Q.Evaluate the definite integral $\int_{\pi/6}^{\pi/3}\frac{\sin x+\cos x}{\sqrt{\sin 2x}}\,dx$

Let $t = 3x - 2$, so $t \to 0$ as $x \to \tfrac{2}{3}$.

Denominator:

Numerator: using $\cos^2 t = 1 - \sin^2 t$,

$\sin\!\big(\pi\cos^2 t\big) = \sin\!\big(\pi - \pi\sin^2 t\big) = \sin\!\big(\pi\sin^2 t\big).$ …

Let $x = s^3$, so $dx = 3s^2\,ds$; limits $x:0\to 8$ give $s:0\to 2$. Then $x^{5/3}=s^5$ and $x^{2/3}=s^2$:

Put $s = 2\sin\theta$, $ds = 2\cos\theta\,d\theta$, $4-s^2 = 4\cos^2\theta$, $s:0\to2 \Rightarrow \theta:0\to\tfrac{\pi}{2}$:

With the Beta/Wallis reduction, …

Substitute $t = x^2 \Rightarrow dt = 2x\,dx$, and $x^3\,dx = x^2\cdot x\,dx = t\cdot\tfrac{1}{2}\,dt$. Limits: $x=0\to t=0$, $x=2\to t=4$.

Substitute again $s = 4-t \Rightarrow t = 4-s,\ dt = -ds$; limits flip $0\to4$:

Integrate.

$\int_{0}^{4}4s^{5/2}\,ds = 4\cdot\frac{2}{7}s^{7/2}\Big|_0^4 = \frac{8}{7}(4^{7/2}) = \frac{8}{7}\cdot128 = \frac{1024}{7},$ …

The concept first. A limit of the form $\frac{0}{0}$ means numerator and denominator share a common vanishing factor; the whole job is to expose it and cancel. Here both vanish at $x = 3/2$ (i.e. $2x = 3$), and the tool for a cube root difference is

applied with $a = u^{1/3},\; b = 3^{1/3}$, which converts the awkward $u^{1/3} - 3^{1/3}$ into the friendly factor $(u-3)$.

Step 1 — Substitute and factorise the numerator.

Let $u = 2x$; as $x \to \tfrac32$, $u \to 3$.

So

using $(u-3)(u^2+3u+9) = u^3 - 27$.

Step 2 — Handle the denominator.

Multiply and divide by $\left(u^{2/3} + u^{1/3}3^{1/3} + 3^{2/3}\right)$:

$\left(u^{1/3} - 3^{1/3}\right)\left(u^{2/3} + u^{1/3}3^{1/3} + 3^{2/3}\right) = u - 3$ …

Critical points in $[0,2]$: $x=\tfrac13$ and $x=1$.

Evaluate at the critical points and the endpoints:

$f(1) = 1 - 2 + 1 - 3 = -3,$ …

We begin by recognizing that both limits are standard trigonometric forms in disguise. The key is to rewrite each expression using known identities so that the limit becomes a familiar one: $\lim_{x \to 0} \frac{\sin x}{x} = 1$ and $\lim_{x \to 0} \frac{\tan x}{x} = 1$.

1. First limit $l$: The numerator $3\sin\theta - 4\sin^3\theta$ is exactly $\sin 3\theta$ (triple-angle identity). So

Multiply numerator and denominator by $3$:

Since $\lim_{u \to 0} \frac{\sin u}{u} = 1$, we get $l = 3$.

2. Second limit $m$: The denominator $\theta(1 - \tan^2\theta)$ suggests the double-angle formula for tangent:

Hence the numerator $2\tan\theta$ divided by $(1 - \tan^2\theta)$ is exactly $\tan 2\theta$.

So

Multiply numerator and denominator by $2$:

$m = \lim_{\theta \to 0} \frac{2\tan 2\theta}{2\theta} = 2 \cdot \lim_{v \to 0} \frac{\tan v}{v} \quad (\text{where } v = 2\theta).$ …

We start by evaluating the definite integral. The expression $\int_{0}^{3} (3x^2 - 4x + 2)\, dx$ is a straightforward polynomial integral. Once we find $k$, we plug it into the equation $3x^2 - 4x + 2 = \dfrac{3k}{5}$ and solve for $x$. The problem asks for an integer root, so we check which of the given options satisfies the equation.

1. Compute $k$ Integrate term by term:

Evaluate from $0$ to $3$:

So $k = 15$.

