Evaluate the integral. ∫tsin2tdt

Question

Evaluate the integral.  ∫tsin2tdt

maul124uk · Accepted Answer

Step 1  the given integral is ∫tsin2tdt  Step 2  let the integral be I.  as we now that sin2t=1−cos⁡2t2  substitute it in the integral I.  I=∫t(1−cos⁡2t2)dt  =∫t2dt−12∫tcos⁡2tdt  =12∫tdt−12∫tcos⁡2tdt  Step 3  now let I1=∫tcos⁡2tdt  now use the integration by parts to slove the integral.  I1=∫tcos⁡2tdt  =t∫cos⁡2tdt−∫(dtdt∫cos⁡2tdt)dt  =t(sin⁡2t2)−∫(1)(sin⁡2t2)dt  =tsin⁡2t2+12∫−sin⁡2tdt  =tsin⁡2t2+12(cos⁡2t2)  =tsin⁡2t2+cos⁡2t4  Step 4  now substitue the value of I1 in the integral I.  therefore,  I=12∫tdt−12(tsin⁡2t2+cos⁡2t4)  =12(t22)−tsin⁡2t4−cos⁡2t8

nghodlokl · Accepted Answer

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=t2−sin⁡(t)⋅cos⁡(t)2  Therefore:  ∫t⋅sin⁡(t)2dt=t⋅(t−sin⁡(2⋅t)2)2−∫(t2−sin⁡(2⋅t)4)dt  We find the integral  ∫t2−sin⁡2t4dt=t24+cos⁡(2⋅t)8  Answer:  ∫t⋅sin⁡(t)2=−t24+t⋅(t−sin⁡(2t)2)2−cos⁡(2t)8+C  or  ∫t⋅sin⁡(t)2=t24−t⋅sin⁡(2t)4−cos⁡(2t)8+C

nick1337 · Accepted Answer

∫t×sin⁡(t)2dt Prepare for integration by parts u=t dv=sin⁡(t)2dt Calculate the differential Determine v du=dt v=12t−sin⁡(2t)4 Substitute the values into the formula t×(12t−sin⁡(2t)4)−∫12t−sin⁡(2t)4dt Use properties of integrals t×(12t−sin⁡(2t)4)−(∫12dt−∫sin⁡(2t)4dt) Evaluate the integrals t×(12t−sin⁡(2t)4)−(t24+cos⁡(2t)8) Simplify 14t2−t×sin⁡(2t)4−cos⁡(2t)8 Add C Solution: 14t2−t×sin⁡(2t)4−cos⁡(2t)8+C

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

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