# \\int_{0}^{\\frac{\\pi}{4}}(1 + \\tan x)^{5}\\sec^{2}xdx
# \\int_{0}^{\\frac{\\pi}{4}}(1 + \\tan x)^{5}\\sec^{2}xdx
Answer
Explanation:
Step1: Substitution
Let (u = 1+\tan x). Then (du=\sec^{2}x dx). When (x = 0), (u=1+\tan(0)=1). When (x=\frac{\pi}{4}), (u=1 + \tan(\frac{\pi}{4})=2). The integral (\int_{0}^{\frac{\pi}{4}}(1 + \tan x)^{5}\sec^{2}x dx) becomes (\int_{1}^{2}u^{5}du).
Step2: Integrate (u^{5})
Using the power - rule (\int u^{n}du=\frac{u^{n + 1}}{n+1}+C) ((n\neq - 1)), for (n = 5), (\int u^{5}du=\frac{u^{6}}{6}+C). So (\int_{1}^{2}u^{5}du=\left[\frac{u^{6}}{6}\right]_{1}^{2}).
Step3: Evaluate the definite integral
(\left[\frac{u^{6}}{6}\right]_{1}^{2}=\frac{2^{6}}{6}-\frac{1^{6}}{6}). Since (2^{6}=64) and (1^{6}=1), we have (\frac{64}{6}-\frac{1}{6}=\frac{64 - 1}{6}=\frac{63}{6}=\frac{21}{2}).
Answer:
(\frac{21}{2})