Evaluate the following integral
$\int {{{\tan }^3}2x{\text{ }}\sec 2x{\text{ }}} dx$

Answer Verified Verified
Hint- Convert the integral in simpler form by the use of trigonometric identity and algebraic terms.

Solving the trigonometric function, in order to make the integral easy
Taking \[{\text{ta}}{{\text{n}}^3}2x\]
\[ \Rightarrow {\text{ta}}{{\text{n}}^3}2x = {\text{ta}}{{\text{n}}^2}2x \times {\text{tan}}2x\]
As we know that
\[{\text{ta}}{{\text{n}}^2}\theta = {\sec ^2}\theta - 1\]
Substituting the identity in above function
\[\therefore {\text{ta}}{{\text{n}}^3}2x = \left( {{{\sec }^2}2x - 1} \right){\text{tan}}2x\]
So now the question becomes
\[\int {\left[ {\left( {{{\sec }^2}2x - 1} \right){\text{.tan}}2x.\sec 2x} \right]dx} \]
Let us assume
\[\sec 2x = t\]
Differentiating both the sides with respect to $x$
   \Rightarrow 2\sec 2x.\tan 2x = \dfrac{{dt}}{{dx}} \\
   \Rightarrow \sec 2x.\tan 2xdx = \dfrac{{dt}}{2} \\
Now in the above problem we have
\[{\text{sec2}}x{\text{ = }}t{\text{& sec2}}x{\text{.tan2}}x{\text{ = }}\dfrac{{dt}}{2}\]
So the problem no becomes
  \int {\left( {{t^2} - 1} \right)\dfrac{{dt}}{2}} \\
   = \dfrac{1}{2}\int {\left( {{t^2} - 1} \right)dt} \\
As we know that
$\left[ {\because \int {{x^n}dx} = \dfrac{{{x^{n + 1}}}}{{n + 1}}} \right]$
So by integration, we get
\[\dfrac{1}{2}\left[ {\dfrac{{{t^3}}}{3} - t} \right]{\text{ + c}}\]
Substituting back the value of $t$ we get
  \because t = \sec 2x \\
   \Rightarrow \dfrac{1}{2}\left[ {\dfrac{{{{\sec }^3}2x}}{3} - \sec 2x} \right]{\text{ + c}} \\
   \Rightarrow \dfrac{{{{\sec }^3}2x}}{6} - \dfrac{{\sec 2x}}{2} + c \\
Hence \[\dfrac{{{{\sec }^3}2x}}{6} - \dfrac{{\sec 2x}}{2} + c\] is the integral of given function.

Note- Whenever we come across complicated trigonometric terms together, we should always try to break them using trigonometric relations and formulas and try to reduce the power. Also sometimes trigonometric terms can be replaced by some algebraic variables but before concluding, it must be converted back to original form.
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