Mathematics: In a triangle, measure of one angle is 60º. The measure of one of the remaining two angles is twice the other. Find the measure of the smallest angle of the triangle. CAN YOU GUYS PLEASE ANSWER THIS FAST! THANKYOU

Step-by-step explanation:Partial fraction:Given integration is (x²-4)= (x+2)(x-2)LetTherefore,A+B=0 ⇒2A+2B=0...(1) and 2B-2A=1.....(2)Adding (1) and (2)2A+2B+2B-2A=0+1⇒4B=1Putting the value of B in equation (1) we getTherefore, [ c is arbitrary constant]Trigonometric substitution:Given integration is Let x = 2 sinθ⇒dx = 2 cosθ dθ [ multiply the numerator and denominator by ]let , [ c is arbitrary constant]Since Putting the value of secθ and tanθ...

In a triangle, measure of one angle is 60º. The measure

Ответ:

igtguith

27.05.2023

Mathematics

$I=\int \frac{5 \ dx}{x^2-4}$ $=-\frac54 ln |x+2|+\frac54 ln |x+2|+c$

Step-by-step explanation:

Partial fraction:

Given integration is

$I=\int \frac{5 \ dx}{x^2-4}$

(x²-4)= (x+2)(x-2)

$\frac{1}{x^2-4}=\frac{1}{(x+2)(x-2)}$

Let

$\frac{1}{(x+2)(x-2)}=\frac A{x+2}+\frac{B}{x-2}$

$\Rightarrow \frac{1}{(x+2)(x-2)}=\frac{ A(x-2)+B(x+2)}{(x+2)(x-2}$

$\Rightarrow \frac{1}{(x+2)(x-2)}=\frac{ Ax-2A+Bx+2B}{(x+2)(x-2)}$

$\Rightarrow \frac{1}{(x+2)(x-2)}=\frac{ (A+B)x-2A+2B}{(x+2)(x-2)}$

$\Rightarrow 1 = (A+B)x-2A+2B$

Therefore,

A+B=0 ⇒2A+2B=0...(1) and 2B-2A=1.....(2)

Adding (1) and (2)

2A+2B+2B-2A=0+1

⇒4B=1

$\Rightarrow B=\frac14$

Putting the value of B in equation (1) we get

$A=-B=-\frac14$

Therefore,

$I=\int \frac{5 \ dx}{x^2-4}$

$=5[\int -\frac{1}{4(x+2)}dx+\int \frac{1}{4(x-2)}dx]$

$=-\frac54 ln |x+2|+\frac54 ln |x+2|+c$ [ c is arbitrary constant]

Trigonometric substitution:

Given integration is

$I=\int \frac{5 \ dx}{x^2-4}$

Let x = 2 sinθ

⇒dx = 2 cosθ dθ

$= \int \frac{5}{(2sin\theta)^2-4}dx$

$=\int \frac{5}{-4(1-sin^2\theta)}(2 cos \theta) d\theta$

$=-\frac52\int \frac{cos \theta}{cos^2\theta}d\theta$

$=-\frac52\int \frac{1}{cos\theta}d\theta$

$=-\frac52\int sec\theta \ d\theta$

$=-\frac52\int sec\theta\times \frac{(sec\theta+tan\theta)}{(sec\theta+tan\theta)}d\theta$ [ multiply the numerator and denominator by $(sec\theta+tan\theta)$ ]