To evaluate the integral \( \int \csc^5 x \, dx \), we use integration by parts. Let
\[ I = \int \csc^3 x \cdot \csc^2 x \, dx. \]
Applying integration by parts, we let:
\[ I = -\cot x \csc^3 x + \int \cot x \cdot (-3 \csc^2 x \cot x \csc x) \, dx. \]
Simplifying, we get:
\[ I = -\cot x \csc^3 x - 3 \int \csc^3 x (\csc^2 x - 1) \, dx, \] \[ I = -\cot x \csc^3 x - 3I + 3 \int \csc^3 x \, dx. \]
Let
\[ I_1 = \int \csc^3 x \, dx = -\csc x \cot x - \int \cot^2 x \csc x \, dx. \]
Using this and simplifying further, we identify values for \( \alpha \) and \( \beta \). After solving, we find:
\[ 8(\alpha + \beta) = 3. \]
If the function \[ f(x) = \begin{cases} \frac{2}{x} \left( \sin(k_1 + 1)x + \sin(k_2 -1)x \right), & x<0 \\ 4, & x = 0 \\ \frac{2}{x} \log_e \left( \frac{2 + k_1 x}{2 + k_2 x} \right), & x>0 \end{cases} \] is continuous at \( x = 0 \), then \( k_1^2 + k_2^2 \) is equal to:
The integral is given by:
\[ 80 \int_{0}^{\frac{\pi}{4}} \frac{\sin\theta + \cos\theta}{9 + 16 \sin 2\theta} d\theta \]
is equals to?
Let \[ I(x) = \int \frac{dx}{(x-11)^{\frac{11}{13}} (x+15)^{\frac{15}{13}}} \] If \[ I(37) - I(24) = \frac{1}{4} \left( b^{\frac{1}{13}} - c^{\frac{1}{13}} \right) \] where \( b, c \in \mathbb{N} \), then \[ 3(b + c) \] is equal to: