The Notes: `int_0^(pi/4) tan^4(t) dt` Evaluate the integral

Thursday, September 4, 2014

$\displaystyle{\int_{{0}}^{{\frac{\pi}{{4}}}}}{{\tan}}^{{4}}{\left({t}\right)}{\left.{d}{t}\right.}$ Evaluate the integral

$\displaystyle{\int_{{0}}^{{\frac{\pi}{{4}}}}}{{\tan}}^{{4}}{\left({t}\right)}{\left.{d}{t}\right.}$

Express the integrand in factor form.

$\displaystyle={\int_{{0}}^{{\frac{\pi}{{4}}}}}{{\tan}}^{{2}}{\left({t}\right)}\cdot{{\tan}}^{{2}}{\left({t}\right)}{\left.{d}{t}\right.}$

Plug-in the trigonometric identity $\displaystyle{{\tan}}^{{2}}{\left({t}\right)}={{\sec}}^{{2}}{\left({t}\right)}-{1}$ to one of the factors.

$\displaystyle={\int_{{0}}^{{\frac{\pi}{{4}}}}}{{\tan}}^{{2}}{\left({t}\right)}\cdot{\left({{\sec}}^{{2}}{\left({t}\right)}-{1}\right)}{\left.{d}{t}\right.}$

$\displaystyle={\int_{{0}}^{{\frac{\pi}{{4}}}}}{\left({{\tan}}^{{2}}{\left({t}\right)}{{\sec}}^{{2}}{\left({t}\right)}-{{\tan}}^{{2}}{\left({t}\right)}\right)}{\left.{d}{t}\right.}$

$\displaystyle={\int_{{0}}^{{\frac{\pi}{{4}}}}}{{\tan}}^{{2}}{\left({t}\right)}{{\sec}}^{{2}}{\left({t}\right)}{\left.{d}{t}\right.}-{\int_{{0}}^{{\frac{\pi}{{4}}}}}{{\tan}}^{{2}}{\left({t}\right)}{\left.{d}{t}\right.}$

Plug-in again the trigonometric identity $\displaystyle{{\tan}}^{{2}}{\left({t}\right)}={{\sec}}^{{2}}{\left({t}\right)}-{1}$ to the second integral.

$\displaystyle={\int_{{0}}^{{\frac{\pi}{{4}}}}}{{\tan}}^{{2}}{\left({t}\right)}{{\sec}}^{{2}}{\left({t}\right)}{\left.{d}{t}\right.}-{\int_{{0}}^{{\frac{\pi}{{4}}}}}{\left({{\sec}}^{{2}}{\left({t}\right)}-{1}\right)}{\left.{d}{t}\right.}$

$\displaystyle={\int_{{0}}^{{\frac{\pi}{{4}}}}}{{\tan}}^{{2}}{\left({t}\right)}{{\sec}}^{{2}}{\left({t}\right)}{\left.{d}{t}\right.}-{\int_{{0}}^{{\frac{\pi}{{4}}}}}{{\sec}}^{{2}}{\left({t}\right)}{\left.{d}{t}\right.}+{\int_{{0}}^{{\frac{\pi}{{4}}}}}{\left.{d}{t}\right.}$

For the first integral, apply the u-substitution method.

$\displaystyle{u}={\tan{{\left({t}\right)}}}$

$\displaystyle{d}{u}={{\sec}}^{{2}}{\left({t}\right)}{\left.{d}{t}\right.}$

>> $\displaystyle\int{{\tan}}^{{2}}{\left({t}\right)}{{\sec}}^{{2}}{\left({t}\right)}{\left.{d}{t}\right.}=\int{{u}}^{{2}}{d}{u}=\frac{{{u}}^{{3}}}{{3}}=\frac{{{{\tan}}^{{3}}{\left({t}\right)}}}{{3}}$

For the second integral, apply the formula $\displaystyle\int{{\sec}}^{{2}}{x}{\left.{d}{x}\right.}={\tan{{x}}}$ .

>> $\displaystyle\int{{\sec}}^{{2}}{\left({t}\right)}{\left.{d}{t}\right.}={\tan{{\left({t}\right)}}}$

And for the third integral, apply the formula $\displaystyle\int{a}{\left.{d}{x}\right.}={a}{x}$ .

$\displaystyle\gt\gt\int{\left.{d}{t}\right.}={t}$

So the integral becomes:

$\displaystyle{\int_{{0}}^{{\frac{\pi}{{4}}}}}{{\tan}}^{{2}}{\left({t}\right)}{{\sec}}^{{2}}{\left({t}\right)}{\left.{d}{t}\right.}-{\int_{{0}}^{{\frac{\pi}{{4}}}}}{{\sec}}^{{2}}{\left({t}\right)}{\left.{d}{t}\right.}+{\int_{{0}}^{{\frac{\pi}{{4}}}}}{\left.{d}{t}\right.}$

$\displaystyle={\left(\frac{{{{\tan}}^{{3}}{\left({t}\right)}}}{{3}}-{\tan{{\left({t}\right)}}}+{t}\right)}{{\mid}_{{0}}^{{\frac{\pi}{{4}}}}}$

$\displaystyle={\left(\frac{{{{\tan}}^{{3}}{\left(\frac{\pi}{{4}}\right)}}}{{3}}-{\tan{{\left(\frac{\pi}{{4}}\right)}}}+\frac{\pi}{{4}}\right)}-{\left(\frac{{{{\tan}}^{{3}}{\left({0}\right)}}}{{3}}-{\tan{{\left({0}\right)}}}+{0}\right)}$