Paganellash

2021-12-03

Trigonometric integrals. Evaluate the following integrals.
${\int }_{0}^{\pi }{\left(1-\mathrm{cos}2x\right)}^{\frac{3}{2}}dx$

Elizabeth Witte

Step 1
To evaluate the given integrals , ${\int }_{0}^{\pi }{\left(1-\mathrm{cos}2x\right)}^{\frac{3}{2}}dx$.
Solution:
Let us denote the given integral by I as,
$I={\int }_{0}^{\pi }{\left(1-\mathrm{cos}2x\right)}^{\frac{3}{2}}dx$
Since we know the identity as, $1-\mathrm{cos}2x={\mathrm{sin}}^{2}x$,
So the integral becomes,
$I={\int }_{0}^{\pi }{\left({\mathrm{sin}}^{2}x\right)}^{\frac{3}{2}}dx$
So further solving the integral gives,
Step 2
$I={\int }_{0}^{\pi }\left({\mathrm{sin}}^{3}x\right)dx$
$={\int }_{0}^{\pi }\left({\mathrm{sin}}^{2}x\right)\mathrm{sin}xdx$
$={\int }_{0}^{\pi }\left(1-{\mathrm{cos}}^{2}x\right)\left(\mathrm{sin}x\right)dx$
Now using Substitution method, Put $cosx=t$, we get $-\mathrm{sin}xdx=dt$
Therefore, the given integral changes with upper limit, $\mathrm{cos}\pi =-1$ and lower limit, $\mathrm{cos}0=1$ as,
$I=-{\int }_{1}^{-1}\left(1-{t}^{2}\right)dt$
$I=-{\left(t-\frac{{t}^{3}}{3}\right)}_{1}^{-1}$
Step 3
Further, solving the integral gives,
$I={\left(\frac{{t}^{3}}{3}-t\right)}_{1}^{-1}$
$=\left(\frac{-1}{3}+1\right)-\left(\frac{1}{3}-1\right)$
$=\frac{2}{3}+\frac{2}{3}$
$=\frac{4}{3}$
Hence, the value of the integral is $\frac{4}{3}$.

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