Problem 36 - Essential Calculus

Question

Evaluate the definite integral. Use a graphing utility to confirm your result. $$ \int_{0}^{\pi / 4} x \sec ^{2} x d x $$

Step-by-Step Solution

Answer

The definite integral \( \int_{0}^{\pi / 4} x \sec ^{2} x d x \) is equal to \(\pi / 4 - \ln ( \sqrt{2} )\).

Step 1: Identifying u and dv

Choose \(x\) as 'u' and \(\sec ^{2} x dx\) as 'dv'. Thus, \(u = x\) and \(dv = \sec^2 x dx\).

Step 2: Computing du and v

Differentiate 'u' to get 'du' and integrate 'dv' to get 'v'. The derivative of \(x\) is \(1\), so \(du = dx\). The integral of \(\sec^2 x\) is \(tan x\), so \(v = tan x\).

Step 3: Apply the Integration by Parts Formula

Insert \(u\), \(v\), \(du\), \(dv\) in the integration by parts formula \(\int u dv = uv - \int v du\). Thus, we obtain \(\int_{0}^{\pi / 4} x \sec^2 x dx = uv - \int_{0}^{\pi / 4} v du = [x \cdot tan x]_{0}^{\pi / 4} - \int_{0}^{\pi / 4} tan x dx\).

Step 4: Evaluate the Remaining Integral

The integral of \(tan x\) is \(\ln| sec x|\). Therefore, \(\int_{0}^{\pi / 4} tan x dx = [\ln| sec x|]_{0}^{\pi / 4}\). Substituting \(\pi / 4\) and \(0\) we get \(\ln| sec (\pi / 4)| - \ln| sec 0| = \ln | \sqrt{2} | - \ln |1| = \ln | \sqrt{2} |\).

Step 5: Putting it all together

Substitute \(\pi / 4\) and \(0\) into the expression obtained in step 3 and the result obtained in step 4 to get the final result. The final integral is \([x \cdot tan x]_{0}^{\pi / 4} - \ln | \sqrt{2} | = [\pi / 4 \cdot 1] - [0 \cdot 0] - \ln | \sqrt{2} | = [\pi / 4 - \ln | \sqrt{2} |]\).

Step 6: Verify the result

Use a graphing utility to confirm the result. Draw the graph of function \(f(x) = x \sec^2 x\), and compute the definite integral from \(0\) to \(\pi / 4\). The result should be the same as the analytical result, \(\pi / 4 - \ln ( \sqrt{2} |\).