Globokim8

2021-08-15

Use the Table of Integrals to evaluate the integral. (Use C for the constant of integration.)
$7{\mathrm{sin}}^{8}\left(x\right)\mathrm{cos}\left(x\right)\mathrm{ln}\left(\mathrm{sin}\left(x\right)\right)dx$
no. 101. $\int {u}^{n}\mathrm{ln}udu=\frac{{u}^{n+1}\left\{{\left(n+1\right)}^{2}\right\}\left[\left(n+1\right)\mathrm{ln}u-1\right]+C}{}$

### Answer & Explanation

unessodopunsep

Step 1
Integral: $=\int 7\left({\mathrm{sin}}^{8}x\right)\cdot \mathrm{cos}x\mathrm{ln}\left(\mathrm{sin}x\right)dx$
Let us $\mathrm{sin}x=u$
$\mathrm{cos}xdx=du$
Substituting it in integral
Integral: $=7\int {u}^{8}\mathrm{ln}udu$
In table of Integral it is in born of integral no. 101. $\int {u}^{n}\mu du=\frac{{U}^{n+1}}{{\left(n+1\right)}^{2}}\left[\left(n+1\right)\mathrm{ln}u-1\right]+c$
$n=8$ Integral: $=7\left[\frac{{u}^{8+1}}{{\left(8+1\right)}^{2}}\left[\left(8-11\right)\mu -1\right]\right]+c$
$=\frac{7{u}^{9}}{81}\left[9\mu -1\right]+c$
$=7\frac{{\mathrm{sin}}^{9}x}{81}\left[9m\left(\mathrm{sin}x\right)-1\right]+c$

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