Question #3cbbc

2 Answers
Leland Adriano Alejandro
May 8, 2016

#int_0^(pi/4) (sin x+cos x)/(3+sin 2x) dx=0.2746530521#

Explanation:

My solution is by Simpson's Rule, the Approximation Formula

#int_a^b y *dx~=#
#h/3(y_0+4*y_1+2*y_2+4*y_3+2*y_4+.....+4*y_(n-1)+y_n)#

Where #h=(b-a)/n# and #b# the upper limit and #a# the lower limit
and #n# any even number (the larger the better)

I chose
#n=20#
given #b=pi/4# and #a=0#
#h=(pi/4-0)/20=pi/80#

This is how to compute. Each #y=(sin x+cos x)/(3+sin 2x)# will use different value

for #y_0#
#x_0=(a+0*h)=(0+0*pi/80)=0#

#y_0=(sin x_0+cos x_0)/(3+sin 2x_0)#
#y_0=(sin (0)+cos (0))/(3+sin 2(0))#
#color(red)(y_0=0.3333333333333)#

for #4*y_1#
#x_1=(a+1*h)=(0+1*pi/80)=pi/80#

#4*y_1=4*(sin x_1+cos x_1)/(3+sin 2x_1)#
#4*y_1=4*(sin (pi/80)+cos (pi/80))/(3+sin (2(pi/80)))#
#color(red)(4*y_1=1.3493618978936)#

for #2*y_2#
#x_2=(a+2*h)=(0+2*pi/80)=2*pi/80#

#2*y_2=2*(sin x_2+cos x_2)/(3+sin 2x_2)#
#2*y_2=2*(sin ((2pi)/80)+cos ((2pi)/80))/(3+sin 2((2pi)/80))#
#color(red)(2*y_2=0.68138682514816)#

for #4*y_3#
#x_3=(a+3*h)=(0+3*pi/80)=3*pi/80#

#4*y_3=4*(sin x_3+cos x_3)/(3+sin 2x_3)#
#4*y_3=4*(sin ((3pi)/80)+cos ((3pi)/80))/(3+sin 2((3pi)/80))#
#color(red)(4*y_3=1.3738977832468)#

for #2*y_4#
#x_4=(a+4*h)=(0+4*pi/80)=4*pi/80#

#2*y_4=4*(sin x_4+cos x_4)/(3+sin 2x_4)#
#2*y_4=4*(sin ((4pi)/80)+cos ((4pi)/80))/(3+sin 2((4pi)/80))#
#color(red)(2*y_4=0.69151824096418)#

the rest are as follows
#color(red)(4*y_5=1.3904648494964)#
#color(red)(2*y_6=0.69821575035862)#
#color(red)(4*y_7=1.4011596185484)#
#color(red)(2*y_8=0.70242415421322)#
#color(red)(4*y_9=1.4076741205702)#
#color(red)(2*y_10=0.70489632049832)#
#color(red)(4*y_11=1.4113400771087)#
#color(red)(2*y_12=0.7062173920012)#
#color(red)(4*y_13=1.4131786935757)#
#color(red)(2*y_14=0.7068293103707)#
#color(red)(4*y_15=1.4139474301694)#
#color(red)(2*y_16=0.70705252678954)#
#color(red)(4*y_17=1.414179352209)#
#color(red)(2*y_18=0.70710341105534)#
#color(red)(4*y_19=1.4142131417552)#
#color(red)(y_20=0.35355339059328)#

The sum of all these #color(red)("sum"=20.98194762)#

#int_0^(pi/4) (sin x+cos x)/(3+sin 2x) dx=(h/3)*"sum"#

#int_0^(pi/4) (sin x+cos x)/(3+sin 2x) dx=((pi/80)/3)*20.98194762#
#int_0^(pi/4) (sin x+cos x)/(3+sin 2x) dx=color(red)(0.2746530521)#

An alternative is to simply use a graphics calculator during when complicated integration arises with a more accurate value
#color(red)(=0.2746530722)#

God bless...I hope the explanation is useful.

sente
May 9, 2016

#int_0^(pi/4)(sin(x)+cos(x))/(3+sin(2x))dx=ln(3)/4#

Explanation:

We will proceed by using substitution. First, we will go through some algebra to get the integrand into a more desirable form.

#3+sin(2x) = 3+2sin(x)cos(x)#

#= 4 + 2sin(x)cos(x) - 1#

#= 4 + 2sin(x)cos(x) - sin^2(x)-cos^2(x)#

#= 4 - (sin(x)-cos(x))^2#

#= (2 + sin(x) - cos(x))(2 - sin(x) + cos(x))#

#=> (sin(x)+cos(x)) / (3 + sin(2x)) = (sin(x)+cos(x))/((2+sin(x)-cos(x))(2-sin(x)+cos(x)))#

#=(4(sin(x)+cos(x)))/(4(2+sin(x)-cos(x))(2-sin(x)+cos(x))#

#= (sin(x)+cos(x))/4 xx#
#xx4/((2+sin(x)-cos(x))(2-sin(x)+cos(x)))#

#= (sin(x)+cos(x))/4 xx#
#xx(1/(2+sin(x)-cos(x))+1/(2-sin(x)+cos(x)))#

#=1/4xx(sin(x)+cos(x))/(2+sin(x)-cos(x))-1/4xx(-sin(x)-cos(x))/(2-sin(x)+cos(x))#

Using that, we can split the integral:

#int_0^(pi/4)(sin(x)+cos(x))/(3+sin(2x))dx =#

#=1/4int_0^(pi/4)(sin(x)+cos(x))/(2+sin(x)-cos(x))dx#

#- 1/4int_0^(pi/4)(-sin(x)-cos(x))/(2-sin(x)+cos(x))dx#

For the first integral, using the substitution #u = 2 + sin(x) - cos(x)# gives us #du = (sin(x) + cos(x))dx# and the bounds of integration change from #0# and #pi/4# to #1# and #2#. Thus, we get

#1/4int_0^(pi/4)(sin(x)+cos(x))/(2+sin(x)-cos(x))dx = int_1^2 1/udu#

#=1/4(ln|u|)_1^2#

#=1/4(ln(2)-ln(1))#

#=1/4ln(2)#

For the second integral, using the substitution #u = 2 - sin(x) + cos(x)# gives us #du = (-sin(x)-cos(x))dx# and the bounds of integration change from #0# and #pi/4# to #3# and #2#. Thus, we get

#-1/4int_0^(pi/4)(-sin(x)-cos(x))/(2-sin(x)+cos(x))dx = -1/4int_3^2 1/udu#

#=1/4int_2^3 1/udu#

#=1/4(ln(3)-ln(2))#

#=1/4(ln(3/2))#

Substituting the values in for the integrals gives us our desired result:

#int_0^(pi/4)(sin(x)+cos(x))/(3+sin(2x))dx =1/4ln(2)+1/4ln(3/2)#

#=1/4(ln(2)+ln(3/2))#

#=1/4ln(2*3/2)#

#=ln(3)/4#

Related questions
See all questions in Integrals of Polynomial functions
Impact of this question
1916 views around the world
You can reuse this answer
Creative Commons License