# Please find the value for this integration. What is the answer?

## ${\int}_{0}^{\frac{\pi}{2}} {\sin}^{3} x \cos x \mathrm{dx}$

Feb 23, 2016

$u = \sin \left(x\right)$

$\mathrm{du} = \cos \left(x\right) \mathrm{dx}$

${\int}_{0}^{1} {u}^{3} \mathrm{du}$

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Feb 23, 2016

$= \frac{1}{4}$

#### Explanation:

Using the Integral

$\int {\sin}^{n} a x \cdot \cos a x \cdot \mathrm{dx} = \frac{1}{a \left(n + 1\right)} {\sin}^{n + 1} a x + C$
for $\left(n \ne - 1\right)$

Given expression is

${\int}_{0}^{\frac{\pi}{2}} {\sin}^{3} x \cdot \cos x \cdot \mathrm{dx}$ ,

here $n = 3 \mathmr{and} a = 1$
We obtain $= {\left[\frac{1}{1 \left(3 + 1\right)} {\sin}^{3 + 1} x + C\right]}_{0}^{\frac{\pi}{2}}$
$= {\left[\frac{1}{4} {\sin}^{4} x + C\right]}_{0}^{\frac{\pi}{2}}$
Inserting the value of $\sin \left(\frac{\pi}{2}\right) = 1 \mathmr{and} \sin 0 = 0$
$= \left[\frac{1}{4} \times {1}^{4} + C\right] - \left[\frac{1}{4} \times {0}^{4} + C\right]$
$= \frac{1}{4}$