Related Formula
⁻¹(y) = ⁻¹((1)/(y))$$\cot^{-1}(y) = \tan^{-1}\left(\frac{1}{y}\right)$$
⁻¹((x - y)/(1 + xy)) = ⁻¹x - ⁻¹y$$\tan^{-1}\left(\frac{x - y}{1 + xy}\right) = \tan^{-1}x - \tan^{-1}y$$
King's Property: ∫₀^a f(x)dx = ∫₀^a f(a-x)dx$$\text{King's Property: } \int_0^a f(x)dx = \int_0^a f(a-x)dx$$
Core Logic
Let I = ∫₀¹ ⁻¹(1-2x+4x²) dx$I = \int_{0}^{1} \cot^{-1}(1-2x+4x^2) dx$.
Convert ⁻¹$\cot^{-1}$ to ⁻¹$\tan^{-1}$:
⁻¹(1 + 2x(2x-1)) = ⁻¹( (1)/(1 + 2x(2x-1)) )$$\cot^{-1}(1 + 2x(2x-1)) = \tan^{-1}\left( \frac{1}{1 + 2x(2x-1)} \right)$$
Notice that 2x - (2x-1) = 1$2x - (2x-1) = 1$. Thus, the integrand is ⁻¹( (2x - (2x-1))/(1 + 2x(2x-1)) )$\tan^{-1}\left( \frac{2x - (2x-1)}{1 + 2x(2x-1)} \right)$.
I = ∫₀¹ ( ⁻¹(2x) - ⁻¹(2x-1) ) dx$$I = \int_0^1 \left( \tan^{-1}(2x) - \tan^{-1}(2x-1) \right) dx$$
Step 1: Apply Definite Integral Properties
Applying King's property to the second term ∫₀¹ ⁻¹(2x-1) dx$\int_0^1 \tan^{-1}(2x-1) dx$:
x → 1-x ∫₀¹ ⁻¹(2(1-x)-1) dx = ∫₀¹ ⁻¹(1-2x) dx = -∫₀¹ ⁻¹(2x-1) dx$$x \to 1-x \implies \int_0^1 \tan^{-1}(2(1-x)-1) dx = \int_0^1 \tan^{-1}(1-2x) dx = -\int_0^1 \tan^{-1}(2x-1) dx$$
Wait, this implies ∫₀¹ ⁻¹(2x-1) dx = 0$\int_0^1 \tan^{-1}(2x-1) dx = 0$!
Thus, I = ∫₀¹ ⁻¹(2x) dx$I = \int_0^1 \tan^{-1}(2x) dx$.
Step 2: Integration by Parts
Solve ∫₀¹ ⁻¹(2x) · 1 dx$\int_0^1 \tan^{-1}(2x) \cdot 1 \, dx$:
I = [ x ⁻¹(2x) ]₀¹ - ∫₀¹ x (2)/(1+4x²) dx$$I = \left[ x \tan^{-1}(2x) \right]_0^1 - \int_0^1 x \frac{2}{1+4x^2} dx$$
I = ⁻¹(2) - (1)/(4) ∫₀¹ (8x)/(1+4x²) dx$$I = \tan^{-1}(2) - \frac{1}{4} \int_0^1 \frac{8x}{1+4x^2} dx$$
Let 1+4x² = t 8x dx = dt$1+4x^2 = t \implies 8x dx = dt$. At x=0, t=1$x=0, t=1$; at x=1, t=5$x=1, t=5$.
I = ⁻¹(2) - (1)/(4) ∫₁⁵ (dt)/(t) = ⁻¹(2) - (1)/(4) ln(5)$$I = \tan^{-1}(2) - \frac{1}{4} \int_1^5 \frac{dt}{t} = \tan^{-1}(2) - \frac{1}{4} \ln(5)$$
Step 3: Compare and Calculate Result
The original integral has a factor of 4$4$:
4I = 4 ⁻¹(2) - ln(5)$$4I = 4\tan^{-1}(2) - \ln(5)$$
Compare with a ⁻¹(2) - b ₑ(5)$a\tan^{-1}(2) - b\log_e(5)$:
a = 4, b = 1$$a = 4, \quad b = 1$$
Therefore, 2a + b = 2(4) + 1 = 9$2a + b = 2(4) + 1 = 9$.
Pattern Recognition
Always convert ⁻¹$\cot^{-1}$ quadratic inputs into ⁻¹(x-y)/(1+xy)$\tan^{-1}\frac{x-y}{1+xy}$ forms. Apply King's property on symmetric limits; often one piece vanishes entirely.
Chapter Mix
Class 12 Maths: Definite Integrals
Class 12 Maths: Inverse Trigonometric Functions