2. Set up the equation The problem gives:

Substitute $k = 15$:

Thus we solve:

3. Solve the quadratic Rearrange:

Factor or use the quadratic formula. Check factors:

Indeed, $(3x - 7)(x + 1) = 3x^2 + 3x - 7x - 7 = 3x^2 - 4x - 7$.

So the roots are:

$x = \frac{7}{3} \quad \text{and} \quad x = -1$ …

Concept and Intuition

When we have a polynomial equation, say $P(x) = 0$, and we want to eliminate a specific term (like the cubic term, $x^3$), we can perform a transformation on the variable $x$. The most common transformation for this purpose is a shift: $x = y + h$, where $h$ is a constant.

Geometrically, this transformation shifts the graph of the polynomial horizontally. If $x_1, x_2, \dots, x_n$ are the roots of $P(x)=0$, then $y_1 = x_1 - h, y_2 = x_2 - h, \dots, y_n = x_n - h$ are the roots of the new polynomial equation $P(y+h)=0$. By carefully choosing $h$, we can make the coefficient of a particular power of $y$ (or $x$, if we rename the variable back) equal to zero.

For a general polynomial $a_0 x^n + a_1 x^{n-1} + \dots + a_n = 0$, if we substitute $x = y+h$, the coefficient of $y^{n-1}$ (the second highest power) in the transformed equation will be $n a_0 h + a_1$. Setting this to zero allows us to find the value of $h$ that eliminates the $(n-1)$-th degree term. Once $h$ is found, we can substitute it back into the expressions for other coefficients to find their values.

Step-by-Step Solution

1. Identify the given equation and coefficients:

   The original equation is $2x^4 - 8x^3 + 3x^2 - 1 = 0$.

   Comparing this with the general quartic equation $a_0 x^4 + a_1 x^3 + a_2 x^2 + a_3 x + a_4 = 0$, we have:

   $a_0 = 2$

   $a_1 = -8$

   $a_2 = 3$

   $a_3 = 0$

   $a_4 = -1$

2. Apply the substitution $x = y+h$:

   Substitute $x = y+h$ into the original equation:

   $2(y+h)^4 - 8(y+h)^3 + 3(y+h)^2 - 1 = 0$

3. Expand the terms and collect coefficients of powers of $y$:

   We use the binomial expansion for each term:

   $(y+h)^4 = y^4 + 4hy^3 + 6h^2y^2 + 4h^3y + h^4$

   $(y+h)^3 = y^3 + 3hy^2 + 3h^2y + h^3$

   $(y+h)^2 = y^2 + 2hy + h^2$

   Substitute these expansions back into the equation:

   $2(y^4 + 4hy^3 + 6h^2y^2 + 4h^3y + h^4)$

   $- 8(y^3 + 3hy^2 + 3h^2y + h^3)$

   $+ 3(y^2 + 2hy + h^2)$

   $- 1 = 0$

   Now, collect the coefficients for each power of $y$:

   • Coefficient of $y^4$: $2$
   • Coefficient of $y^3$: $2(4h) - 8(1) = 8h - 8$
   • Coefficient of $y^2$: $2(6h^2) - 8(3h) + 3(1) = 12h^2 - 24h + 3$
   • Coefficient of $y^1$: $2(4h^3) - 8(3h^2) + 3(2h) = 8h^3 - 24h^2 + 6h$
   • Constant term: $2h^4 - 8h^3 + 3h^2 - 1$

4. Determine the value of $h$ to eliminate the cubic term:

   The problem states that the term containing the cubic power of $x$ (which is $y$ in our transformed equation) must be absent. This means its coefficient must be zero.

   Set the coefficient of $y^3$ to zero:

   $8h - 8 = 0$

   $8h = 8$

   $h = 1$

5. Calculate the coefficient of $y^2$ (which corresponds to $b$):

   The transformed equation is given as $2x^4 + bx^2 + cx + d = 0$. This means the coefficient of $x^2$ in the transformed equation is $b$. In our $y$-equation, this corresponds to the coefficient of $y^2$.

   The coefficient of $y^2$ is $12h^2 - 24h + 3$. …

The problem asks us to evaluate the limit of a sum as $n$ approaches infinity. This is a classic scenario where the sum can be converted into a definite integral using the concept of Riemann sums.

Concept: Riemann Sums and Definite Integrals

A definite integral $\int_a^b f(x) dx$ represents the area under the curve of $f(x)$ from $x=a$ to $x=b$. This area can be approximated by summing the areas of many thin rectangles. As the number of rectangles approaches infinity (and their width approaches zero), the sum of their areas converges to the exact definite integral.

