# Active Calculus

## Section9.3Integration with Trigonometric Functions

### Subsection9.3.1Preview Activity

#### Question9.3.1.

We already know we can do the following problem using u-substitution, with $$u = \sin(8 x )$$
$$\displaystyle \int \sin^{6}\!\left(8x\right)\cos\!\left(8x\right) \,dx \ = \$$
We see that $$\displaystyle \int \sin^{6} (8 x) \cos^3(8 x) \,dx \$$ is solved similarly, because
• the exponent of $$\sin$$ is odd.
• the exponent of $$\cos$$ is odd.
• None of the above
So we’ll still use u-substitution with $$u = \sin(8 x )\text{,}$$ after first rewriting $$\cos^3 (8 x) = \cos^2(8 x) \cos(8 x )$$ and using a trig identity.
Select the trigonometric identity needed to solve this problem.
• $$\displaystyle \displaystyle \sin^2 u + \cos^2 u = 1$$
• $$\displaystyle \displaystyle \cos^2 u = \frac{1+\cos(2 u)}{2}$$
• $$\displaystyle \displaystyle \sin^2 u = \frac{1-\cos(2 u)}{2}$$
• $$\displaystyle \displaystyle \tan^2 u +1 = \sec^2 u$$
• $$\displaystyle \displaystyle 1+ \cot^2 u = \csc^2 u$$
• None of the above
Finally, put it all together to calculate $$\displaystyle \int \sin^{6}\!\left(8x\right)\cos^{3}\!\left(8x\right) \,dx \ = \$$
Hint.
The integral involves $$\sin u$$ and $$\cos u$$ where the exponent of $$\cos u$$ is an odd, positive integer, namely, 3. So we factor out a copy of $$\cos(8 x)$$ and use the identity $$\sin^2 u + \cos^2 u = 1$$ to turn the remaining $$\cos^2(8 x)$$ into an expression involving $$\sin^2(8 x)\text{.}$$
Thus the integral can be rewritten as
\begin{equation*} \int \sin^{6}\!\left(8x\right)\cos^{3}\!\left(8x\right) \, dx = \int \sin^{6} (8 x) \cos^2(8 x) \cos(8 x) \, dx \end{equation*}
\begin{equation*} = \int \sin^{6} (8 x) \bigg(1-\sin^2(8 x)\bigg) \cos(8 x) \, dx \end{equation*}
and we use u-substitution.
Letting $$u= \sin (8 x)$$ so $$du = 8 \cos (8 x) \, dx$$ and thus $$\displaystyle dx = \frac{ du}{8 \cos (8 x)}\text{.}$$
This turns the integral into
\begin{equation*} \int u^{6} (1-u^2) \cos(8 x) \, \frac{ du}{8 \cos (8 x)} = \frac{1}{8}\int u^{6} (1-u^2) \, du \end{equation*}
which can be multiplied out, and then each part integrated by power rule.

#### Question9.3.2.

Consider the integral $$\displaystyle \int 9\sin^{2}\!\left(11x\right) \,dx\text{.}$$
In order to integrate, the key feature is that
• the exponent of $$\sin$$ is odd.
• the exponent of $$\cos$$ is odd.
• the exponents of $$\sin$$ and $$\cos$$ are both even (remember that 0 is even)
• None of the above
Then the trigonometric identity needed to solve this problem is
• $$\displaystyle \displaystyle \tan^2 u +1 = \sec^2 u$$
• $$\displaystyle \displaystyle 1+ \cot^2 u = \csc^2 u$$
• $$\displaystyle \displaystyle \sin^2 u = \frac{1-\cos(2 u)}{2}$$
• $$\displaystyle \displaystyle \cos^2 u = \frac{1+\cos(2 u)}{2}$$
• $$\displaystyle \displaystyle \sin^2 u + \cos^2 u = 1$$
• None of the above
And we use this identity to rewrite the integral as $$\displaystyle \int 9\sin^{2}\!\left(11x\right) \, dx = \ \int$$ $$dx$$
And this can be split into two basic integrals and integrated, though I’m not asking for you to do that here.

#### Question9.3.3.

Select all the key features we could look for to evaluate trig integrals
• the exponent of $$\sin$$ is odd.
• the exponent of $$\sec$$ is odd.
• the exponent of $$\sec$$ is even.
• the exponent of $$\tan$$ is even.
• the exponent of $$\sin$$ is even.
• the exponent of $$\tan$$ is odd.
• the exponent of $$\cos$$ is odd.
• the exponents of both $$\sin$$ and $$\cos$$ are even.
• the exponent of $$\cos$$ is even.
• None of the above
• All of the above
You should be sure to have this information written down so that you can use it to solve problems in class.

#### Question9.3.4.

Consider the integral $$\displaystyle \int 9\tan^{3}\!\left(12x\right)\sec^{3}\!\left(12x\right) \,dx\text{.}$$
In order to integrate, the key feature is that
• the exponent of $$\tan$$ is even.
• the exponent of $$\tan$$ is odd.
• the exponent of $$\sec$$ is even.
• the exponent of $$\sec$$ is odd.
• None of the above
Then the trigonometric identity needed to solve this problem is
• $$\displaystyle \displaystyle \sin^2 u + \cos^2 u = 1$$
• $$\displaystyle \displaystyle \cos^2 u = \frac{1+\cos(2 u)}{2}$$
• $$\displaystyle \displaystyle \tan^2 u +1 = \sec^2 u$$
• $$\displaystyle \displaystyle \sin^2 u = \frac{1-\cos(2 u)}{2}$$
• $$\displaystyle \displaystyle 1+ \cot^2 u = \csc^2 u$$
• None of the above
So to integrate, we would use u-substitution with $$u= \sec(12 x)\text{.}$$ But first we’d have to rewrite the integral in order to have a copy of $$u$$’s derivative $$\sec(12 x) \tan(12 x)\text{,}$$ and use a trigonometric identity to turn the remaining even powers of $$\tan(12 x)$$ into an expression of $$\sec^2(12 x)\text{.}$$

