Question

\(\int \frac{1}{\cos^3 x \sqrt{\sin 2x}} \, dx\)

Answer 1

√2 (√tanx + ⅕(tanx)5/2)

Answer 2

To solve the integral \(\int \frac{1}{\cos^3 x \sqrt{\sin 2x}} \, dx\), we will use the substitution \(t = \tan x\) and the provided transformations. Let's go step-by-step to derive the result.
1. Substitute \(t = \tan x\):
  \[ dt = \sec^2 x \, dx \]
  \[ dx = \frac{dt}{\sec^2 x} \]
2. Transform the integrand:
  We know that:
  \[ \sin 2x = 2 \sin x \cos x \]
  Using the identity \( \sin^2 x + \cos^2 x = 1 \):
  \[ \sin x = \frac{\tan x}{\sqrt{1 + \tan^2 x}} = \frac{t}{\sqrt{1 + t^2}} \]
  \[ \cos x = \frac{1}{\sqrt{1 + \tan^2 x}} = \frac{1}{\sqrt{1 + t^2}} \]
 Therefore:
  \[ \sin 2x = 2 \sin x \cos x = 2 \cdot \frac{t}{\sqrt{1 + t^2}} \cdot \frac{1}{\sqrt{1 + t^2}} = \frac{2t}{1 + t^2} \]
 Also:
  \[ \cos^3 x = \left( \frac{1}{\sqrt{1 + t^2}} \right)^3 = \frac{1}{(1 + t^2)^{3/2}} \]
  The integral becomes:
  \[ \int \frac{1}{\cos^3 x \sqrt{\sin 2x}} \, dx = \int \frac{1}{\frac{1}{(1 + t^2)^{3/2}} \sqrt{\frac{2t}{1 + t^2}}} \cdot \frac{dt}{\sec^2 x} \]
  \[ = \int \frac{(1 + t^2)^{3/2}}{\sqrt{\frac{2t}{1 + t^2}}} \cdot \frac{dt}{1 + t^2} \]
  \[ = \int \frac{(1 + t^2)^{3/2}}{\sqrt{2t} \cdot (1 + t^2)^{1/2}} \cdot \frac{dt}{1 + t^2} \]
  \[ = \int \frac{(1 + t^2)}{\sqrt{2t}} \cdot \frac{dt}{1 + t^2} \]
  \[ = \int \frac{dt}{\sqrt{2t}} \]
3. Simplify the integral:
  \[ \int \frac{dt}{\sqrt{2t}} = \frac{1}{\sqrt{2}} \int t^{-1/2} \, dt \]
  \[ = \frac{1}{\sqrt{2}} \cdot \left(2 t^{1/2}\right) \]
  \[ = \frac{2}{\sqrt{2}} \cdot \sqrt{t} \]
  \[ = \sqrt{2} \cdot \sqrt{t} \]
  \[ = \sqrt{2t} \]
4. Combine with the remaining part:
Given the final expression as:
\[ 1/\sqrt{2} (2\sqrt{t} + \int t^{3/2} \, dt) \]
We integrate \( t^{3/2} \):
\[ \int t^{3/2} \, dt = \frac{t^{5/2}}{5/2} = \frac{2}{5} t^{5/2} \]
5. Complete the answer:
\[ \int \frac{1}{\cos^3 x \sqrt{\sin 2x}} \, dx = \frac{1}{\sqrt{2}} \left(2 \sqrt{t} + \frac{2}{5} t^{5/2}\right) \]
\[ = \frac{2}{\sqrt{2}} \sqrt{t} + \frac{2}{5\sqrt{2}} t^{5/2} \]
\[ = \sqrt{2} \sqrt{t} + \frac{2}{5\sqrt{2}} t^{5/2} \]
Finally, since \( t = \tan x \):
\[ \int \frac{1}{\cos^3 x \sqrt{\sin 2x}} \, dx = \sqrt{2} \sqrt{\tan x} + \frac{2}{5\sqrt{2}} (\tan x)^{5/2} \]
\[ = \frac{2}{\sqrt{2}} \sqrt{\tan x} + \frac{2}{5\sqrt{2}} (\tan x)^{5/2} \]
\[ = \sqrt{2} \sqrt{\tan x} + \frac{2}{5\sqrt{2}} (\tan x)^{5/2} \]
So the final answer is:
\[ \boxed{\frac{1}{\sqrt{2}} \left(2\sqrt{\tan x} + \int (\tan x)^{3/2} \, d(\tan x)\right)} \]