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

Evaluate the integral.
$\int t{\mathrm{sin}}^{2}tdt$
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maul124uk
Step 1
the given integral is $\int t{\mathrm{sin}}^{2}tdt$
Step 2
let the integral be I.
as we now that ${\mathrm{sin}}^{2}t=\frac{1-\mathrm{cos}2t}{2}$
substitute it in the integral I.
$I=\int t\left(\frac{1-\mathrm{cos}2t}{2}\right)dt$
$=\int \frac{t}{2}dt-\frac{1}{2}\int t\mathrm{cos}2tdt$
$=\frac{1}{2}\int tdt-\frac{1}{2}\int t\mathrm{cos}2tdt$
Step 3
now let ${I}_{1}=\int t\mathrm{cos}2tdt$
now use the integration by parts to slove the integral.
${I}_{1}=\int t\mathrm{cos}2tdt$
$=t\int \mathrm{cos}2tdt-\int \left(\frac{dt}{dt}\int \mathrm{cos}2tdt\right)dt$
$=t\left(\frac{\mathrm{sin}2t}{2}\right)-\int \left(1\right)\left(\frac{\mathrm{sin}2t}{2}\right)dt$
$=\frac{t\mathrm{sin}2t}{2}+\frac{1}{2}\int -\mathrm{sin}2tdt$
$=\frac{t\mathrm{sin}2t}{2}+\frac{1}{2}\left(\frac{\mathrm{cos}2t}{2}\right)$
$=\frac{t\mathrm{sin}2t}{2}+\frac{\mathrm{cos}2t}{4}$
Step 4
now substitue the value of ${I}_{1}$ in the integral I.
therefore,
$I=\frac{1}{2}\int tdt-\frac{1}{2}\left(\frac{t\mathrm{sin}2t}{2}+\frac{\mathrm{cos}2t}{4}\right)$
$=\frac{1}{2}\left(\frac{{t}^{2}}{2}\right)-\frac{t\mathrm{sin}2t}{4}-\frac{\mathrm{cos}2t}{8}$

nghodlokl
The formula for integration by parts:
$\int U\left(t\right)\cdot dV\left(t\right)=U\left(t\right)\cdot V\left(t\right)-\int V\left(t\right)\cdot dU\left(t\right)$
We put
U = t
$dV={\mathrm{sin}\left(t\right)}^{2}dt$
Then:
dU = dt
$V=\frac{t}{2}-\mathrm{sin}\left(t\right)\cdot \frac{\mathrm{cos}\left(t\right)}{2}$
Therefore:
$\int t\cdot {\mathrm{sin}\left(t\right)}^{2}dt=\frac{t\cdot \left(t-\frac{\mathrm{sin}\left(2\cdot t\right)}{2}\right)}{2}-\int \left(\frac{t}{2}-\frac{\mathrm{sin}\left(2\cdot t\right)}{4}\right)dt$
We find the integral
$\int \frac{t}{2}-\frac{\mathrm{sin}2t}{4}dt=\frac{{t}^{2}}{4}+\frac{\mathrm{cos}\left(2\cdot t\right)}{8}$
$\int t\cdot {\mathrm{sin}\left(t\right)}^{2}=-\frac{{t}^{2}}{4}+\frac{t\cdot \left(t-\frac{\mathrm{sin}\left(2t\right)}{2}\right)}{2}-\frac{\mathrm{cos}\left(2t\right)}{8}+C$
or
$\int t\cdot {\mathrm{sin}\left(t\right)}^{2}=\frac{{t}^{2}}{4}-t\cdot \frac{\mathrm{sin}\left(2t\right)}{4}-\frac{\mathrm{cos}\left(2t\right)}{8}+C$

nick1337

$\int t×\mathrm{sin}\left(t{\right)}^{2}dt$
Prepare for integration by parts
u=t
$dv=\mathrm{sin}\left(t{\right)}^{2}dt$
Calculate the differential
Determine v
du=dt
$v=\frac{1}{2}t-\frac{\mathrm{sin}\left(2t\right)}{4}$
Substitute the values into the formula
$t×\left(\frac{1}{2}t-\frac{\mathrm{sin}\left(2t\right)}{4}\right)-\int \frac{1}{2}t-\frac{\mathrm{sin}\left(2t\right)}{4}dt$
Use properties of integrals
$t×\left(\frac{1}{2}t-\frac{\mathrm{sin}\left(2t\right)}{4}\right)-\left(\int \frac{1}{2}dt-\int \frac{\mathrm{sin}\left(2t\right)}{4}dt\right)$
Evaluate the integrals
$t×\left(\frac{1}{2}t-\frac{\mathrm{sin}\left(2t\right)}{4}\right)-\left(\frac{{t}^{2}}{4}+\frac{\mathrm{cos}\left(2t\right)}{8}\right)$
Simplify
$\frac{1}{4}{t}^{2}-\frac{t×\mathrm{sin}\left(2t\right)}{4}-\frac{\mathrm{cos}\left(2t\right)}{8}$
Solution:
$\frac{1}{4}{t}^{2}-\frac{t×\mathrm{sin}\left(2t\right)}{4}-\frac{\mathrm{cos}\left(2t\right)}{8}+C$