3.2 Trigonometric integrals  (Page 3/6)

Since the power on $\text{sec}x$ is even, rewrite ${\text{sec}}^{4}x={\text{sec}}^{2}x{\text{sec}}^{2}x$ and use ${\text{sec}}^{2}x={\text{tan}}^{2}x+1$ to rewrite the first ${\text{sec}}^{2}x$ in terms of $\text{tan}x.$ Thus,

Since the power on $\text{tan}x$ is odd, begin by rewriting ${\text{tan}}^{5}x{\text{sec}}^{3}x={\text{tan}}^{4}x{\text{sec}}^{2}x\text{sec}x\text{tan}x.$ Thus,

Begin by rewriting ${\text{tan}}^{3}x=\text{tan}x{\text{tan}}^{2}x=\text{tan}x\left({\text{sec}}^{2}x-1\right)=\text{tan}x{\text{sec}}^{2}x-\text{tan}x.$ Thus,

For the first integral, use the substitution $u=\text{tan}x.$ For the second integral, use the formula.

This integral requires integration by parts. To begin, let $u=\text{sec}x$ and $dv={\text{sec}}^{2}x.$ These choices make $du=\text{sec}x\text{tan}x$ and $v=\text{tan}x.$ Thus,

We now have

Since the integral ${{\displaystyle \int }}^{}{\text{sec}}^{3}xdx$ has reappeared on the right-hand side, we can solve for ${{\displaystyle \int }}^{}{\text{sec}}^{3}xdx$ by adding it to both sides. In doing so, we obtain

Dividing by 2, we arrive at

By applying the first reduction formula, we obtain

Applying the reduction formula for ${{\displaystyle \int }}^{}{\text{tan}}^{4}xdx$ we have