Q. 53 - Calculus

Question

Solve each of the integrals in Exercises 21–70. Some integrals require substitution, and some do not. (Exercise 69 involves a hyperbolic function.)

$\int \sin x{\cos }^{7}x\mathrm{d}x$

Step-by-Step Solution

Answer

The solution of the given integral is $\int {\mathrm{sinxcos}}^7\mathrm{xdx}=-\frac{1}{8}{\cos }^8\mathrm{x}+\mathrm{C}$.

Step 1. Given Information

Solving the given integrals.

$\int \sin x{\cos }^{7}x\mathrm{d}x$

Step 2. Using the substitution method.

Let

$$\begin{gathered} u=\cos x \\ \frac{du}{dx}=-\sin x \\ du=-\sin xdx \\ -du=\sin xdx \end{gathered}$$

Step 3. This substitution changes the integral into

$$\begin{gathered} \int \sin x{\cos }^{7}x\mathrm{d}x=-\int u^7\mathrm{du} \\ \int {\mathrm{sinxcos}}^7\mathrm{xdx}=-\left[\frac{{\mathrm{u}}^{7+1}}{7+1}\right]+\mathrm{C} \\ \int {\mathrm{sinxcos}}^7\mathrm{xdx}=-\frac{{\mathrm{u}}^8}{8}+\mathrm{C} \\ \int {\mathrm{sinxcos}}^7\mathrm{xdx}=-\frac{1}{8}{\mathrm{u}}^8+\mathrm{C} \\ \int {\mathrm{sinxcos}}^7\mathrm{xdx}=-\frac{1}{8}{\cos }^8\mathrm{x}+\mathrm{C} \end{gathered}$$