Evaluate the integrals. ∫xsin⁡x2cos8x2dx

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

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