Question

Find the integral of the following expression
${{\tan }^{3}}2x\sec 2x$ with respect to x.

Answer 1

Hint: Let I $=\int{{{\tan }^{3}}2x\sec 2xdx}$. Put 2x = t and hence prove that the given integral is equal to $\int{\dfrac{{{\tan }^{3}}t\sec t}{2}dt}$. Take $\sec t\tan t$ as the second function and \[\dfrac{{{\tan }^{2}}t}{2}\] as first function and integrate by parts. Finally, use ${{\sec }^{2}}t=1+{{\tan }^{2}}t$ and hence express the latter integral formed by integration by parts in terms of I. Hence find the value of I.

Complete step-by-step answer:
Let $I=\int{{{\tan }^{3}}2x\sec 2xdx}$
Put 2x = t
Differentiating both sides, we get
2dx = dt
Dividing both sides by 2, we get
$dx=\dfrac{dt}{2}$
Hence, we get
$I=\int{{{\tan }^{3}}t\sec t\dfrac{dt}{2}}=\int{\dfrac{{{\tan }^{2}}t}{2}\sec t\tan tdt}$
We know that if $\int{g\left( x \right)dx=v\left( x \right)}$ and $\dfrac{d}{dx}f\left( x \right)=u\left( x \right)$, then $\int{f\left( x \right)g\left( x \right)dx}=f\left( x \right)v\left( x \right)-\int{u\left( x \right)v\left( x \right)dx}$
This is known as integration by parts.
Hence, if we take $f\left( t \right)=\dfrac{{{\tan }^{2}}t}{2}$ and $g\left( t \right)=\sec t\tan t$, we have $v\left( t \right)=\int{\sec t\tan tdt}=\sec t$ and $u\left( t \right)=\dfrac{d}{dt}\left( \dfrac{{{\tan }^{2}}t}{2} \right)=\tan t{{\sec }^{2}}t$
Hence, by integration by parts rule, we have
$\int{\dfrac{{{\tan }^{2}}t}{2}\sec t\tan tdt}=\dfrac{{{\tan }^{2}}t}{2}\sec t-\int{\sec t\tan t{{\sec }^{2}}tdt}$
Hence, we have
$I=\dfrac{\sec t{{\tan }^{2}}t}{2}-\int{\tan t{{\sec }^{3}}tdt}$
We know that
${{\sec }^{2}}t=1+{{\tan }^{2}}t$
Hence, we have
${{\sec }^{3}}t\tan t=\sec t\left( 1+{{\tan }^{2}}t \right)\tan t=\sec t\tan t+\sec t{{\tan }^{3}}t$
Hence, we have
$I=\dfrac{\sec t{{\tan }^{2}}t}{2}-\int{\left( \sec t\tan t+\sec t{{\tan }^{3}}t \right)dt}$
We know that \[\int{\left( f\left( x \right)+g\left( x \right) \right)dx}=\int{f\left( x \right)dx}+\int{g\left( x \right)dx}\]
Hence, we have
$I=\dfrac{\sec t{{\tan }^{2}}t}{2}-\int{\sec t\tan tdt}-\int{\sec t{{\tan }^{3}}tdt}$
Now, we know that $I=\int{\dfrac{\sec t{{\tan }^{3}}t}{2}dt}$
Hence, we have
$2I=\int{\sec t{{\tan }^{3}}tdt}$
Hence, we have
$I=\dfrac{\sec t{{\tan }^{2}}t}{2}-\int{\sec t\tan tdt-2I}$
We know that $\int{\sec x\tan xdx}=\sec x$
Hence, we have
$I=\dfrac{\sec t{{\tan }^{2}}t}{2}-\sec t-2I$
Adding 2I on both sides, we get
$3I=\dfrac{\sec t{{\tan }^{2}}t}{2}-\sec t$
Dividing both sides by 3, we get
$I=\dfrac{\sec t{{\tan }^{2}}t}{6}-\dfrac{\sec t}{3}+C$, where C is the constant of integration.
Reverting to the original variable, we have
$I=\dfrac{\sec 2x{{\tan }^{2}}2x}{6}-\dfrac{\sec 2x}{3}+C$

Note: Verification:
We have
$\begin{align}
  & \dfrac{d}{dx}\left( \dfrac{\sec 2x{{\tan }^{2}}2x}{6}-\dfrac{\sec 2x}{3} \right)=\dfrac{1}{6}\sec 2x\tan 2x\left( 2 \right){{\tan }^{2}}2x+\dfrac{1}{6}\sec 2x2\tan 2x{{\sec }^{2}}2x\left( 2 \right)-\dfrac{\sec 2x\tan 2x}{3}\left( 2 \right) \\
 & =\dfrac{1}{3}\sec 2x{{\tan }^{3}}2x+\dfrac{2}{3}\tan 2x{{\sec }^{3}}2x-\dfrac{2\sec 2x\tan 2x}{3} \\
\end{align}$
Taking sec2xtan2x common, we get
$\dfrac{dI}{dx}=\sec 2x\tan 2x\left( \dfrac{{{\tan }^{2}}x}{3}+\dfrac{2{{\sec }^{2}}x}{3}-\dfrac{2}{3} \right)$
We know that ${{\sec }^{2}}x=1+{{\tan }^{2}}x$
Hence, we have
$\dfrac{dI}{dx}=\dfrac{\sec 2x\tan 2x}{3}\left( {{\tan }^{2}}2x+2{{\tan }^{2}}2x+2-2 \right)$
Hence, we have
$\dfrac{dI}{dx}=\sec 2x{{\tan }^{3}}2x$
Hence our answer is verified to be correct.