Question

solve the following

$\int \: {sec}^{4}x \: tanx \: dx \:$
​

Answer 1

$\large\underline{\sf{Solution-}}$

Given integral is

$\rm \: \displaystyle\int\rm {sec}^{4}x \: tanx \: dx$

can be rewritten as

$\rm \: = \: \displaystyle\int\rm {sec}^{2}x \: \times \: {sec}^{2}x \times \: tanx \: dx$

can be further rewritten as

$\rm \: = \: \displaystyle\int\rm {sec}^{2}x \: \times \: (1 + {tan}^{2}x) \times \: tanx \: dx$

can be further rearrange as

$\rm \: = \: \displaystyle\int\rm tanx \: \times \: (1 + {tan}^{2}x) \times \: {sec}^{2}x \: dx$

To evaluate this integral, we use method of Substitution,

So, substitute

$\rm \: tanx \: = \: y$

$\rm \: {sec}^{2}x \: dx \: = \: dy$

So, on substituting these values, we get

$\rm \: = \: \displaystyle\int\rm y(1 + {y}^{2}) \: dy$

$\rm \: = \: \displaystyle\int\rm (y + {y}^{3})\: dy$

We know,

$\color{green}\boxed{ \rm{ \:\displaystyle\int\rm {x}^{n} \: dx \: = \: \frac{ {x}^{n + 1} }{n + 1} + c \: }}$

So, using this result, we get

$\rm \: = \: \dfrac{ {y}^{1 + 1} }{1 + 1} + \dfrac{ {y}^{3 + 1} }{3 + 1} + c$

$\rm \: = \: \dfrac{ {y}^{2} }{2} + \dfrac{ {y}^{4} }{4} + c$

$\rm \: = \: \dfrac{ {tan}^{2}x }{2} + \dfrac{ {tan}^{4}x }{4} + c$

Hence,

$\color{green}\rm\implies \:\boxed{ \rm{ \:\displaystyle\int\rm {sec}^{4}x \: tanx \: dx = \: \dfrac{ {tan}^{2}x }{2} + \dfrac{ {tan}^{4}x }{4} + c \: }}$

$\rule{190pt}{2pt}$

Additional Information :-

$\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \displaystyle \int \rm \:f(x) \: dx\\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf kx + c \\ \\ \sf sinx & \sf - \: cosx+ c \\ \\ \sf cosx & \sf \: sinx + c\\ \\ \sf {sec}^{2} x & \sf tanx + c\\ \\ \sf {cosec}^{2}x & \sf - cotx+ c \\ \\ \sf secx \: tanx & \sf secx + c\\ \\ \sf cosecx \: cotx& \sf - \: cosecx + c\\ \\ \sf tanx & \sf logsecx + c\\ \\ \sf \dfrac{1}{x} & \sf logx+ c\\ \\ \sf {e}^{x} & \sf {e}^{x} + c\end{array}} \\ \end{gathered}\end{gathered}$

Answer 2

$\color{darkcyan} \underline{ \begin{array}{ || |l| || } \hline \color{magenta} \\ \hline \boxed{ \text{ \tt \: Solution:-} } \end{array}}$

Step-by-step explanation:

$\red{ \begin{aligned} \tt \int \sec ^{4} x \tan x d x & \tt=\int \sec ^{2} x \cdot \sec ^{2} x \tan x d x \\ \\ & \tt=\int\left(1+\tan ^{2} x\right) \sec ^{2} x \tan x d x \\ \\ & \tt=\int\left(1+t^{2}\right) t d t, \text { where } \tan x=t \\ \\ & \tt=\int t d t+\int t^{3} d t=\frac{t^{2}}{2}+\frac{t^{4}}{4}+C \\ \\ & \tt=\frac{1}{2} \tan ^{2} x+\frac{1}{4} \tan ^{4} x+C . \end{aligned} }$