Let f(x)$f(x)$ be a a positive function and I₁ = ∫-(1)/(2)¹ 2xf(2x(1 - 2x)) dx$I_1 = \int_{-\frac{1}{2}}^{1} 2xf(2x(1 - 2x)) \, dx$ and I₂ = ∫₋₁² f(x(1 - x)) dx$I_2 = \int_{-1}^{2} f(x(1 - x)) \, dx$. Then the value of (I₂)/(I₁)$\frac{I_2}{I_1}$ is equal to

Related Formula

Core Logic

Let I = ∫-π/6π/6 π+4x¹¹1- (|x|+π/6)dx$I = \int_{-\pi/6}^{\pi/6}\frac{\pi+4x^{11}}{1-\sin(|x|+\pi/6)}dx$. The denominator 1 - (|x| + π/6)$1 - \sin(|x| + \pi/6)$ is an even function. The numerator can be split into an even part (π$\pi$) and an odd part (4x¹¹$4x^{11}$).

Step 1: Simplify to Even Integral

We are left with the even part:

Using even function property ∫₋ₐa f(x) dx = 2 ∫₀a f(x) dx$\int_{-a}^{a} f(x) dx = 2 \int_{0}^{a} f(x) dx$:

Step 2: Substitution

Let t = x + (π)/(6) ⇒ dt = dx$t = x + \frac{\pi}{6} \Rightarrow dt = dx$. Limits: when x = 0 ⇒ t = π/6$x = 0 \Rightarrow t = \pi/6$, when x = π/6 ⇒ t = π/3$x = \pi/6 \Rightarrow t = \pi/3$.

Step 3: Solve the Integral

Rationalize the denominator:

Integrate directly:

Evaluate limits: Upper limit (π/3$\pi/3$): (π/3) + (π/3) = √(3) + 2$\tan(\pi/3) + \sec(\pi/3) = \sqrt{3} + 2$ Lower limit (π/6$\pi/6$): (π/6) + (π/6) = 1√(3) + 2√(3) = 3√(3) = √(3)$\tan(\pi/6) + \sec(\pi/6) = \frac{1}{\sqrt{3}} + \frac{2}{\sqrt{3}} = \frac{3}{\sqrt{3}} = \sqrt{3}$

Pattern Recognition

Symmetric limits [-a, a]$[-a, a]$ instantly demand testing for odd/even parity. Any mixed polynomial like c + k xodd$c + k x^{\text{odd}}$ over an even denominator guarantees the odd power term strictly vanishes, halving calculation time.

Chapter Mix

Class 12 Maths: Definite Integration Class 11 Maths: Trigonometric Functions

Related Formula

Core Logic

Let I = 6∫₀π| 3x + x + 2x| dx$I = 6\int_{0}^{\pi}|\sin 3x + \sin x + \sin 2x| dx$. Apply sum-to-product on 3x + x$\sin 3x + \sin x$: 3x + x = 2 (2x) (x)$\sin 3x + \sin x = 2 \sin(2x) \cos(x)$

So the expression becomes: |2 (2x) x + 2x| = | 2x (2 x + 1)| = |2 x x (2 x + 1)|$|2 \sin(2x) \cos x + \sin 2x| = |\sin 2x (2 \cos x + 1)| = |2 \sin x \cos x (2 \cos x + 1)|$ Since x in [0, π]$x \in [0, \pi]$, x ≥ 0$\sin x \geq 0$. We can pull it out of the modulus. I = 12 ∫₀π x |2 ² x + x| dx$I = 12 \int_{0}^{\pi} \sin x |2 \cos^2 x + \cos x| dx$

Step 1: Coordinate Substitution

Substitute t = x$t = \cos x$, then dt = - x dx$dt = -\sin x dx$. Limits: when x = 0$x = 0$, t = 1$t = 1$. When x = π$x = \pi$, t = -1$t = -1$.

Step 2: Resolve Modulus Intervals

The roots of 2t² + t = 0$2t^2 + t = 0$ are t = 0$t = 0$ and t = -1/2$t = -1/2$. The quadratic 2t² + t$2t^2 + t$ is negative in the interval (-1/2, 0)$(-1/2, 0)$ and positive elsewhere. Split the integral:

Step 3: Evaluate Integrals

Anti-derivative: F(t) = (2t³)/(3) + (t²)/(2)$F(t) = \frac{2t^3}{3} + \frac{t^2}{2}$. F(1) = 2/3 + 1/2 = 7/6$F(1) = 2/3 + 1/2 = 7/6$ F(0) = 0$F(0) = 0$ F(-1/2) = 2(-1/8)/3 + 1/8 = -1/12 + 1/8 = 1/24$F(-1/2) = 2(-1/8)/3 + 1/8 = -1/12 + 1/8 = 1/24$ F(-1) = -2/3 + 1/2 = -1/6$F(-1) = -2/3 + 1/2 = -1/6$

Evaluate each segment:

