y^2} dx, x > 0$, $y(1) = 0$, be $y = y(x)$. Then $y(3)$ is equal to
- The area of the region $A = \{(x, y) : 4x^2 + y^2 \leq 8 \text{ and } y^2 \leq 4x\}$ is :
If $y = y(x)$ satisfies the differential equation
$16(\sqrt{x+ 9\sqrt{x}})(4 + \sqrt{9 + \sqrt{x}}) \cos y \, dy = (1 + 2 \sin y) dx, x > 0$ and $y(256) = \frac{\pi}{2}, y(49) = \alpha$, then $2 \sin \alpha$ is equal to :- Let a differentiable function $f$ satisfy the equation $\int_{0}^{36} f\left(\frac{tx}{36}\right) dt = 4\alpha f(x)$. If $y = f(x)$ is a standard parabola passing through the points (2, 1) and (– 4, $\beta$), then $\beta^\alpha$ is equal to ______.
- Let $f(\alpha)$ denote the area of the region in the first quadrant bounded by $x = 0, x = 1, y^2 = x$ and $y = |\alpha x - 5| - |1 - \alpha x| + \alpha x^2$. Then $(f(0) + f(1))$ is equal to
