If ∫( x)(-11)/(2)( x)(-5)/(2)dx= - p₁q₁( x)(9)/(2)- p₂q₂( x)(5)/(2)- p₃q₃( x)(1)/(2)+ p₄q₄( x)(-3)/(2)+C$\int(\sin x)^{\frac{-11}{2}}(\cos x)^{\frac{-5}{2}}dx= -\frac{p_{1}}{q_{1}}(\cot x)^{\frac{9}{2}}-\frac{p_{2}}{q_{2}}(\cot x)^{\frac{5}{2}}-\frac{p_{3}}{q_{3}}(\cot x)^{\frac{1}{2}}+\frac{p_{4}}{q_{4}}(\cot x)^{\frac{-3}{2}}+C$, where pᵢ$p_{i}$ and qᵢ$q_{i}$ are positive integers with (pᵢ, qᵢ)=1$\gcd(p_{i}, q_{i})=1$ for i=1, 2, 3, 4 and C is the constant of integration, then 15p₁p₂p₃p₄q₁q₂q₃q₄$\frac{15p_{1}p_{2}p_{3}p_{4}}{q_{1}q_{2}q_{3}q_{4}}$ is equal to ____.

Core Logic

Let Iᵣ = ∫₀r x | π x| dx$I_r = \int_{0}^{r} x |\sin\pi x| dx$. Apply King's property of definite integrals: ∫₀a f(x) dx = ∫₀a f(a-x) dx$\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx$.

Since r$r$ is an integer, | π(r - x)| = | (π r - π x)| = |(-1)^r (π x)| = | π x|$|\sin\pi(r - x)| = |\sin(\pi r - \pi x)| = |(-1)^r \sin(\pi x)| = |\sin\pi x|$.

Adding the original and the transformed integral:

Step 1: Integral of Modulus

The function | π x|$|\sin\pi x|$ is periodic with period 1. Using the property of periodic functions ∫₀nT f(x)dx = n ∫₀T f(x)dx$\int_{0}^{nT} f(x)dx = n \int_{0}^{T} f(x)dx$:

For x in [0, 1]$x \in [0, 1]$, π x ≥ 0$\sin\pi x \geq 0$.

So, Iᵣ = (r)/(2) ( r · (2)/(π) ) = (r²)/(π)$I_r = \frac{r}{2} \left( r \cdot \frac{2}{\pi} \right) = \frac{r^2}{\pi}$.

Step 2: Evaluate the Sum

The expression inside the summation is:

So the total sum is:

Chapter Mix

Class 12 Mathematics: Integrals Class 11 Mathematics: Sequences and Series

Core Logic

Split the integral I$I$ at integer points and trigonometric step thresholds between -π/2 ≈ -1.57$-\pi/2 \approx -1.57$ and π/2 ≈ 1.57$\pi/2 \approx 1.57$: Intervals are: (-π/2, -1), (-1, 0), (0, 1), (1, π/2)$(-\pi/2, -1), (-1, 0), (0, 1), (1, \pi/2)$. For all x in (-π/2, π/2)$x \in (-\pi/2, \pi/2)$ excluding exactly 0$0$, [ x] = 0$[\cos x] = 0$. [ x] = -1$[\sin x] = -1$ for x in (-π/2, 0)$x \in (-\pi/2, 0)$ and [ x] = 0$[\sin x] = 0$ for x in (0, π/2)$x \in (0, \pi/2)$.

Execution

Evaluate in pieces:

• x in (-π/2, -1)$x \in (-\pi/2, -1)$:

[x] = -2, [ x] = -1, [ x] = 0$[x] = -2, [\sin x] = -1, [\cos x] = 0$. Integrand = (12(3 - 2))/(3 - 1 + 0) = (12(1))/(2) = 6$\frac{12(3 - 2)}{3 - 1 + 0} = \frac{12(1)}{2} = 6$.

• x in (-1, 0)$x \in (-1, 0)$:

[x] = -1, [ x] = -1, [ x] = 0$[x] = -1, [\sin x] = -1, [\cos x] = 0$. Integrand = (12(3 - 1))/(2) = (24)/(2) = 12$\frac{12(3 - 1)}{2} = \frac{24}{2} = 12$.

