Week 0 Discussion Notes

Since this discussion was before the first lecture, we just reviewed some trigonometry and integration.

Calculate $\displaystyle \int \sec^3 x \tan x \,\diff{x}$ .

We will solve this in two different ways.

Method 1: $u = \sec x \implies \diff{u} = \sec x \tan x \,\diff{x}$ .

\begin{aligned} \int \sec^3 x \tan x \,\diff{x} &= \int \p{\sec^2 x} \p{\sec x \tan x} \,\diff{x} \\ &= \int u^2 \,\diff{u} \\ &= \frac{1}{3} u^3 + C \\ &= \boxed{\frac{1}{3} \sec^3 x + C} \end{aligned}

Method 2: $u = \cos x \implies \diff{u} = -\sin x \,\diff{x}$ .

\begin{aligned} \int \sec^3 x \tan x \,\diff{x} &= \int \frac{\sin x}{\cos^4 x} \,\diff{x} \\ &= -\int \frac{\diff{u}}{u^4} \\ &= -\p{-\frac{1}{3}u^{-3}} + C \\ &= \frac{1}{3} \frac{1}{\cos^3 x} + C \\ &= \boxed{\frac{1}{3} \sec^3 x + C} \end{aligned}

Calculate $\displaystyle \int \sec^4 x \,\diff{x}$ .

Recall the identity $\tan^2 x + 1 = \sec^2 x$ . This gives

\int \sec^4 x \,\diff{x} = \int \p{\tan^2 x + 1} \sec^2 x \,\diff{x}.

Let $u = \tan x \implies \diff{u} = \sec^2 x \,\diff{x}$ . Then

\begin{aligned} \int \sec^4 x \,\diff{x} &= \int \p{\tan^2 x + 1} \sec^2 x \,\diff{x} \\ &= \int \p{u^2 + 1} \,\diff{u} \\ &= \frac{1}{3} u^3 + u + C \\ &= \boxed{\frac{1}{3} \tan^3 x + \tan x + C}. \end{aligned}

Calculate $\displaystyle \int \tan^2 x \,\diff{x}$ .

Using the same identity as in Example 2, we get

\int \tan^2 x \,\diff{x} = \int \sec^2 x - 1 \,\diff{x} = \boxed{\tan x - x + C}.