Integrating the Trigonometric Functions (HSC SSCE Mathematics Advanced): Revision Notes

Deriving the Cosine Integral

We know that $\frac{d}{dx} \sin x = \cos x$

Working backwards from this differentiation result:

$\int \cos x \, dx = \sin x + C$

This is our first standard integral.

Deriving the Sine Integral

We know that $\frac{d}{dx} \cos x = -\sin x$

Reversing this:

$\int (-\sin x) \, dx = \cos x + C$

Multiplying both sides by $-1$:

$\int \sin x \, dx = -\cos x + C$

Notice the negative sign that appears in the final result.

Deriving the Secant Squared Integral

We know that $\frac{d}{dx} \tan x = \sec^2 x$

Reversing this:

$\int \sec^2 x \, dx = \tan x + C$

Finding the Area Under the First Arch of y = sin x

Let's find the area of the first arch of the curve $y = \sin x$.

Because the region is entirely above the x-axis, we can calculate:

$\text{area} = \int_0^{\pi} \sin x \, dx$

$= [-\cos x]_0^{\pi}$

$= -\cos \pi + \cos 0$

$= -(-1) + 1$

$= 2 \text{ square units}$

Note that $\cos \pi = -1$ and $\cos 0 = 1$ from the graph of $y = \cos x$.

Evaluating Definite Integrals with Trigonometric Functions

a) Evaluate $\int_0^{\pi} \cos x \, dx$

$\int_0^{\pi} \cos x \, dx = [\sin x]_0^{\pi}$

$= \sin \pi - \sin 0$

$= 0 - 0 = 0$

The result is zero because the positive and negative areas cancel out.

b) Evaluate $\int_0^{\frac{\pi}{3}} \sec^2 x \, dx$

$\int_0^{\frac{\pi}{3}} \sec^2 x \, dx = [\tan x]_0^{\frac{\pi}{3}}$

$= \tan \frac{\pi}{3} - \tan 0$

$= \sqrt{3} - 0 = \sqrt{3}$

c) Why is $\int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \sec^2 x \, dx$ meaningless?

This integral is meaningless because it crosses the asymptote at $x = \frac{\pi}{2}$, where $\tan x$ (and therefore $\sec^2 x$) is undefined.

Deriving the Cosine Formula for ax + b

We know that $\frac{d}{dx} \sin(ax + b) = a \cos(ax + b)$

Reversing this:

$\int a \cos(ax + b) \, dx = \sin(ax + b)$

Dividing by $a$:

$\int \cos(ax + b) \, dx = \frac{1}{a} \sin(ax + b) + C$

Notice we multiply by 1/a, the reciprocal of the coefficient.

Deriving the Sine Formula for ax + b

We know that $\frac{d}{dx} \cos(ax + b) = -a \sin(ax + b)$

Reversing this:

$\int -a \sin(ax + b) \, dx = \cos(ax + b)$

Dividing by $-a$:

$\int \sin(ax + b) \, dx = -\frac{1}{a} \cos(ax + b) + C$

Deriving the Secant Squared Formula for ax + b

We know that $\frac{d}{dx} \tan(ax + b) = a \sec^2(ax + b)$

Reversing this:

$\int a \sec^2(ax + b) \, dx = \tan(ax + b)$

Dividing by $a$:

$\int \sec^2(ax + b) \, dx = \frac{1}{a} \tan(ax + b) + C$

Example 1: Evaluating $\int_0^{\frac{\pi}{6}} \cos 3x \, dx$

$\int_0^{\frac{\pi}{6}} \cos 3x \, dx = \frac{1}{3}[\sin 3x]_0^{\frac{\pi}{6}}$

$= \frac{1}{3}(\sin \frac{\pi}{2} - \sin 0)$

$= \frac{1}{3}(1 - 0) = \frac{1}{3}$

Here, the coefficient is $3$, so we multiply by 1/3.

Example 2: Evaluating $\int_{\pi}^{2\pi} \sin \frac{1}{4}x \, dx$

Since the coefficient of $x$ is $\frac{1}{4}$, the reciprocal is 4:

$\int_{\pi}^{2\pi} \sin \frac{1}{4}x \, dx = -4[\cos \frac{1}{4}x]_{\pi}^{2\pi}$

$= -4 \cos \frac{\pi}{2} + 4 \cos \frac{\pi}{4}$

$= -4(0) + 4 \times \frac{\sqrt{2}}{2}$

$= 2\sqrt{2}$

Example 3: Evaluating $\int_0^{\frac{\pi}{8}} \sec^2(2x + \pi) \, dx$

$\int_0^{\frac{\pi}{8}} \sec^2(2x + \pi) \, dx = \frac{1}{2}[\tan(2x + \pi)]_0^{\frac{\pi}{8}}$

$= \frac{1}{2}(\tan \frac{5\pi}{4} - \tan \pi)$

$= \frac{1}{2}(1 - 0)$

$= \frac{1}{2}$

Note: $\frac{5\pi}{4}$ is in quadrant 3 with related angle $\frac{\pi}{4}$, so $\tan \frac{5\pi}{4} = 1$.

