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Integration Techniques: (lesson 3 of 4)

Integrals Involving Trigonometric Functions

Integrals of the form $\int \sin^m x \cdot \cos^n x dx$

Case 1: $m$ is an odd integer :

Step 1: Write $\color{blue}{\sin^m x}$ as $\color{blue}{\sin^{m-1} x \cdot \sin x}$.

Step 2: Apply identity: $\color{blue}{\sin^2 x = 1 - \cos^2 x}$

Step 3: Use the substitution $\color{blue}{u = \cos x}$.

Example 1: Evaluate the following integral

$$\int sin^3 x \cdot \cos^2 xdx$$

Solution:

In this example $m = 3$ and $n = 2$. Because $m$ is an odd integer we have:

Step 1:

$$\int \color{blue}{sin^5 x} \cdot cos^2 xdx = \int \color{blue}{sin^4 x \cdot \sin x} \cdot cos^2 xdx$$

Step 2: Apply: $\color{blue}{\sin^2 x = 1 - cos^2 x}$, then continue:

$$\int \sin^4 x \cdot \sin x \cdot \cos^2 xdx = \int (\color{blue}{\sin^2 x})^2 \cdot \sin x \cdot \cos^2 xdx = \int (\color{blue}{1 - \cos^2 x})^2 \cdot \cos^2 x \cdot \sin xdx$$

Step 3: Use the substitution $\color{blue}{u = \cos x}$.

If $\color{blue}{u = \cos x}$, then $\color{blue}{- \sin xdx}$, so:

\begin{aligned} \int {{{(1 - {{\cos }^2}x)}^2}} &\cdot {\cos ^2}x \cdot \sin xdx \color{blue}{\mathop { = = = = = = }\limits_{du = - \sin xdx}^{u = \cos x}} \int {{{(1 - {u^2})}^2}} \cdot {u^2}du = \\ &= \int {(1 - 2{u^2} + {u^4}) \cdot {u^2}du} = \int {{u^2} - 2{u^4}} + {u^6}du = \\ &= \frac{{{u^3}}}{3} - 2\frac{{{u^5}}}{5} + \frac{{{u^7}}}{7} + C = \frac{1}{3}{(\cos x)^3} - \frac{2}{5}{(\cos x)^5} + \frac{1}{7}{(\cos x)^7} + C \end{aligned}

 Try yourself $$\color{blue}{\int \sin^3 x dx = }$$ $\frac{1}{3} \cos^3 x + \cos x + C$ $\frac{1}{3} \sin^3 x + \cos x + C$ $\frac{1}{3} \cos^3 x - \cos x + C$ $\frac{1}{3} \sin^3 x - \sin x + C$

Case 2: $n$ is an odd integer :

Step 1: Write $\color{blue}{\cos^m x}$ as $\color{blue}{\cos^{m-1} x \cdot \cos x}$.

Step 2: Apply identity: $\color{blue}{\cos^2 x = 1 - \sin^2 x}$

Step 3: Use the substitution $\color{blue}{u = \sin x}$.

Example 2: Evaluate the following integral

$$\int \sin^6 x \cdot \cos^7 xdx$$

Solution:

In this example $m = 6$ and $n = 7$. Because $n$ is an odd integer we have:

Step 1:

$\int \sin^6 x \cdot \color{blue}{\cos^7 x} dx = \int sin^6 x \color{blue}{\cos^6 x \cdot \cos x} dx$

Step 2: Apply identity: $\color{blue}{cos^2 x = 1 - \sin^2 x}$.

\begin{aligned} &\int \sin^6 x \cdot \cos^6 x \cos xdx = \int \sin^6 x \cdot (\color{blue}{\cos^2 x})^3 \cdot \cos x dx = \\ &= \int \sin^6 x \cdot (\color{blue}{1-sin^2 x})^3 \cdot \cos x dx \end{aligned}

Step 3: Use the substitution $\color{blue}{u = \sin x}$.

If $\color{blue}{u = \sin x}$, then $\color{blue}{du = \cos x \cdot dx}$, so:

\begin{aligned} \int {{\sin }^6}x{{(1 - {{\sin }^2}x)}^3} &\cdot \cos xdx \color{blue}{\mathop { = = = = = = }\limits_{du = \cos xdx}^{u = \sin x}} \int {{u^6}{{(1 - {u^2})}^3}} \cdot du = \\ &= \int {{u^6}(1 - 3{u^2} + 3{u^4} - {u^6})du} = \int {{u^6} - 3{u^8}} + 3{u^{10}} - {u^{12}}du = \\ &= \frac{{{u^7}}}{7} - 3\frac{{{u^9}}}{9} + 3\frac{{{u^{11}}}}{{11}} - \frac{{{u^{13}}}}{{13}} + C = \\ &= \frac{1}{7}{(\sin x)^7} - \frac{1}{3}{(\sin x)^9} + \frac{3}{{11}}{(\sin x)^{11}} - \frac{1}{{13}}{(\sin x)^{13}} + C \end{aligned}

 Try yourself $$\color{blue}{\int \sin^2 x \cdot \cos^3 x dx = }$$ $\sin^5 x - \sin^3 x + C$ $\frac{1}{3} \sin^3 x + C$ $\frac{1}{5} \sin^5 x - \frac{1}{3} \sin^3 x + C$ $-\frac{1}{5} \sin^5 x - \frac{1}{3} \sin^3 x + C$

Case 3: If both $m$ and $n$ are even, the following identities are used:

$\color{blue}{\sin^2 x = \frac{1 - \cos 2x}{2} \ \ \cos^2 x = \frac{1 + \cos 2x}{2} \ \ \sin x \cos x = \frac{\sin 2x}{2}}$

Example 3: Evaluate the following integral

$$\int \sin^2 x \cdot \cos^2 x dx$$

Solution:

In this example $m = 2$ and $n = 2$. Because both $m$ and $n$ are even, the above noted identities are used.

\begin{aligned} \int \color{blue}{\sin^2 x} \cdot \color{red}{cos^2 x} dx &= \int \color{blue}{\frac{1 - \cos 2x}{2}} \cdot \color{red}{\frac{1 + \cos 2x}{2}}dx = \\ &= \frac{1}{4} \int(1 - \cos 2x)(1 + \cos 2x) dx \mathop { = = = = = = }\limits_{du = \frac{1}{2}du}^{2x = u} = \frac{1}{8} \int (1 - \cos u)(1 + \cos u) du = \\ &= \frac{1}{8} \int 1 - \color{red}{\cos^2 u} du = \frac{1}{8} \int 1 - \color{red}{\frac{1 + cos 2u}{2}} du = \frac{1}{16} \int 1 - \cos 2u du \mathop { = = = = = = }\limits_{du = \frac{1}{2}dt}^{2u = t} \frac{1}{32} \int 2 - \cos t dt \\ &= \frac{1}{32} (2t - \sin t) + C = \frac{1}{32} (2 \cdot 2u - \sin 2u) + C = \frac{1}{32} (4 \cdot 2x - \sin 4x) + C = \frac{1}{32} (8x - sin 4x) + C \end{aligned}

 Try yourself $$\color{blue}{\int \cos^2 x dx = }$$ $\frac{1}{2} x + \frac{1}{4} \sin 2x$ $\frac{1}{2} x - \frac{1}{4} \sin 2x$ $\frac{1}{2} x - \frac{1}{4} \cos 2x$ $\frac{1}{2} x + \frac{1}{4} \cos 2x$