Mathematics Test - 14

\(\int \frac{\sin ^{8} x-\cos ^{8} x}{1-2 \sin ^{2} x \cos ^{2} x} d x\)
\(=\int \frac{\left(\sin ^{2} x-\cos ^{2} x\right)\left(\sin ^{4} x+\cos ^{4} x\right)}{1-2 \sin ^{2} x \cos ^{2} x}\)
\(=\int-\cos 2 x d x=-\frac{1}{2} \sin 2 x+C\)
Standard Integrals
(1) Algebric Function
(i) \(\int x^{n} \mathrm{dx}=\frac{x^{n+1}}{n+1}+c, n \neq-1\)
(ii) \(\int(\mathrm{ax}+b)^{n} \mathrm{dx}=\frac{1}{a} \cdot \frac{(\mathrm{ax}+b)^{n+1}}{n+1}+c \cdot n \neq-1\)
(2) Fraction Function
(i) \(\int \frac{1}{x} \mathrm{dx}=\log |x|+c\)
(ii) \(\int \frac{1}{\operatorname{ax}+b} \mathrm{dx}=\frac{1}{a}(\log |\mathrm{ax}+b|)+c\)
(3) \(\int \mathrm{e}^{x} \mathrm{dx}=\mathrm{e}^{x}+c\)
(4) \(\int a^{x} \mathrm{dx}=\frac{a^{x}}{\log _{e} a}+c\)
(5) \(\int \sin x d x=-\cos x+c\)
(6) \(\int \cos x d x=\sin x+c\)
(7) \(\int \sec ^{2} x d x=\tan x+c\)
(8) \(\int \operatorname{cosec}^{2} x d x=-\cot x+c\)
(9) \(\int \sec x \tan x d x=\sec x+c\)
\((10) \int \operatorname{cosec} x \cot x d x=-\operatorname{cosec} x+c\)
(11)\(\int \tan x d x=-\log |\cos x|+c=\log |\sec x|+c\)
(12)\(\int_{+c}^{(1)} \cot x \mathrm{d} \mathrm{x}=\log |\sin x|+c=-\log |\operatorname{cosec} x|\)
\(+c\)
In any of the fundamental integration formulae, if \(x\) is replaced by \(a x+b,\) then the same formulae is applicable but we must divide by coefficient of \(x\) or derivative of \((a x+b)\) i.e.
a. In general, if
\(\int f(x) \mathrm{d} \mathrm{x}=\phi(x)+c,\) then \(\int f(\mathrm{ax}+b) \mathrm{d} \mathrm{x}\)
\(=\frac{1}{a} \phi(\mathrm{ax}+b)+c\)
\(\int \sin (a x+b) \mathrm{d} \mathrm{x}=\frac{-1}{a} \cos (\mathrm{ax}+b)+c\)
\(\int \sec (\mathrm{ax}+b) \mathrm{dx}=\frac{1}{a}\)
\(\log |\sec (\mathrm{ax}+b)+\tan (\mathrm{ax}+b)|+c\)
Some more results:
(i) \(\int \frac{1}{x^{2}-a^{2}}=\frac{1}{2 a} \log \left|\frac{x-a}{x+a}\right|+c\)
(ii) \(\int \frac{1}{a^{2}-x^{2}} \mathrm{dx}=\frac{1}{2 a} \log \left|\frac{a+x}{a-x}\right|+c\)
(iii) \(\int \frac{\mathrm{d} \mathrm{x}}{\sqrt{x^{2}-a^{2}}}=\log \{|x+\sqrt{x^{2}-a^{2}}|\}+c\)
(iv) \(\int \frac{\mathrm{d} x}{\sqrt{x^{2}+a^{2}}}=\log \{|x+\sqrt{x^{2}+a^{2}}|\}+c\)
(v) \(\int \sqrt{a^{2}-x^{2}} \mathrm{dx}=\frac{1}{2} x \sqrt{a^{2}-x^{2}}+\frac{1}{2} a^{2}\)
\(\sin ^{-1}\left(\frac{x}{a}\right)+c\)
(vi) \(\int \sqrt{x^{2}-a^{2}} \mathrm{dx}=\frac{1}{2} x \sqrt{x^{2}-a^{2}}-\frac{1}{2} a^{2} \log\)
\(|x+\sqrt{x^{2}+a^{2}}|+c\)
(vii) \(\int \sqrt{x^{2}+a^{2}} \mathrm{dx}=\frac{1}{2} x \sqrt{x^{2}+a^{2}}+\frac{1}{2} a^{2} \log\)
\(|x+\sqrt{x^{2}+a^{2}}|+c\)

