Evaluate the integral. \int t\sin^{2}tdt

Charles Kingsley 2021-12-18 Answered
Evaluate the integral.
tsin2tdt
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Expert Answer

maul124uk
Answered 2021-12-19 Author has 35 answers
Step 1
the given integral is tsin2tdt
Step 2
let the integral be I.
as we now that sin2t=1cos2t2
substitute it in the integral I.
I=t(1cos2t2)dt
=t2dt12tcos2tdt
=12tdt12tcos2tdt
Step 3
now let I1=tcos2tdt
now use the integration by parts to slove the integral.
I1=tcos2tdt
=tcos2tdt(dtdtcos2tdt)dt
=t(sin2t2)(1)(sin2t2)dt
=tsin2t2+12sin2tdt
=tsin2t2+12(cos2t2)
=tsin2t2+cos2t4
Step 4
now substitue the value of I1 in the integral I.
therefore,
I=12tdt12(tsin2t2+cos2t4)
=12(t22)tsin2t4cos2t8

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nghodlokl
Answered 2021-12-20 Author has 33 answers
The formula for integration by parts:
U(t)dV(t)=U(t)V(t)V(t)dU(t)
We put
U = t
dV=sin(t)2dt
Then:
dU = dt
V=t2sin(t)cos(t)2
Therefore:
tsin(t)2dt=t(tsin(2t)2)2(t2sin(2t)4)dt
We find the integral
t2sin2t4dt=t24+cos(2t)8
Answer:
tsin(t)2=t24+t(tsin(2t)2)2cos(2t)8+C
or
tsin(t)2=t24tsin(2t)4cos(2t)8+C

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nick1337
Answered 2021-12-28 Author has 575 answers

t×sin(t)2dt
Prepare for integration by parts
u=t
dv=sin(t)2dt
Calculate the differential
Determine v
du=dt
v=12tsin(2t)4
Substitute the values into the formula
t×(12tsin(2t)4)12tsin(2t)4dt
Use properties of integrals
t×(12tsin(2t)4)(12dtsin(2t)4dt)
Evaluate the integrals
t×(12tsin(2t)4)(t24+cos(2t)8)
Simplify
14t2t×sin(2t)4cos(2t)8
Add C
Solution:
14t2t×sin(2t)4cos(2t)8+C

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