91影视

The given integral is$鈭�{\sin }^{13}\left(x\right){\cos }^5\left(x\right)dx$.

• Use the trigonometric identities to rewrite the integral as follows:

$鈭�{\sin }^{13}\left(x\right){\cos }^5\left(x\right)dx=鈭�{\left(1-{\cos }^2\left(x\right)\right)}^6\sin \left(x\right){\cos }^5\left(x\right)dx$

• Consider role="math" localid="1649012126134" $\cos x=u$, so $-\sin xdx=du$.
• Substitute and simplify the integral.

$\begin{array}{rcl}鈭�{\left(1-{\cos }^2\left(x\right)\right)}^6\sin \left(x\right){\cos }^5\left(x\right)dx& =& 鈭�鈥�-u^5{\left(1-u^2\right)}^{6}du\\ & =& -鈭�\left(1-6u^2+15u^4-20u^6+15u^8-6u^{10}+u^{12}\right)u^{5}du\\ & =& -鈭�\left(u^5-6u^7+15u^9-20u^{11}+15u^{13}-6u^{15}+u^{17}\right)du\end{array}$

• Apply the sum rule on the obtained integral.

$=-鈭�\left(u^5-6u^7+15u^9-20u^{11}+15u^{13}-6u^{15}+u^{17}\right)du==-鈭�鈥�u^{5}du+鈭�鈥�6u^{7}du-鈭�鈥�15u^{9}du+鈭�鈥�20u^{11}du-鈭�鈥�15u^{13}du+鈭�鈥�6u^{15}du-鈭�鈥�u^{17}du$

• Perform integration on the obtained integral.

$-鈭�鈥�u^{5}du+鈭�鈥�6u^{7}du-鈭�鈥�15u^{9}du+鈭�鈥�20u^{11}du-鈭�鈥�15u^{13}du+鈭�鈥�6u^{15}du-鈭�鈥�u^{17}du=-\frac{u^6}{6}+\frac{3u^8}{4}-\frac{3u^{10}}{2}+\frac{5u^{12}}{3}-\frac{15u^{14}}{14}+\frac{3u^{16}}{8}-\frac{u^{18}}{18}$

• Substitute $u=\cos x$into the obtained integral to find the solution.
• So, the value of the integral is$-\frac{{\cos }^6\left(x\right)}{6}+\frac{3{\cos }^8\left(x\right)}{4}-\frac{3{\cos }^{10}\left(x\right)}{2}+\frac{5{\cos }^{12}\left(x\right)}{3}-\frac{15{\cos }^{14}\left(x\right)}{14}+\frac{3{\cos }^{16}\left(x\right)}{8}-\frac{{\cos }^{18}\left(x\right)}{18}$