# Integrate \int \tan x (1+\sec^{4} x)^{3/2} \; dx

## Question:

Integrate {eq}\displaystyle \int \tan x (1 + \sec^{4} x)^{3/2} \; dx {/eq}

## Integration by Substitution:

Integration by substitution is the best methods to find an integration if an integral is in the form of {eq}I = \int f(g(x))g'(x)dx {/eq}

Here, we can put {eq}g(x) = t {/eq}

{eq}\implies g'(x)dx = dt {/eq}

And in this way, the integral becomes {eq}I = \int f(t)dt {/eq}

Given

• An integral {eq}I = \int \tan x (1+\sec^{4} x)^{3/2} \; dx {/eq}

We have,

{eq}\begin{align} I &= \int \tan x (1+\sec^{4} x)^{3/2} \; dx\\ &= \int 4 \sec^4 x \tan x \dfrac{(1+\sec^{4} x)^{3/2}}{4\sec^4 x} \; dx\\ \end{align} {/eq}

Now, put {eq}1+\sec^{4} x = t^2 {/eq}

Differentiating both side of the above equation

{eq}\begin{align} 4\sec^3 x (\sec x \tan x)\ dx &= 2t dt\\ \implies 4\sec^4 x \tan x\ dx &= 2t dt\\ \end{align} {/eq}

Therefore, the integral becomes;

{eq}\begin{align} I &= \int \dfrac{(1+\sec^{4} x)^{3/2}}{4\sec^4 x} 4 \sec^4 x \tan x \; dx\\ &= \int \dfrac{t^3}{4(t^2-1)}\ 2t dt\\ &= \dfrac{1}{2}\int \dfrac{t^4}{t^2-1}\ dt\\ &= \dfrac{1}{2}\left \{ \int \dfrac{t^4-1}{t^2-1}\ dt + \int \dfrac{1}{t^2-1}\ dt \right \}\\ &= \dfrac{1}{2}\left \{ \int (t^2+1)\ dt + \dfrac{1}{2}\int \dfrac{1}{t-1}\ dt- \dfrac{1}{2}\int \dfrac{1}{t+1}\ dt \right \}\\ &= \dfrac{1}{2}\left \{\dfrac{t^3}{3}+t + \dfrac{1}{2}\ln |(t-1)|- \dfrac{1}{2}\ln |(t+1)| \right \}\\ &= \dfrac{1}{2}\left \{\dfrac{t^3}{3}+t + \dfrac{1}{2}\ln |\dfrac{t-1}{t+1}| \right \}\\ &= \dfrac{1}{2}\left \{\dfrac{(1+\sec^{4} x)^{3/2}}{3}+(1+\sec^{4} x)^{1/2} + \dfrac{1}{2}\ln |\dfrac{(1+\sec^{4} x)^{1/2}-1}{(1+\sec^{4} x)^{1/2}+1}| \right \}\\ \end{align} {/eq} 