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Question

Describe the technique for finding sec5xtan7xdx\int \sec ^{5} x \tan ^{7} x dx. Do not integrate.

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To solve\textbf{To solve} the given integral we will apply identity:

tan2x=sec2x1\color{#c34632}{\tan^2{x}=\sec^2{x}-1}

This leads\textbf{This leads} to:

sec5xtan7x dx=sec4x(sec3x1)3secxtanx dx\begin{align*} \int{\sec^5{x}\tan^7{x}\ dx}&=\int{\sec^4{x}(\sec^3{x}-1)^3\cdot \sec{x}\cdot \tan{x}\ dx}\\ \end{align*}

Now\textbf{Now}, we will apply the substitution:

u=secxdu=secxtanx dx\color{#c34632}{u=\sec{x}\Rightarrow du=\sec{x}\tan{x}\ dx}

Now\textbf{Now}, we have:

sec4x(sec3x1)3secxtanx dx=u4(u21)3 du\begin{align*} \int{\sec^4{x}(\sec^3{x}-1)^3\cdot \sec{x}\cdot \tan{x}\ dx}&=\int{u^4(u^2-1)^3\ du}\\ \end{align*}

Last integral\textbf{Last integral} is simple and we will not it solve since we are not asked to ...

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