Question

If \( \int_0^1 4 \cot^{-1}(1 - 2x + 4x^2) dx = a \tan^{-1}(2) - b \ln(5) \), where \( a, b \in \mathbb{N} \), then \( (2a + b) \) is equal to _________.

Hint:

Whenever you see a cubic or quadratic in the denominator of an inverse trig function, look for the \( \tan^{-1}X - \tan^{-1}Y \) pattern in the form \( 1 + XY \).

Answer 1

Correct Answer: 13

Step 1: Understanding the Concept:
We simplify the integrand using the property \( \cot^{-1}z = \tan^{-1}(1/z) \) and the formula \( \tan^{-1}A - \tan^{-1}B = \tan^{-1}\left(\frac{A - B}{1 + AB}\right) \).

Step 2: Key Formula or Approach:
Integrand: \( \cot^{-1}(1 - 2x + 4x^2) = \tan^{-1}\left(\frac{1}{1 + 2x(2x-1)}\right) \).
Recognize the numerator as \( 2x - (2x-1) \):
\[ \tan^{-1}\left(\frac{2x - (2x-1)}{1 + (2x)(2x-1)}\right) = \tan^{-1}(2x) - \tan^{-1}(2x-1) \]

Step 3: Detailed Explanation:
The integral becomes:
\[ I = 4 \int_0^1 [\tan^{-1}(2x) - \tan^{-1}(2x-1)] dx \]
Evaluate the second part: \( \int_0^1 \tan^{-1}(2x-1) dx \). Let \( 2x-1 = t \implies dx = dt/2 \).
\[ \frac{1}{2} \int_{-1}^{1} \tan^{-1}(t) dt = 0 \quad (\text{since } \tan^{-1}t \text{ is an odd function over a symmetric interval}) \]
So, \( I = 4 \int_0^1 \tan^{-1}(2x) dx \). Let \( 2x = u \implies dx = du/2 \):
\[ I = 2 \int_0^2 \tan^{-1} u du \]
Integrating by parts (\( \int \tan^{-1} u du = u \tan^{-1} u - \frac{1}{2}\ln(1+u^2) \)):
\[ I = 2 \left[ u \tan^{-1} u - \frac{1}{2}\ln(1+u^2) \right]_0^2 \]
\[ I = 2 \left[ (2 \tan^{-1} 2 - \frac{1}{2}\ln 5) - (0 - 0) \right] = 4 \tan^{-1} 2 - \ln 5 \]
Comparing with \( a \tan^{-1} 2 - b \ln 5 \), we get \( a = 4 \) and \( b = 1 \).
Required value: \( 2a + b = 2(4) + 1 = 9 \).

Step 4: Final Answer:
The value of \( 2a + b \) is 9.