Show that, for \(n> 0\),
\[
\int_0^{\frac14\pi} \tan^n x \,\sec^2 x \, \d x =
\frac 1 {n+1} \;
\text{ and }
\int_0^{\frac14\pi} \!\! \sec ^n\! x \, \tan x \,
\d x = \frac{(\sqrt 2)^n - 1}n \,.
\]
Evaluate the following integrals:
\[
\displaystyle
\int_0^{\frac14\pi}
\!\! x\, \sec ^4 \! x\, \tan x \, \d x \,
\text{ and }
\int_0^{\frac14\pi}
\!\! x^2 \sec ^2 \! x\, \tan x \, \d x \,.
\]
Solution:
\begin{align*}
u = \tan x, \d u = \sec^2 x \d x: &&\int_0^{\pi/4} \tan^n x \sec^2 x \d x &= \int_0^1 u^n \d u \\
&&&= \frac{1}{n+1}
\end{align*}
\begin{align*}
u = \sec x, \d u = \sec x \tan x \d x: &&\int_0^{\pi/4} \sec^n x \tan x \d x &= \int_{u=1}^{u=\sqrt{2}} u^{n-1} \d u \\
&&&= \left [ \frac{u^n}{n}\right] \\
&&&= \frac{(\sqrt{2})^n - 1}n
\end{align*}
\begin{align*}
&&\int_0^{\frac14\pi}
x \sec ^4 x \tan x \d x &= \left [x \frac{1}{4} \sec^4 x \right]_0^{\frac14\pi} - \frac14 \int_0^{\frac14\pi} \sec^4 x \d x \\
&&&= \frac{\pi}{4} - \frac14 \int_0^{\frac14\pi} \sec^2 x(1+ \tan^2 x) \d x \\
&&&= \frac{\pi}{4} - \frac14 \left [ \tan x+ \frac13 \tan^3 x \right] _0^{\frac14\pi} \\
&&&= \frac{\pi}{4} - \frac{1}{3}
\end{align*}
\begin{align*}
\int_0^{\frac14\pi}
\!\! x^2 \sec ^2 \! x\, \tan x \, \d x &= \left [x^2 \frac12 \tan^2 x \right]_0^{\frac14\pi} - \int_0^{\frac14\pi} x \tan^2 x \d x\\
&= \frac{\pi^2}{32} - \int_0^{\frac14\pi} x (\sec^2 x - 1) \d x\\
&= \frac{\pi^2}{32} - \left [x (-x + \tan x) \right]_0^{\frac14\pi} + \int_0^{\frac14\pi}-x + \tan x \d x \\
&= \frac{\pi^2}{32} - \frac{\pi}{4} (-\frac{\pi}{4} + 1) -\frac{\pi^2}{32} + \left [ -\ln \cos x \right]_0^{\pi/4} \\
&= \frac{\pi^2}{16}- \frac{\pi}{4} + \frac12 \ln 2
\end{align*}