Anabella Gilbert

2022-12-23

How to find the integral of $\int {\mathrm{sin}}^{3}xdx$ ?

Raymond Patel

Expert

$\int {\mathrm{sin}}^{3}xdx=\int \mathrm{sin}x\left(1-{\mathrm{cos}}^{2}x\right)dx\left(\because {\mathrm{sin}}^{2}x=1-{\mathrm{cos}}^{2}x\right)⇒\int {\mathrm{sin}}^{3}xdx=\int \mathrm{sin}xdx-\int \mathrm{sin}x.{\mathrm{cos}}^{2}xdx$
Solve the first integral.
$\int \mathrm{sin}xdx=\mathrm{cos}x+ C$
Solve the second integral.
Let's suppose $\mathrm{cos}x=u$
hence, $-\mathrm{sin}xdx=du$
replace the value,
$-\int \mathrm{sin}x.{\mathrm{cos}}^{2}xdx=\int {u}^{2}du⇒-\int \mathrm{sin}x.{\mathrm{cos}}^{2}xdx=\frac{{u}^{3}}{3}+C⇒-\int \mathrm{sin}x.{\mathrm{cos}}^{2}xdx=\frac{1}{3}{\mathrm{cos}}^{3}x+C$
Combining the aforementioned,$\int {\mathrm{sin}}^{3}xdx=\mathrm{cos}x+\frac{1}{3}{\mathrm{cos}}^{3}x+C$
Therefore the required value of $\int {\mathrm{sin}}^{3}xdx$ is $\mathrm{cos}x+\frac{1}{3}{\mathrm{cos}}^{3}x+C$

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