Deriving the Integral of cot x

Using the ratio formula $\cot x = \frac{\cos x}{\sin x}$:

$\int \cot x \, dx = \int \frac{\cos x}{\sin x} \, dx$

Let $u = \sin x$, then $u' = \cos x$

This matches the pattern u'/u, so:

$\int \cot x \, dx = \log_e |\sin x| + C$

Deriving the Integral of tan x

Using the ratio formula $\tan x = \frac{\sin x}{\cos x}$:

$\int \tan x \, dx = \int \frac{\sin x}{\cos x} \, dx$

We need to rewrite this to match the pattern $\frac{u'}{u}$:

$= -\int \frac{-\sin x}{\cos x} \, dx$

Let $u = \cos x$, then $u' = -\sin x$

Therefore:

$\int \tan x \, dx = -\log_e |\cos x| + C$

This can also be written as:

$\int \tan x \, dx = \log_e |\sec x| + C$

Finding a Function from its Derivative and Initial Condition

The derivative of a function is $y' = \cos x$, and the graph has $y$-intercept $(0, 3)$. Find the original function $f(x)$ and then find $f(\frac{\pi}{2})$.

Step 1: Integrate the derivative

$y' = \cos x$

Taking the primitive:

$y = \sin x + C$

Step 2: Use the initial condition

When $x = 0$, $y = 3$, so:

$3 = \sin 0 + C$

$3 = 0 + C$

$C = 3$

Step 3: Write the complete function

$y = \sin x + 3$

Finding the requested value:

When $x = \frac{\pi}{2}$:

$y = \sin \frac{\pi}{2} + 3 = 1 + 3 = 4$

The answer is 4.

Finding and Evaluating a Function

Given that $f'(x) = \sin 2x$ and $f(\pi) = 1$:

a) Find the function $f(x)$

$f'(x) = \sin 2x$

Taking the primitive:

$f(x) = -\frac{1}{2} \cos 2x + C$

Using the condition $f(\pi) = 1$:

$1 = -\frac{1}{2} \cos 2\pi + C$

$1 = -\frac{1}{2} \times 1 + C$

$1 = -\frac{1}{2} + C$

$C = 1\frac{1}{2}$

Therefore:

$f(x) = -\frac{1}{2} \cos 2x + 1\frac{1}{2}$

b) Find $f(\frac{\pi}{4})$

Substituting $x = \frac{\pi}{4}$:

$f(\frac{\pi}{4}) = -\frac{1}{2} \cos \frac{\pi}{2} + 1\frac{1}{2}$

$= -\frac{1}{2}(0) + 1\frac{1}{2}$

$= 1\frac{1}{2}$

Using Differentiation to Find an Integral

Use the chain rule to differentiate $\cos^5 x$, then find $\int_0^{\pi} \sin x \cos^4 x \, dx$

Part 1: Differentiation

Let $y = \cos^5 x$

Let $u = \cos x$, then $y = u^5$

By the chain rule:

$\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}$

$\frac{du}{dx} = -\sin x \text{ and } \frac{dy}{du} = 5u^4$

Therefore:

$\frac{dy}{dx} = 5u^4 \times (-\sin x) = -5 \sin x \cos^4 x$

Part 2: Finding the integral

From Part 1, we know:

$\frac{d}{dx}(\cos^5 x) = -5 \sin x \cos^4 x$

Reversing this:

$\int (-5 \sin x \cos^4 x) \, dx = \cos^5 x$

Dividing by $-5$:

$\int \sin x \cos^4 x \, dx = -\frac{1}{5} \cos^5 x + C$

Therefore:

$\int_0^{\pi} \sin x \cos^4 x \, dx = -\frac{1}{5}[\cos^5 x]_0^{\pi}$

$= -\frac{1}{5}((-1)^5 - 1^5)$

$= -\frac{1}{5}(-1 - 1)$

$= -\frac{1}{5}(-2) = \frac{2}{5}$

Direct Application of the Reverse Chain Rule

Use the reverse chain rule to find $\int \sin x \cos^4 x \, dx$

We need to identify the pattern $u^n \frac{du}{dx}$.

Let $u = \cos x$, then $u' = -\sin x$

Rewriting the integral:

$\int \sin x \cos^4 x \, dx = -\int (-\sin x) \cos^4 x \, dx$

$= -\int u^4 \frac{du}{dx} \, dx$

$= -\frac{u^5}{5} + C$

$= -\frac{1}{5} \cos^5 x + C$

This matches our previous result.