If ((1-5^2x)/(^5x^2x))dx = f(x)+C, where C is the constant of integration, then f(6)-f(4) …

Question

If $\displaystyle\int\left(\frac{1-5\cos^2x}{\sin^5x\cos^2x}\right)dx = f(x)+C$, where $C$ is the constant of integration, then $f\left(\dfrac\pi6\right)-f\left(\dfrac\pi4\right)$ is equal to:

Accepted Answer

Split: $\dfrac{1-5\cos^2x}{\sin^5x\cos^2x} = \dfrac{\sec^2x}{\sin^5x} - 5\csc^5x$. One verifies $\dfrac{d}{dx}\left(\dfrac{\tan x}{\sin^5x}\right)$ equals this expression, so

$$f(x) = \frac{\tan x}{\sin^5x} = \frac{1}{\sin^4x\cos x}$$

$$f\left(\frac\pi6\right) = \frac{1}{(1/2)^4(\sqrt3/2)} = \frac{32}{\sqrt3},\qquad f\left(\frac\pi4\right) = \frac{1}{(1/\sqrt2)^4(1/\sqrt2)} = 4\sqrt2$$

$$f\left(\frac\pi6\right)-f\left(\frac\pi4\right) = \frac{32}{\sqrt3}-4\sqrt2 = \frac{4}{\sqrt3}\left(8-\sqrt6\right)$$

Solutions · 1