The standard form for converting a limit of a sum to a definite integral is:

The key is to identify $f(x)$ and the limits of integration by matching the terms in the given sum with the general form. We look for a term of the form $\frac{k}{n}$ or $a + \frac{k}{n}$ inside the function $f$.

Step-by-step Solution

1. Identify the general term of the sum: The given limit is:

Let's examine the arguments of the sine function:
*   First term: $\frac{\pi}{4}$
*   Second term: $\frac{\pi}{12} \left( 3 + \frac{1}{n} \right)$
*   Third term: $\frac{\pi}{12} \left( 3 + \frac{2}{n} \right)$
*   ...
*   Last term: $\frac{\pi}{3}$

We can rewrite the first term to fit the pattern:
$\frac{\pi}{4} = \frac{3\pi}{12} = \frac{\pi}{12} \times 3 = \frac{\pi}{12} \left( 3 + \frac{0}{n} \right)$.
Similarly, let's see if the last term fits the pattern for some $k$.
$\frac{\pi}{3} = \frac{4\pi}{12} = \frac{\pi}{12} \times 4 = \frac{\pi}{12} \left( 3 + 1 \right) = \frac{\pi}{12} \left( 3 + \frac{n}{n} \right)$.

So, the general term inside the sine function is $\frac{\pi}{12} \left( 3 + \frac{k}{n} \right)$, where $k$ ranges from $0$ to $n$.
The sum can be written as:

2. Convert the sum into a definite integral: We have the form $\lim_{n \to \infty} \frac{1}{n} \sum_{k=0}^{n} f\left(\frac{k}{n}\right)$. We can identify $x = \frac{k}{n}$ and $f(x) = \sin \left( \frac{\pi}{12} (3 + x) \right)$. The limits of integration will be from $x=0$ (when $k=0$) to $x=1$ (when $k=n$). …

The problem asks us to evaluate a definite integral. Integrals of this form, where the integrand involves a function $f(x)$ and $f(a+b-x)$ in a symmetric way, are often solved using a powerful property of definite integrals, commonly known as King's property.

Concept: King's Property for Definite Integrals

King's property states that for a continuous function $f(x)$ on the interval $[a, b]$:

Let's apply this concept to the given integral.

1. Identify the integral and its structure: Let the given integral be $I$:

The limits of integration are $a=5$ and $b=9$.
Let $f(x) = \log 3x^2$.
We need to check if the second term in the denominator, $\log(588 - 84x + 3x^2)$, can be expressed as $f(a+b-x)$.
First, calculate $a+b-x$:
$a+b-x = 5+9-x = 14-x$.
Now, let's find $f(a+b-x) = f(14-x)$:
$f(14-x) = \log 3(14-x)^2$.
Expand the argument of the logarithm:
$3(14-x)^2 = 3(14^2 - 2 \cdot 14 \cdot x + x^2)$
$3(14-x)^2 = 3(196 - 28x + x^2)$
$3(14-x)^2 = 588 - 84x + 3x^2$.
This exactly matches the argument of the second logarithm in the denominator of the original integral.
So, the integral is indeed of the form:

where $f(x) = \log 3x^2$, $a=5$, and $b=9$.

2. Apply King's Property:

Using the property $\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$, we replace $x$ with $a+b-x = 14-x$ in the integral $I$:

From our calculation in step 1, we know that $3(14-x)^2 = 588 - 84x + 3x^2$.
And similarly, $588 - 84(14-x) + 3(14-x)^2 = 3x^2$.
Substituting these back into the transformed integral: …

The problem asks us to find the value of $a-b$ for a given cubic function $f(x) = x^3 + a x^2 + b x$, using two conditions. The core idea is to translate these conditions into algebraic equations involving $a$ and $b$, and then solve the resulting system of equations.

1. Understand the function and its derivative: The given function is $f(x) = x^3 + a x^2 + b x$. Since $f(x)$ is a polynomial, it is continuous and differentiable everywhere. We will need its derivative, $f'(x)$:

2. Use the first condition: $f(1) - f(3) = 0$ This condition means $f(1) = f(3)$. Let's calculate the values of $f(x)$ at $x=1$ and $x=3$:

Setting $f(1) = f(3)$:

Rearrange the terms to form a linear equation in $a$ and $b$:

Dividing the entire equation by 2 simplifies it: