zagonek34

2021-12-29

Evaluate the following integral.
$\int {x}^{3}\mathrm{sin}\left({x}^{4}\right){\mathrm{cos}}^{5}\left({x}^{4}\right)dx$

### Answer & Explanation

Joseph Fair

Step 1
Consider the given integral
$\int {x}^{3}\mathrm{sin}\left({x}^{4}\right){\mathrm{cos}}^{5}\left({x}^{4}\right)dx$
Step 2
Use substitution method to evaluate the integral
Let $u=\mathrm{cos}\left({x}^{4}\right)$
Then $du=-4{x}^{3}\mathrm{sin}\left({x}^{4}\right)dx$
$⇒{x}^{3}\mathrm{sin}\left({x}^{4}\right)dx=-\frac{1}{4}du$
Step 3
Then the integral becomes
$\int {x}^{3}\mathrm{sin}\left({x}^{4}\right){\mathrm{cos}}^{5}\left({x}^{4}\right)dx=\int {u}^{5}\left(-\frac{1}{4}\right)du$
$=-\frac{1}{4}\int {u}^{5}du$
$=-\frac{1}{4}\left[\frac{{u}^{6}}{6}\right]$
$=-\frac{{u}^{6}}{24}$
Substituting $u=\mathrm{cos}\left({x}^{4}\right)$, we get $-\frac{{\mathrm{cos}}^{6}\left({x}^{4}\right)}{24}$
$\therefore \int {x}^{3}\mathrm{sin}\left({x}^{4}\right){\mathrm{cos}}^{5}\left({x}^{4}\right)dx=-\frac{{\mathrm{cos}}^{6}\left({x}^{4}\right)}{24}+C$,
where C is the constant of integration.

Heather Fulton

$\int {x}^{3}\mathrm{sin}\left({x}^{4}\right){\mathrm{cos}}^{5}\left({x}^{4}\right)dx$
$4\cdot {x}^{3}dx=d\left({x}^{4}\right),t={x}^{4}$
$\int \mathrm{sin}\left(t\right)\cdot \frac{{\mathrm{cos}\left(t\right)}^{5}}{4}dt$
$\left(-\mathrm{sin}\left(x\right)\right)dx=d\left(\mathrm{cos}\left(x\right)\right),u=\mathrm{cos}\left(x\right)$
$\int \left(-\frac{{u}^{5}}{4}\right)du$
$\int -\frac{{u}^{5}}{4}du=-\frac{{u}^{6}}{24}+C$
$-\frac{{\mathrm{cos}\left(t\right)}^{6}}{24}+C$
$-\frac{{\mathrm{cos}\left({x}^{4}\right)}^{6}}{24}+C$

Vasquez