Math Question and Answers


Question and Answers Forum

All Questions Topic List
Integration Questions
Previous in All Question Next in All Question
Previous in Integration Next in Integration
Question Number 118318 by mnjuly1970 last updated on 16/Oct/20

	Answered by Bird last updated on 17/Oct/20
	$A=∫_0 ^∞ ((sin^4 x)/x^2 )dx by parts A =[−((sin^4 x)/x)]_0 ^∞ +∫_0 ^∞ ((4sin^3 x cosx)/x)dx =4 ∫_0 ^∞ ((sin^3 x cosx)/x)dx sin^3 x =sinx((1−cos(2x))/2) =(1/2)sinx−(1/2)sinx cos(2x) sinx cos(2x)=cos(2x)cos((π/2)−x) =(1/2){cos(x+(π/2))+cos(3x−(π/2))} =(1/2){sin(3x)−sinx} ⇒ sin^3 x=(1/2)sinx−(1/4)sin(3x)+(1/4)sinx =(3/4)sin(x)−(1/4)sin(3x) ⇒ I =4∫_0 ^∞ (3/(4x))sin(x)cosxdx −∫_0 ^∞ ((sin(3x)cosx)/x)dx =(3/2)∫_0 ^∞ ((sin(2x))/x)dx−∫_0 ^∞ ((sin(3x)cosx)/x)dx but sin(3x)cos(x) =cos((π/2)−3x)cos(x) =(1/2){cos((π/2)−2x)+cos((π/2)−4x)} =(1/2){sin(2x)−sin(4x)} ⇒ I =(3/2)∫_0 ^∞ ((sin(2x))/x)dx−(1/2)∫_0 ^∞ ((sin(2x))/x)dx +(1/2)∫_0 ^∞ ((sin(4x))/x)dx =∫_0 ^∞ ((sin(2x))/x)dx+(1/2)∫_0 ^∞ ((sin(4x))/x)dx we have ∫_0 ^∞ ((sin(2x))/x)dx =_(2x=t) ∫_0 ^∞ ((sint)/(t.2^(−1) ))2^(−1) dt =(π/2) ∫_0 ^∞ ((sin(4x))/x)dx =_(4x=t) ∫_0 ^∞ ((sint)/(4^(−1) t))4^(−1) dt =(π/2) ⇒ I =(π/2) +(π/4) =((3π)/4)$
	$A = \int_{0}^{\infty} \frac{{s i n}^{4} x}{x^{2}} d x b y p a r t s$ $A = {[- \frac{{s i n}^{4} x}{x}]}_{0}^{\infty} + \int_{0}^{\infty} \frac{4 {s i n}^{3} x c o s x}{x} d x$ $= 4 \int_{0}^{\infty} \frac{{s i n}^{3} x c o s x}{x} d x$ ${s i n}^{3} x = s i n x \frac{1 - c o s (2 x)}{2}$ $= \frac{1}{2} s i n x - \frac{1}{2} s i n x c o s (2 x)$ $s i n x c o s (2 x) = c o s (2 x) c o s (\frac{π}{2} - x)$ $= \frac{1}{2} {c o s (x + \frac{π}{2}) + c o s (3 x - \frac{π}{2})}$ $= \frac{1}{2} {s i n (3 x) - s i n x} \Rightarrow$ ${s i n}^{3} x = \frac{1}{2} s i n x - \frac{1}{4} s i n (3 x) + \frac{1}{4} s i n x$ $= \frac{3}{4} s i n (x) - \frac{1}{4} s i n (3 x) \Rightarrow$ $I = 4 \int_{0}^{\infty} \frac{3}{4 x} s i n (x) c o s x d x$ $- \int_{0}^{\infty} \frac{s i n (3 x) c o s x}{x} d x$ $= \frac{3}{2} \int_{0}^{\infty} \frac{s i n (2 x)}{x} d x - \int_{0}^{\infty} \frac{s i n (3 x) c o s x}{x} d x$ $b u t s i n (3 x) c o s (x)$ $= c o s (\frac{π}{2} - 3 x) c o s (x)$ $= \frac{1}{2} {c o s (\frac{π}{2} - 2 x) + c o s (\frac{π}{2} - 4 x)}$ $= \frac{1}{2} {s i n (2 x) - s i n (4 x)} \Rightarrow$ $I = \frac{3}{2} \int_{0}^{\infty} \frac{s i n (2 x)}{x} d x - \frac{1}{2} \int_{0}^{\infty} \frac{s i n (2 x)}{x} d x$ $+ \frac{1}{2} \int_{0}^{\infty} \frac{s i n (4 x)}{x} d x$ $= \int_{0}^{\infty} \frac{s i n (2 x)}{x} d x + \frac{1}{2} \int_{0}^{\infty} \frac{s i n (4 x)}{x} d x$ $w e h a v e \int_{0}^{\infty} \frac{s i n (2 x)}{x} d x$ $=_{2 x = t} \int_{0}^{\infty} \frac{s i n t}{t . 2^{- 1}} 2^{- 1} d t = \frac{π}{2}$ $\int_{0}^{\infty} \frac{s i n (4 x)}{x} d x =_{4 x = t} \int_{0}^{\infty} \frac{s i n t}{4^{- 1} t} 4^{- 1} d t$ $= \frac{π}{2} \Rightarrow I = \frac{π}{2} + \frac{π}{4} = \frac{3 π}{4}$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum