Question

Trigonometric integrals: The integrals that follow can be solved by using algebra to write the integrands in the form $f\text{'}\left(u\right(x\left)\right)u\text{'}\left(x\right)$so that $u-$substitution will apply.

Solve$\int {\sin }^{5}x\mathrm{d}x$ by using the Pythagorean identity ${\sin }^{2}x+{\cos }^{2}x=1$ to rewrite the integrand as role="math" localid="1649173494208" ${(1-{\cos }^{2}x)}^2\sin x$ and then applying substitution with $u=\cos x$.

Short answer

The value of integral is $\int {\sin }^5\mathrm{xdx}=-\cos x+\frac{2}{3}{\cos }^{3}x-\frac{1}{5}{\cos }^{5}x+C$.

Step-by-step solution

Step 1. Given Information 

Solve$\int {\sin }^{5}x\mathrm{d}x$ by using the Pythagorean identity ${\sin }^{2}x+{\cos }^{2}x=1$ to rewrite the integrand as ${(1-{\cos }^{2}x)}^2\sin x$ and then applying substitution with $u=\cos x$.

Step 2. The given integral is ∫sin5xdx.

We can write as

$\int {\sin }^{5}x\mathrm{d}x=\int {(1-{\cos }^{2}x)}^2\sin x\mathrm{d}x$

Let

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

Step 3. Now the integral is

$$\begin{gathered} \int {\sin }^{5}x\mathrm{d}x=-\int {(1-u^2)}^2\mathrm{du} \\ \int {\sin }^5\mathrm{xdx}=-\int {\{{\left(1\right)}^2-2\times 1\times {\mathrm{u}}^2+({\mathrm{u}}^2)}^2\}\mathrm{du} \\ \int {\sin }^5\mathrm{xdx}=-\int (1-2{\mathrm{u}}^2+{\mathrm{u}}^4)\mathrm{du} \\ \int {\sin }^5\mathrm{xdx}=-\left[\int \mathrm{du}-2\int {\mathrm{u}}^2\mathrm{du}+\int {\mathrm{u}}^4\mathrm{du}\right]+C \\ \int {\sin }^5\mathrm{xdx}=-\left[\left(\frac{{\mathrm{u}}^1}{1}\right)-2\left(\frac{{\mathrm{u}}^{2+1}}{2+1}\right)+\left(\frac{{\mathrm{u}}^{4+1}}{4+1}\right)\right]+C \\ \int {\sin }^5\mathrm{xdx}=-\left[\mathrm{u}-2\left(\frac{{\mathrm{u}}^3}{3}\right)+\left(\frac{{\mathrm{u}}^5}{5}\right)\right]+C \\ \int {\sin }^5\mathrm{xdx}=-\left[\mathrm{u}-\frac{2}{3}{\mathrm{u}}^3+\frac{1}{5}{\mathrm{u}}^5\right]+C \\ \int {\sin }^5\mathrm{xdx}=-\mathrm{u}+\frac{2}{3}{\mathrm{u}}^3-\frac{1}{5}{\mathrm{u}}^5+C \\ \int {\sin }^5\mathrm{xdx}=-\cos x+\frac{2}{3}{\cos }^{3}x-\frac{1}{5}{\cos }^{5}x+C \end{gathered}$$