petrusrexcs

2021-12-29

Evaluate the integrals.
$\int x{\mathrm{sin}x}^{2}{\mathrm{cos}}^{8}{x}^{2}dx$

Bertha Jordan

Step 1
Given- $\int x{\mathrm{sin}x}^{2}{\mathrm{cos}}^{8}{x}^{2}dx$
To find- The value of the above integral.
Concept Used- The above integral can be solved by substitution method.
Step 2
Explanation- Rewrite the given expression as,
$I=\int x{\mathrm{sin}x}^{2}{\mathrm{cos}}^{8}{x}^{2}dx$
Now substituting $\mathrm{cos}{x}^{2}=t$ and differentiating both sides w.r.t. x, we get,
$\mathrm{cos}{x}^{2}=t$
$-{\mathrm{sin}x}^{2}\cdot 2xdx=dt$
$x{\mathrm{sin}x}^{2}dx=\frac{-dt}{2}$
So, from the above integral, we can write as,
$I=\int {t}^{8}\cdot \left(\frac{-dt}{2}\right)$
$=\frac{-1}{2}\cdot \int {t}^{8}dt$
$=\frac{-1}{2}\cdot \frac{{t}^{9}}{9}+C$
$=\frac{-{t}^{9}}{18}+C$
Step 3
As previously, we have substituted $\mathrm{cos}{x}^{2}=t$, so the solution of the integeral can be written as,
$I=\frac{-{\left({\mathrm{cos}x}^{2}\right)}^{9}}{18}+C$
$=\frac{-{\mathrm{cos}}^{9}{x}^{2}}{18}+C$, where C is arbitrary constant.
So, the solution of the integral is $\frac{-{\mathrm{cos}}^{9}{x}^{2}}{18}+C$.
Answer- Hence, the solution of the integral $\int x{\mathrm{sin}x}^{2}{\mathrm{cos}}^{8}{x}^{2}dx$ is $\frac{-{\mathrm{cos}}^{9}{x}^{2}}{18}+C$, where C is an arbitrary constant.

Neil Dismukes

$\int x{\mathrm{cos}}^{8}\left({x}^{2}\right)\mathrm{sin}\left({x}^{2}\right)dx$
$=-\frac{1}{2}\int {u}^{8}du$
$\int {u}^{8}du$
$=\frac{{u}^{9}}{9}$
$-\frac{1}{2}\int {u}^{8}du$
$=-\frac{{u}^{9}}{18}$
$=-\frac{{\mathrm{cos}}^{9}\left({x}^{2}\right)}{18}$
$\int x{\mathrm{cos}}^{8}\left({x}^{2}\right)\mathrm{sin}\left({x}^{2}\right)dx$
$=-\frac{{\mathrm{cos}}^{9}\left({x}^{2}\right)}{18}+C$

karton

$\begin{array}{}\int x×\mathrm{sin}\left({x}^{2}\right)\mathrm{cos}\left({x}^{2}{\right)}^{8}dx\\ \int -\frac{{t}^{8}}{2}dt\\ -\frac{1}{2}×\int {t}^{8}dt\\ -\frac{1}{2}×\frac{{t}^{9}}{9}\\ -\frac{1}{2}×\frac{\mathrm{cos}\left({x}^{2}{\right)}^{9}}{9}\\ -\frac{\mathrm{cos}\left({x}^{2}{\right)}^{9}}{18}\end{array}$