Calculate the area bounded by the curve y = tan x

Calculate the area bounded by the curve $y = \tan x$, the line $x = \frac{\pi}{4}$, and the $x$-axis.

We will evaluate the integral directly using a substitution for the antiderivative of $\tan x$.

Step 1: Identify the region

The region is bounded by:

• The curve: $y = \tan x$
• The line: $x = \frac{\pi}{4}$
• The $x$-axis: $y = 0$
• (Implicitly, starting from $x = 0$ where $\tan 0 = 0$)

Therefore, we integrate from $x = 0$ to $x = \frac{\pi}{4}$.

Step 2: Verify the curve is above the $x$-axis

For $x \in \left[0, \frac{\pi}{4}\right]$:

\[\tan x \geq 0\]

So the curve is above or on the $x$-axis throughout this interval. ✓

Step 3: Set up the integral

\[A = \int_0^{\pi/4} \tan x \, dx\]

Step 4: Find the antiderivative of $\tan x$

Recall that:

\[\tan x = \frac{\sin x}{\cos x}\]

Let $u = \cos x$, so $du = -\sin x \, dx$, which gives $\sin x \, dx = -du$.

\[\int \tan x \, dx = \int \frac{\sin x}{\cos x} \, dx = \int \frac{-du}{u} = -\ln|u| + C\]

\[= -\ln|\cos x| + C\]

Alternatively, we can write:

\[\int \tan x \, dx = -\ln|\cos x| + C = \ln|\sec x| + C\]

Since $\ln\frac{1}{|\cos x|} = -\ln|\cos x|$.

Step 5: Evaluate the definite integral

\[A = \left[-\ln|\cos x|\right]_0^{\pi/4}\]

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

\[-\ln\left|\cos\frac{\pi}{4}\right| = -\ln\frac{1}{\sqrt{2}} = -\ln\frac{\sqrt{2}}{2}\]

\[= -\left(\ln\sqrt{2} - \ln 2\right) = -\left(\frac{\ln 2}{2} - \ln 2\right) = -\left(-\frac{\ln 2}{2}\right) = \frac{\ln 2}{2}\]

Or more directly:

\[-\ln\frac{1}{\sqrt{2}} = \ln\sqrt{2} = \frac{1}{2}\ln 2\]

At $x = 0$:

\[-\ln|\cos 0| = -\ln 1 = 0\]

Therefore:

\[A = \frac{\ln 2}{2} - 0 = \frac{\ln 2}{2}\]

\[\boxed{A = \frac{\ln 2}{2} = \frac{1}{2}\ln 2 \text{ square units}}\]

Step 6: Numerical value

\[A = \frac{\ln 2}{2} \approx \frac{0.6931}{2} \approx 0.3466 \text{ square units}\]

Verification

Let's verify using the alternative form $\ln|\sec x|$:

\[A = \left[\ln|\sec x|\right]_0^{\pi/4} = \ln\left|\sec\frac{\pi}{4}\right| - \ln|\sec 0|\]

\[= \ln\sqrt{2} - \ln 1 = \frac{\ln 2}{2} - 0 = \frac{\ln 2}{2} \quad \checkmark\]

Summary

The area bounded by the curve $y = \tan x$, the line $x = \frac{\pi}{4}$, and the $x$-axis is:

\[\boxed{A = \frac{\ln 2}{2} \approx 0.347 \text{ square units}}\]

Key steps:

1. Identify region: from $x = 0$ to $x = \frac{\pi}{4}$
2. Set up integral: $A = \int_0^{\pi/4} \tan x \, dx$
3. Find antiderivative: $\int \tan x \, dx = -\ln|\cos x| + C = \ln|\sec x| + C$
4. Evaluate: $-\ln\cos\frac{\pi}{4} - (-\ln\cos 0) = \frac{\ln 2}{2}$

Key insight:

The antiderivative of $\tan x$ is $-\ln|\cos x|$ (or equivalently $\ln|\sec x|$), which comes from the substitution $u = \cos x$.