#### Question9.3.5.

Consider the integral $$\displaystyle \int 2\tan^{4}\!\left(14x\right)\sec^{4}\!\left(14x\right) \,dx\text{.}$$
In order to integrate, the key feature is that
• the exponent of $$\tan$$ is even.
• the exponent of $$\sec$$ is even.
• the exponent of $$\tan$$ is odd.
• the exponent of $$\sec$$ is odd.
• None of the above
Then the trigonometric identity needed to solve this problem is
• $$\displaystyle \displaystyle 1+ \cot^2 u = \csc^2 u$$
• $$\displaystyle \displaystyle \cos^2 u = \frac{1+\cos(2 u)}{2}$$
• $$\displaystyle \displaystyle \tan^2 u +1 = \sec^2 u$$
• $$\displaystyle \displaystyle \sin^2 u = \frac{1-\cos(2 u)}{2}$$
• $$\displaystyle \displaystyle \sin^2 u + \cos^2 u = 1$$
• None of the above
So to integrate, we would use u-substitution with $$u= \tan(14 x)\text{.}$$ But first we’d have to rewrite the integral in order to have a copy of $$u$$’s derivative $$\sec^2(14 x)\text{,}$$ and use a trigonometric identity to turn the remaining even powers of $$\sec(14 x)$$ into an expression involving $$\tan^2(14 x)\text{.}$$

### Exercises9.3.2Exercises

#### 1.

Evaluate the indefinite integral.
$$\displaystyle \int 7\tan^{3}\!\left(-15x\right)\sec^{3}\!\left(-15x\right) \,dx \ = \$$
Select the trigonometric identity you used in solving this problem.
• $$\displaystyle \displaystyle \tan^2 u +1 = \sec^2 u$$
• $$\displaystyle \displaystyle \sin^2 u + \cos^2 u = 1$$
• $$\displaystyle \displaystyle 1+ \cot^2 u = \csc^2 u$$
• $$\displaystyle \displaystyle \sin^2 u = \frac{1-\cos(2 u)}{2}$$
• $$\displaystyle \displaystyle \cos^2 u = \frac{1+\cos(2 u)}{2}$$
• None of the above

#### 2.

Evaluate the indefinite integral.
$$\displaystyle \int 12\tan^{4}\!\left(3x\right)\sec^{4}\!\left(3x\right) \,dx \ = \$$
Select the trigonometric identity you used in solving this problem.
• $$\displaystyle \displaystyle 1+ \cot^2 u = \csc^2 u$$
• $$\displaystyle \displaystyle \tan^2 u +1 = \sec^2 u$$
• $$\displaystyle \displaystyle \sin^2 u = \frac{1-\cos(2 u)}{2}$$
• $$\displaystyle \displaystyle \sin^2 u + \cos^2 u = 1$$
• $$\displaystyle \displaystyle \cos^2 u = \frac{1+\cos(2 u)}{2}$$
• None of the above

#### 3.

Evaluate the indefinite integral.
$$\displaystyle \int 5\cot^{3}\!\left(4x\right)\csc^{3}\!\left(4x\right) \,dx \ = \$$
Select the trigonometric identity you used in solving this problem.
• $$\displaystyle \displaystyle \tan^2 u +1 = \sec^2 u$$
• $$\displaystyle \displaystyle \cos^2 u = \frac{1+\cos(2 u)}{2}$$
• $$\displaystyle \displaystyle \sin^2 u + \cos^2 u = 1$$
• $$\displaystyle \displaystyle \sin^2 u = \frac{1-\cos(2 u)}{2}$$
• $$\displaystyle \displaystyle 1+ \cot^2 u = \csc^2 u$$
• None of the above

#### 4.

Evaluate the indefinite integral.
$$\displaystyle \int 64 \cos^4(8 x) dx \ =$$

#### 5.

Evaluate the indefinite integral.
$$\displaystyle\int \tan^{7}\!\left(x\right)\sec^{4}\!\left(x\right) \, dx$$ =

#### 6.

Evaluate the integral:
$$\displaystyle \int {\frac{6 \tan^3(x)}{\cos^4(x)}}\, dx\ =$$

#### 7.

Evaluate the integral $$\displaystyle \int {-7 \cot^5(x) \sin^4(x)}\, dx\ =$$

#### 8.

Evaluate the indefinite integral.
$$\displaystyle\int \sin(7 x)\cos(15 x)\ dx=$$ $$+C$$

#### 9.

Evaluate the indefinite integral.
\begin{equation*} \int \sec^3 x \,dx \end{equation*}
Answer: $$\ + \ C$$

#### 10.

Match each of the trigonometric expressions below with the equivalent non-trigonometric function from the following list.
 $$\cos(\theta)$$ where $$x = 9 \sin \theta$$ $$\cos(\theta)$$ where $$x = 9 \tan \theta$$ $$\tan(\theta)$$ where $$x = 9 \sin \theta$$ $$\sin(\theta)$$ where $$x = 9 \tan \theta$$ $$\displaystyle \displaystyle \frac{9}{\sqrt{81 + x^2}}$$ $$\displaystyle \displaystyle \frac{x}{\sqrt{81 + x^2}}$$ $$\displaystyle \displaystyle \frac{\sqrt{81 - x^2}}{9}$$ $$\displaystyle \displaystyle \frac{x}{\sqrt{81 - x^2}}$$ None of the above