The equation of all the straight lines passing through origin (0, 0) is

y=mx ...(i)

Hence, required differential equation of all such lines is

y=( dy / dx )x ( ∵m=dy / dx )

Formation of Differential Equations by Eliminating Arbitrary Constants

Properties:

• The order of differential equation is equal to the number of arbitrary constants in the given relation.

• The differential equation is consistent with the relation.

• The differential equation is free from arbitrary constants.

• Formulating a differential equation from a given equation representing a family of curves means finding a differential equation whose solution is the given equation. The equation so obtained is the differential equation of order n for the family of given curves.

• Algorithm for formation of differential equations

• Step (i): Write the given equation involving independent variable x(say), dependent variable y (say) and the arbitrary constants.

• Step (ii): Obtain the number of arbitrary constants in step (i). Let there be n arbitrary constants.

• Steps (iii): Differentiate the relation in step (i) n times with respect to x.

• Step (iv): Eliminate arbitrary constant with the help of n equations involving differential coefficients obtained in step (iii) and an equation in step (i). The equation so obtained is the desired differential equation.

Given, p: Ravi races, q: Ravi wins

∴ The statement of given proposition ∼(p∨(∼q)) is

"It is not true that Ravi races or that Ravi does not win."

Negation of a statement in mathematical reasoning

The denial of a statement p is called its negation, written as ~ p~ p. Negation of any statement p is formed by writing "It is not the case that ..... "or " It is false that......." before p or, if possible by inserting in p the word "not".

Negation is called a connective although it does not combine two or more statements. In fact, it only modifies a statement.

Negation of compound statements:

We have learnt about negation of a simple statement. Writing the negation of compound statements having conjunction, disjunctions, implication, equivalence, etc, is not very simple. So, let us discuss the negation of compound statement.

(i) Negation of conjunction:If p and q are two statements, then ∼ (p∧q )≡(∼p∨∼q)∼ p∧q ≡(∼p∨∼q)

(ii) Negation of disjunction:If p and q are two statements, then ∼ (p∨q )≡(∼p∧ ∼q)∼ p∨q ≡(∼p∧ ∼q)

(iii) Negation of implication:If p and q are two statements, then ∼(p⇒q)≡∼p⇒q≡∼(∼p∨q)≡∼(∼p∨q)≡(p ∧ ∼q)p ∧ ∼q

(iv) Negation of biconditional statement:If p and q are two statements, then∼(p ⇔q)≡(p ∧ ∼q) ∨(q ∧ ∼p) ≡ ∼p⇔q≡p⇔∼q

​The area bounded by a cartesian curve y = f (x), x - axis and ordinates x = a and x = b is given by Area = ∫ a b y dx = ∫ a b f x dx

Given equation is:

\(\frac{d y}{d x}=\frac{x \log x^{2}+x}{\sin y+y \cos y}\)

\(\Rightarrow (\sin y+y \cos y) d y=\left(x \log x^{2}+x\right) d x\)

On intearatina both sides, we get:

\( \Rightarrow -\cos y+y \sin y+\cos y=\frac{x^{2}}{2} \log x^{2}-\int \frac{x^{2}}{2} \cdot \frac{1}{x^{2}} \cdot 2 x d x+\int x d x+C\)

\(\Rightarrow y \sin y=\frac{x^{2}}{2} \cdot 2 \log x-\int x d x+\int x d x+C\)

\(\Rightarrow y \sin y=x^{2} \log x+C\)

t₁ = 1 + 1/1 = 2

t₂ = 1 + ½ = 3/2

t₁ x t₂ = 2 x 3/2 = 3

t₁ x t₂ x t₃ = 3 x 4/3 = 4

Similarly, t₁ x t₂ …….. txn = n + 1

Thus, t₁ x t₂ x t₃ ……. x t₂₅ = 25 + 1 = 26