• ∫₋₁-1/2 = F(-1/2) - F(-1) = 1/24 - (-1/6) = 1/24 + 4/24 = 5/24$\int_{-1}^{-1/2} = F(-1/2) - F(-1) = 1/24 - (-1/6) = 1/24 + 4/24 = 5/24$
• -∫-1/2⁰ = -(F(0) - F(-1/2)) = -(0 - 1/24) = 1/24$-\int_{-1/2}^{0} = -(F(0) - F(-1/2)) = -(0 - 1/24) = 1/24$
• ∫₀¹ = F(1) - F(0) = 7/6 - 0 = 28/24$\int_{0}^{1} = F(1) - F(0) = 7/6 - 0 = 28/24$

Sum of parts inside bracket: (5)/(24) + (1)/(24) + (28)/(24) = (34)/(24) = (17)/(12)$\frac{5}{24} + \frac{1}{24} + \frac{28}{24} = \frac{34}{24} = \frac{17}{12}$

Step 4: Final Output

Pattern Recognition

Whenever an integral features a cascading sum of sine frequencies like (kx)$\sin(kx)$, pair the highest and lowest frequencies first. The resulting common factor often matches the middle term, instantly yielding a clean polynomial substitution under t = x$t = \cos x$.

Chapter Mix

Class 12 Maths: Definite Integration Class 11 Maths: Trigonometric Functions

Related Formula

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$\tan^{-1}$:

Notice that 2x - (2x-1) = 1$2x - (2x-1) = 1$. Thus, the integrand is ⁻¹( (2x - (2x-1))/(1 + 2x(2x-1)) )$\tan^{-1}\left( \frac{2x - (2x-1)}{1 + 2x(2x-1)} \right)$.

Step 1: Apply Definite Integral Properties

Applying King's property to the second term ∫₀¹ ⁻¹(2x-1) dx$\int_0^1 \tan^{-1}(2x-1) dx$:

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$\int_0^1 \tan^{-1}(2x) \cdot 1 \, dx$:

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$x=1, t=5$.

Step 3: Compare and Calculate Result

The original integral has a factor of 4$4$:

Compare with a ⁻¹(2) - b ₑ(5)$a\tan^{-1}(2) - b\log_e(5)$:

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

Related Formula

Core Logic

The integral bounds are from -π/2 ≈ -1.57$-\pi/2 \approx -1.57$ to π/2 ≈ 1.57$\pi/2 \approx 1.57$. We must split the integral at every integer point between these bounds.

Intervals:

• [-π/2, -1) [x] = -2$[-\pi/2, -1) \implies [x] = -2$
• [-1, 0) [x] = -1$[-1, 0) \implies [x] = -1$
• [0, 1) [x] = 0$[0, 1) \implies [x] = 0$
• [1, π/2) [x] = 1$[1, \pi/2) \implies [x] = 1$

Step 1: Splitting the Integral

Step 2: Evaluating Sub-integrals

Evaluate the limits for each constant integral:

Step 3: Simplifying the Expression

Group the ( (π)/(2) - 1 )$\left( \frac{\pi}{2} - 1 \right)$ terms:

Find a common denominator for the constants (LCD is 60):

Rewrite (7π)/(20)$\frac{7\pi}{20}$ with denominator 60:

So,

Pattern Recognition

Integration of step functions always transforms into a simple sum of rectangle areas (cᵢ × Δ xᵢ)$(c_i \times \Delta x_i)$. Immediately break the bounds at integers, turning a calculus problem into elementary arithmetic.

Chapter Mix

Class 12 Maths: Definite Integration

Related Formula

Integration by parts formula: ∫ u dv = uv - ∫ v du$\int u \, dv = uv - \int v \, du$.

Core Logic

Let I(x) = ∫₀x t² (x - t) dt$I(x) = \int_{0}^{x} t^2 \sin(x - t) dt$. Use integration by parts. Let u = t²$u = t^2$ and dv = (x - t) dt v = (x - t)$dv = \sin(x - t) dt \implies v = \cos(x - t)$.

Step 1: Second Integration by Parts

Now integrate ∫₀x 2t (x - t) dt$\int_{0}^{x} 2t \cos(x - t) dt$ by parts again: Let u = 2t$u = 2t$ and dv = (x - t) dt v = - (x - t)$dv = \cos(x - t) dt \implies v = -\sin(x - t)$.

Step 2: Equating and Solving

Substitute this back into the first equation:

We are given I(x) = x²$I(x) = x^2$. Therefore:

The solutions for x = 1$\cos x = 1$ are x = 2nπ$x = 2n\pi$, where n$n$ is an integer. We need the roots in the interval [0, 100]$[0, 100]$. 0 ≤ 2nπ ≤ 100 0 ≤ n ≤ (100)/(2π) ≈ (100)/(6.28) ≈ 15.92$0 \leq 2n\pi \leq 100 \implies 0 \leq n \leq \frac{100}{2\pi} \approx \frac{100}{6.28} \approx 15.92$.

Since n$n$ must be an integer, n$n$ takes values 0, 1, 2, , 15$0, 1, 2, \dots, 15$. The total number of solutions is 16$16$.

Pattern Recognition

Convoluted-looking definite integrals containing a shift (x-t)$(x-t)$ often rapidly unpack through successive integration by parts where the polynomial variable diminishes until exhaustion.

Chapter Mix

Class 12 Maths: Definite Integration Class 11 Maths: Trigonometric Equations