Question

The value of the integral \(\int \frac{dx}{\cos 2x + \sin 2x + 2\sin^2 x}\) is:

• A. \( -\frac{1}{1 + \tan x} + c \)
• B. \( -\frac{\tan x}{1 + \tan x} + c \)
• C. \( -\cot x + c \)
• D. \( -\tan x + c \)

Hint:

When an integral contains terms like \(\sin^2 x, \cos^2 x,\) and \(\sin x \cos x\), dividing the numerator and denominator by \(\cos^2 x\) is a standard strategy to transform the integral into a form involving \(\tan x\) and \(\sec^2 x\).

Answer 1

The Correct Option is A

Step 1: Simplify the denominator.
Using $\cos2x=\cos^2x-\sin^2x$ and $\sin2x=2\sin x\cos x$, the denominator is \[ (\cos^2x-\sin^2x)+2\sin x\cos x+2\sin^2x=\cos^2x+\sin^2x+2\sin x\cos x. \]
Step 2: Recognise the perfect square.
This equals $1+2\sin x\cos x=(\sin x+\cos x)^2$.
Step 3: Convert to tangent.
Multiply top and bottom by $\sec^2x$. The integral becomes \[ \int\frac{\sec^2x}{\sec^2x+2\tan x}\,dx. \]
Step 4: Complete the square in $\tan x$.
Since $\sec^2x=1+\tan^2x$, the denominator is $1+\tan^2x+2\tan x=(1+\tan x)^2$, so \[ I=\int\frac{\sec^2x}{(1+\tan x)^2}\,dx. \]
Step 5: Substitute.
Let $u=1+\tan x$, so $du=\sec^2x\,dx$: \[ I=\int u^{-2}\,du=-\frac{1}{u}+c. \]
Step 6: Back-substitute.
\[ I=-\frac{1}{1+\tan x}+c. \]
\[ \boxed{-\dfrac{1}{1+\tan x}+c} \]