Trigonometric integrals. Evaluate the following integrals. ∫sin3xcos5xdx

Step 1
Given: The integral ${\displaystyle \int {\sin }^{3}x{\cos }^{5}xdx}$
To evaluate: the integration of given trigonometric integral.
Step 2
Explanation:
Let ${\displaystyle I=\int {\sin }^{3}x{\cos }^{5}xdx}$
This can be re-written as
${\displaystyle I=\int \sin x{\sin }^{2}x{\cos }^{5}xdx}$
${\displaystyle \Rightarrow I=\int \sin x(1-{\cos }^{2}x){\cos }^{5}xdx[\because {\sin }^{2}x+{\cos }^{2}x=1]}$
${\displaystyle \Rightarrow I=\int \sin x({\cos }^{5}x-{\cos }^{7}x)dx}$
Now substituting $\cos x=t$
${\displaystyle \Rightarrow -\sin xdx=dt}$
${\displaystyle \Rightarrow \sin xdx=-dt}$
Hence
${\displaystyle I=\int (t^5-t^7)(-dt)}$
${\displaystyle \Rightarrow I=-\int (t^5-t^7)dt}$
${\displaystyle \Rightarrow I=-(\frac{t^6}{6}-\frac{t^8}{8})+C}$
${\displaystyle \Rightarrow I=\frac{t^8}{8}-\frac{t^6}{6}+C}$
Now as ${\displaystyle t=\cos x}$
Therefore
${\displaystyle I=\frac{{\cos }^{8}x}{8}-\frac{{\cos }^{6}x}{6}+C}$ where C is arbitrary constant and also known as "constant of integration"
Answer: ${\displaystyle \int {\sin }^{3}x{\cos }^{5}xdx=\frac{{\cos }^{8}x}{8}-\frac{{\cos }^{6}x}{6}+C}$ where C is constant of integration.

$\begin{array}{}\int \sin (x{)}^3\cos (x{)}^{5}dx\\ \int t^3-2t^5+t^{7}dt\\ \int t^{3}dt-\int 2t^{5}dt+\int t^{7}dt\\ \frac{t^4}{4}-\frac{t^6}{3}+\frac{t^8}{8}\\ \frac{\sin (x{)}^4}{4}-\frac{\sin (x{)}^6}{3}+\frac{\sin (x{)}^8}{8}\\ \frac{\sin (x{)}^4}{4}-\frac{\sin (x{)}^6}{3}+\frac{\sin (x{)}^8}{8}+C\end{array}$