Question

Integration = x³ tan⁴ x⁴ sec² x⁴

Answer 1

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

Given integral is

$\displaystyle\int\rm \: {x}^{3} \: {tan}^{4} {x}^{4} \: {sec}^{2} {x}^{4} \: dx \:$

can be re-arranged as

$\rm \: = \: \displaystyle\int\rm \: {tan}^{4} {x}^{4} \: \red{ {sec}^{2} {x}^{4} \: {x}^{3}} \: dx \:$

can be further rewritten as

$\rm \: = \: \displaystyle\int\rm \: {( {tanx}^{4})}^{4} \: \red{ {sec}^{2} {x}^{4} \: {x}^{3}} \: dx \:$

To evaluate this integral we use Method of Substitution

So, Substitute

$\rm \: {tanx}^{4} = y$

$\rm \: \dfrac{d}{dx}{tanx}^{4} = \dfrac{dy}{dx}$

$\rm \: {sec}^{2} {x}^{4} \: ( {4x}^{3}) \: dx \: = \: dy$

$\rm\implies \:\rm \: {x}^{3} {sec}^{2} {x}^{4} \: dx \: = \: \dfrac{dy}{4}$

So, above integral can be rewritten as

$\rm \: = \: \dfrac{1}{4} \displaystyle\int {y}^{4} \: dy$

$\rm \:  =  \: \dfrac{1}{4} \times \dfrac{ {y}^{4 + 1} }{4 + 1} + c$

$\rm \:  =  \: \dfrac{ {y}^{5} }{20} + c$

$\rm \:  =  \: \dfrac{ {tan}^{5} {x}^{4} }{20} + c$

Hence,

$\rm\implies \:\boxed{\tt{ \displaystyle\int\rm \: {x}^{3} \: {tan}^{4} {x}^{4} \: {sec}^{2} {x}^{4} \: dx \: = \frac{ {tan}^{5} {x}^{4} }{20} + c}}$

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FORMULA USED

$\boxed{\tt{ \dfrac{d}{dx}tanx = {sec}^{2}x \: }}$

$\boxed{\tt{ \dfrac{d}{dx} {x}^{n} = {nx}^{n - 1} \: }}$

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

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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

Question:-

Integration = x³ tan⁴ x⁴ sec² x⁴.

Given:-

Integral is x³ tan⁴ x⁴ sec² x⁴.

To Find:-

Integration of given function w.r.t.x = ?.

Solution:-

$\sf \large \int x {}^{3} .tan {}^{4} .x {}^{4} .sec {}^{2} .x {}^{4} \: dx$

$\sf \large Let,x {}^{4} = t ⇒ 4x {}^{3} \: dx = dt⇒x {}^{3} \: dx = \frac{1}{4} \: dt$

$\sf \large = \frac{1}{4} \int tan {}^{ {}^{4} } \: t.sec {}^{2} \: t \: \: dt$

$\sf \large Now, let \: tan \: t = u ⇒ sec² \: t . dt = du$

$\sf \large = \frac{1}{4} \int u {}^{4} .du$

$\sf \large = \frac{1}{4} . \frac{u {}^{5} }{5} + c$

$\sf \large = \frac{1}{4} \times \frac{1}{5} (tan \: t) {}^{5} + c$

$\sf \large = \frac{1}{20} tan {}^{5} \: x {}^{4} + c$

Answer:-

$\sf \large \blue{Hence, the \: Integration \: of \: given \: } \\ \sf \large \blue{ function \: is \: \frac{1}{20} tan {}^{5} \: x {}^{4} + c.}$

Hope you have satisfied. ⚘