Question

Evaluate \[ \int \frac{1}{\left(\sin x+\cos x+\sqrt{2\sin 2x}\right)^2}\,dx= \]

• A. \(-\dfrac{1+3\sqrt{\tan x}}{(3+\tan^2x)^3}+C\)
• B. \(-\dfrac{1+3\sqrt{\tan x}}{3(1+\sqrt{\tan x})^3}+C\)
• C. \(-\dfrac{1+\sqrt{\tan x}}{3(1+3\sqrt{\tan x})^2}+C\)
• D. \(-\dfrac{1}{(1+3\sqrt{\tan x})^3}+C\)

Hint:

When an integrand contains \[ \sin x+\cos x+2\sqrt{\sin x\cos x}, \] rewrite it as \[ (\sqrt{\sin x}+\sqrt{\cos x})^2 \] and then use \(t=\sqrt{\tan x}\).

Answer 1

The Correct Option is B

Step 1: Simplify the denominator.
Given integral: \[ I=\int \frac{1}{\left(\sin x+\cos x+\sqrt{2\sin 2x}\right)^2}\,dx \] Since \[ \sin 2x=2\sin x\cos x, \] we get \[ 2\sin 2x=4\sin x\cos x \] Therefore, \[ \sqrt{2\sin 2x}=2\sqrt{\sin x\cos x} \] So, \[ \sin x+\cos x+\sqrt{2\sin 2x} = \sin x+\cos x+2\sqrt{\sin x\cos x} \] \[ = (\sqrt{\sin x}+\sqrt{\cos x})^2 \] Hence, \[ \left(\sin x+\cos x+\sqrt{2\sin 2x}\right)^2 = (\sqrt{\sin x}+\sqrt{\cos x})^4 \]

Step 2: Put \(t=\sqrt{\tan x}\).
Let \[ t=\sqrt{\tan x} \] Then, \[ \tan x=t^2 \] Also, \[ \sin x=\frac{t^2}{\sqrt{1+t^4}}, \quad \cos x=\frac{1}{\sqrt{1+t^4}} \] So, \[ \sqrt{\sin x}=\frac{t}{(1+t^4)^{1/4}} \] and \[ \sqrt{\cos x}=\frac{1}{(1+t^4)^{1/4}} \] Thus, \[ (\sqrt{\sin x}+\sqrt{\cos x})^4 = \left(\frac{t+1}{(1+t^4)^{1/4}}\right)^4 \] \[ = \frac{(1+t)^4}{1+t^4} \]

Step 3: Transform \(dx\).
Since \[ \tan x=t^2, \] differentiate: \[ \sec^2x\,dx=2t\,dt \] But \[ \sec^2x=1+\tan^2x=1+t^4 \] Hence, \[ dx=\frac{2t}{1+t^4}\,dt \]

Step 4: Substitute in the integral.
\[ I=\int \frac{1}{\frac{(1+t)^4}{1+t^4}}\cdot \frac{2t}{1+t^4}\,dt \] \[ I=\int \frac{1+t^4}{(1+t)^4}\cdot \frac{2t}{1+t^4}\,dt \] \[ I=\int \frac{2t}{(1+t)^4}\,dt \]

Step 5: Integrate.
Let \[ u=1+t \] Then, \[ t=u-1 \] and \[ dt=du \] So, \[ I=\int \frac{2(u-1)}{u^4}\,du \] \[ I=\int \left(\frac{2}{u^3}-\frac{2}{u^4}\right)\,du \] \[ I=2\int u^{-3}\,du-2\int u^{-4}\,du \] \[ I=2\left(\frac{u^{-2}}{-2}\right)-2\left(\frac{u^{-3}}{-3}\right) \] \[ I=-\frac{1}{u^2}+\frac{2}{3u^3} \] \[ I=\frac{-3u+2}{3u^3} \] \[ I=-\frac{3u-2}{3u^3} \] Since \[ u=1+t, \] we get \[ I=-\frac{3(1+t)-2}{3(1+t)^3} \] \[ I=-\frac{1+3t}{3(1+t)^3} \] Now substituting \[ t=\sqrt{\tan x}, \] we get \[ I=-\frac{1+3\sqrt{\tan x}}{3(1+\sqrt{\tan x})^3}+C \]

Step 6: Final conclusion.
Therefore, \[ \boxed{-\frac{1+3\sqrt{\tan x}}{3(1+\sqrt{\tan x})^3}+C} \]