For xin(-(π)/(2),(π)/(2))$x\in(-\frac{\pi}{2},\frac{\pi}{2})$, if y(x)=∫ cosec x+ xcosec x x+ x ² xdx$y(x)=\int\frac{\operatorname{cosec} x+\sin x}{\operatorname{cosec} x \sec x+\tan x \sin^2 x}dx$ and xarrow((π)/(2))^-y(x)=0$\lim_{x\rightarrow(\frac{\pi}{2})^-}y(x)=0$ then y((π)/(4))$y(\frac{\pi}{4})$ is equal to

Related Formula

Integration by Parts formula:

Core Logic

Given F(x) = xf(x)$F(x) = xf(x)$ and f(2) = 1 F(2) = 2f(2) = 2$f(2) = 1 \implies F(2) = 2f(2) = 2$.

Let's apply Integration by Parts to the first given integral:

Step 1: Simplify Second Integral via Integration by Parts

Now look at the second integral:

Applying Integration by parts (taking u = x²$u = x^2$ and v = F''(x)$v = F''(x)$):

We already know ∫₀² x F'(x) dx = 6$\int_{0}^{2} x F'(x) dx = 6$:

Step 2: Sum the Values

We need to find F'(2) + ∫₀² F(x) dx$F'(2) + \int_{0}^{2} F(x) dx$: 13 + (-2) = 11$13 + (-2) = 11$

Pattern Recognition

Notice how the definition of f(x)$f(x)$ is mostly a distraction to find F(2)=2$F(2)=2$. The problem is fundamentally testing consecutive applications of integration by parts to reduction structures.

Chapter Mix

Class 12 Mathematics: Definite Integration

Related Formula

Newton-Leibnitz Theorem for differentiation under integral sign:

Core Logic

Given:

Differentiating both sides with respect to x$x$:

We are given g(x³) = x⁶ + x⁷$g(x^3) = x^6 + x^7$. Let y = x³ x = y1/3$y = x^3 \implies x = y^{1/3}$. Substituting this into the expression for g$g$:

Thus, replacing y$y$ back with x$x$:

Step 1: Differentiate g(x) to find f(x)

Now find f(x)$f(x)$:

Step 2: Evaluate the Summation

We need to find Σr=1¹⁵ f(r³)$\sum_{r=1}^{15} f(r^3)$:

Now, compute the summation from r=1$r=1$ to 15$15$:

Pattern Recognition

Converting g(x³)$g(x^3)$ directly into a function of variable y=x³$y=x^3$ prevents multi-layer chain rule complications when applying differentiation immediately.

Chapter Mix

Class 11 Mathematics: Sequences and Series Class 12 Mathematics: Definite Integration

Related Formula

Core Logic

Let θ - θ = t$\sin\theta - \cos\theta = t$. Then ( θ + θ)dθ = dt$(\cos\theta + \sin\theta)d\theta = dt$. Transform limits: When θ = 0 t = 0 - 1 = -1$\theta = 0 \implies t = 0 - 1 = -1$ When θ = (π)/(4) t = 1√(2) - 1√(2) = 0$\theta = \frac{\pi}{4} \implies t = \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} = 0$

Step 1: Perform the algebraic substitution

Express the denominator base:

The integral transforms to:

Step 2: Execute Integral Calculation

Pattern Recognition

Whenever ( θ + θ)$(\sin\theta + \cos\theta)$ sits inside the numerator, instantly choose t = θ - θ$t = \sin\theta - \cos\theta$ as your core linear substitution engine to clean denominators.

Chapter Mix

Class 12 Mathematics: Definite Integrals

Related Formula

Core Logic

Perform variable substitution inside the integral: let λ x = t dλ = (1)/(x) dt$\lambda x = t \implies d\lambda = \frac{1}{x} dt$. When λ = 0 t = 0$\lambda = 0 \implies t = 0$; when λ = 1 t = x$\lambda = 1 \implies t = x$. The equation transforms to:

Step 1: Differentiate with respect to x

Using Leibniz rule and product rule:

Integrating both sides yields:

Step 2: Calculate Constants using boundaries

Given f(1) = 1 C = 1$f(1) = 1 \implies C = 1$. Given f(16) = (1)/(8) (1)/(8) = (16)(1-a)/(a) 2⁻³ = (2⁴)(1-a)/(a)$f(16) = \frac{1}{8} \implies \frac{1}{8} = (16)^{\frac{1-a}{a}} \implies 2^{-3} = (2^4)^{\frac{1-a}{a}}$

Therefore, power exponent = (1-4)/(4) = -(3)/(4) f(x) = x-(3)/(4)$= \frac{1-4}{4} = -\frac{3}{4} \implies f(x) = x^{-\frac{3}{4}}$.

Step 3: Evaluate target derivative value

Find the derivative:

Substitute x = (1)/(16)$x = \frac{1}{16}$:

Final requested computation calculation:

Pattern Recognition

Scaling inputs inside functional definite integrals tracks closely to homogenous Euler equation properties. Converting integrations quickly to local power functions reduces processing parameters.

Chapter Mix

Class 12 Mathematics: Definite Integrals Class 12 Mathematics: Differential Equations

Related Formula

King's Property of Definite Integrals:

Core Logic

Let the given integral be I$I$:

Substitute 2x = t 2dx = dt dx = (1)/(2)dt$2x = t \implies 2dx = dt \implies dx = \frac{1}{2}dt$. When x = 0 t = 0$x = 0 \implies t = 0$, and when x = (π)/(4) t = (π)/(2)$x = \frac{\pi}{4} \implies t = \frac{\pi}{2}$.

Step 1: Apply King's Property

Applying the property ∫₀a f(t) dt = ∫₀a f(a-t) dt$\int_{0}^{a} f(t) \, dt = \int_{0}^{a} f(a-t) \, dt$:

Adding equations (1) and (2):

Step 2: Substitution and Algebraic Limits

Let t = y ²t dt = dy$\tan t = y \implies \sec^{2}t \, dt = dy$. When t = 0 y = 0$t = 0 \implies y = 0$, and when t = (π)/(2) y = ∞$t = \frac{\pi}{2} \implies y = \infty$.

Now put y - (1)/(y) = p (1 + 1y²) dy = dp$y - \frac{1}{y} = p \implies \left(1 + \frac{1}{y^{2}}\right) dy = dp$. When y → 0^+ p → -∞$y \to 0^+ \implies p \to -\infty$, and when y → ∞ p → ∞$y \to \infty \implies p \to \infty$. Also, y² + 1y² = p² + 2 = p² + (√(2))²$y^{2} + \frac{1}{y^{2}} = p^{2} + 2 = p^{2} + (\sqrt{2})^{2}$.