• x in (0, 1)$x \in (0, 1)$:

[x] = 0, [ x] = 0, [ x] = 0$[x] = 0, [\sin x] = 0, [\cos x] = 0$. Integrand = (12(3 + 0))/(3) = 12$\frac{12(3 + 0)}{3} = 12$.

• x in (1, π/2)$x \in (1, \pi/2)$:

[x] = 1, [ x] = 0, [ x] = 0$[x] = 1, [\sin x] = 0, [\cos x] = 0$. Integrand = (12(3 + 1))/(3) = (48)/(3) = 16$\frac{12(3 + 1)}{3} = \frac{48}{3} = 16$.

Integrate piece by piece:

I = 11π + 2$I = 11\pi + 2$

Pattern Recognition

For integrals combining integer functions and trigonometric boundaries, breaking the domain purely at integer values (-1, 0, 1)$(-1, 0, 1)$ generally perfectly matches trigonometric boundaries since the ranges of $\sin$ and $\cos$ in (-π/2, π/2)$(-\pi/2, \pi/2)$ hover near these integers.

Chapter Mix

Class 12 Maths: Definite Integration

Core Logic

To eliminate fractional powers, use substitution. The LCM of denominators 3 and 2 is 6. Put x = t⁶ ⇒ dx = 6t⁵ dt$x = t^6 \Rightarrow dx = 6t^5 \, dt$.

Execution

Rewrite the numerator using polynomial division or algebraic manipulation:

Substitute t = x1/6$t = x^{1/6}$:

Evaluate constant C$C$ using f(0)$f(0)$: f(0) = 24 ln 2 + C = -26 + 24 ln 2 ⇒ C = -26$f(0) = 24 \ln 2 + C = -26 + 24 \ln 2 \Rightarrow C = -26$.

Find f(1)$f(1)$:

Given f(1) = a + b ln 3$f(1) = a + b \ln 3$, we map a = -35$a = -35$ and b = 24$b = 24$. a + b = -35 + 24 = -11$a + b = -35 + 24 = -11$.

Pattern Recognition

For integrals with multiple roots x1/m$x^{1/m}$ and x1/n$x^{1/n}$, substituting x = tLCM(m,n)$x = t^{LCM(m,n)}$ strictly normalizes the algebraic fraction into simple polynomial synthetic division.

Chapter Mix

Class 12 Maths: Indefinite Integration

Related Formula

Core Logic

First, we find the coordinates of intersection in the second quadrant to determine the integration limits a$a$ and b$b$.

Step 1: Find points of intersection

Substitute y = (x²)/(2)$y = \frac{x^2}{2}$ into the line equation y - 2x - 6 = 0$y - 2x - 6 = 0$:

Since the point (a, b)$(a, b)$ lies in the second quadrant, x$x$ must be negative:

a = -2$a = -2$

Thus, the integration limits are a = -2$a = -2$ and b = 2$b = 2$.

Step 2: Solve the Integral using King's Property

The integral is:

Apply King's property, substituting x → -x$x \to -x$ (since -2 + 2 - x = -x$-2 + 2 - x = -x$):

Adding equations (1) and (2):

Since 9x²$9x^2$ is an even function:

Pattern Recognition

Whenever you see an exponential denominator like 1 + c^x$1 + c^x$ inside a symmetric interval definite integral [-a, a]$[-a, a]$, applying King's property will almost always cancel the exponential factor cleanly when the remaining numerator is even.

Chapter Mix

Class 12 Mathematics: Integral Calculus

Related Formula

Core Logic

We first rationalise the denominator to split the integral into two standard integration terms.

Step 1: Rationalise the integrand

Multiply the numerator and denominator by √(3+x²) - √(1+x²)$\sqrt{3+x^2} - \sqrt{1+x^2}$:

Thus, the integral expression simplifies to:

Step 2: Evaluate the integrals

For the first integral:

For the second integral:

Step 3: Sum the terms

Now compile all terms:

Pattern Recognition

Integration of roots of quadratics: Always look for algebraic rationalisation when dealing with sum of root denominators. It directly reduces complex fractions into standard integrable functions.

Chapter Mix

Class 12 Mathematics: Integral Calculus