Question

# Use the Table of Integrals to evaluate the integral. (Use C for the constant of integration.)7\sin^{8}(x)\cos(x)\ln(\sin(x))dxno. 101. \int u^{n}\ln u du=\frac{u^{n+1}{(n+1)^{2}}[(n+1)\ln u-1]+C

Integrals

Use the Table of Integrals to evaluate the integral. (Use C for the constant of integration.)
$$7\sin^{8}(x)\cos(x)\ln(\sin(x))dx$$

no. 101. $$\displaystyle\int{u}^{{{n}}} \ln{{u}}{d}{u}={\frac{{{u}^{{{n}+{1}}}}}{{{\left({n}+{1}\right)}^{{{2}}}}}}{\left[{\left({n}+{1}\right)} \ln{{u}}-{1}\right]}+{C}$$

2021-05-18
Step 1
Integral: $$=\int7(\sin^{8}x)\cdot\cos x\ln(\sin x) dx$$
Let us $$\sin x=u$$
$$\cos x dx=du$$
Substituting it in integral
Integral: $$=7\int u^{8}\ln u du$$
In table of Integral it is in born of integral no. 101. $$\int u^{n} mudu=\frac{U^{n+1}}{(n+1)^{2}}[(n+1)\ln u-1]+c$$
$$n=8$$ Integral: $$=7\left[\frac{u^{8+1}}{(8+1)^{2}}[(8-11)mu-1]\right]+c$$
$$=\frac{7u^{9}}{81}[9mu-1]+c$$
$$=7\frac{\sin^{9}x}{81}[9m(\sin x)-1]+c$$