Question

integrate sqrt (tanx)*sec^4 x dx​

Answer 1

AnswEr :

Given Integrand,

$\displaystyle \sf \int \sqrt{tan \: x} {sec}^{4} x .dx$

Let t = tan x

$\longrightarrow \sf \: \dfrac{dt}{dx} = {sec}^{2} x \\ \\ \longrightarrow \sf \: dx = \dfrac{dt}{ {sec}^{2}x }$

Now,

$\implies \displaystyle \sf \int \sqrt{t} \: {sec}^{4} x . \dfrac{dt}{ {sec}^{2}x } \\ \\ \implies \displaystyle \sf \int \sqrt{t} \: {sec}^{2} x .dt \\ \\ \implies \displaystyle \sf \int \sqrt{t} ( {t}^{2} + 1)dt \\ \\ \implies \displaystyle \sf \int \sqrt{t} dt + \int t^2 \sqrt{t} dt \\ \\ \implies \sf \: \dfrac{2 \sqrt{t {}^{3} } }{3} + \dfrac{2 \sqrt{t} {}^{7} }{7} + c \\ \\ \implies \boxed{ \boxed{ \sf \: \dfrac{2 \sqrt{(tan \: x) {}^{3} } }{3} + \dfrac{2 \sqrt{(tan \: x) {}^{7} } }{7} + c}}$

Answer 2

♣ Qᴜᴇꜱᴛɪᴏɴ :

$\boxed{\bf{Integrate\:\::\int \sqrt{tanx}\:sec^4xdx}}$

♣ ᴀɴꜱᴡᴇʀ :

$\boxed{\boxed{\bf{\int \sqrt{\tan \left(x\right)}\sec ^4\left(x\right)dx=\tfrac{2}{3}\tan ^{\tfrac{3}{2}}\left(x\right)+\tfrac{2}{7}\tan ^{\tfrac{7}{2}}\left(x\right)+C}}}$

♣ ᴄᴀʟᴄᴜʟᴀᴛɪᴏɴꜱ :

$\sf{\int \sqrt{\tan \left(x\right)}\sec ^4\left(x\right)dx=\int \sec ^2\left(x\right)\sec ^2\left(x\right)\sqrt{\tan \left(x\right)}dx}$

How ?

$\sf{Apply\:\: the \:\:following\:\: algebraic\:\: property\:\:: x^{a}=x^{a-2} \cdot x^{2}}$

$\mathrm{Use\:the\:following\:identity}:\quad \sec ^2\left(x\right)=1+\tan ^2\left(x\right)$

$=\int \left(1+\tan ^2\left(x\right)\right)\sec ^2\left(x\right)\sqrt{\tan \left(x\right)}dx$

$\sf{Apply\: u\: -\: substitution\:: u=\tan (x)}$

Integral Substitution definition :

$\int \:f\left(g\left(x\right)\right)\cdot \:g^'\left(x\right)dx=\int \:f\left(u\right)du,\:\quad \:u=g\left(x\right)$

$\mathrm{Substitute:}\:u=\tan \left(x\right)$

$\dfrac{du}{dx}=\sec ^2\left(x\right)$

      $\Rightarrow \:du=\sec ^2\left(x\right)dx$

      $\Rightarrow \:dx=\frac{1}{\sec ^2\left(x\right)}du$

$=\int \left(1+u^2\right)\sec ^2\left(x\right)\sqrt{u}\dfrac{1}{\sec ^2\left(x\right)}du$

$\sf{\left(1+u^2\right)\sec ^2\left(x\right)\sqrt{u}\dfrac{1}{\sec ^2\left(x\right)} : \sqrt{u}\left(1+u^2\right)}$

$=\int \sqrt{u}\left(1+u^2\right)du$

$\sf{Expand\:\sqrt{u}\left(1+u^{2}\right): \sqrt{u}+u^{\dfrac{5}{2}}}$

$\text { Apply the distributive law: } a(b+c)=a b+a c$

$a=\sqrt{u},\:b=1,\:c=u^2$

$=\sqrt{u}\cdot \:1+\sqrt{u}u^2$

$=1\cdot \sqrt{u}+u^2\sqrt{u}$

$\mathrm{Apply\:the\:Sum\:Rule}:\quad \int f\left(x\right)\pm g\left(x\right)dx=\int f\left(x\right)dx\pm \int g\left(x\right)dx$

$=\int \sqrt{u}du+\int \:u^{\frac{5}{2}}du$

$\int \sqrt{u}du = \dfrac{2}{3}u^\frac{3}{2}$

$\text { Apply radical rule: } \sqrt{a}=a^{\frac{1}{2}}$

$=\int u^{\frac{1}{2}} d u$

$\text { Apply the Power Rule: } \int x^{a} d x=\frac{x^{a+1}}{a+1}, \quad a \neq-1$

$=\dfrac{u^{\dfrac{1}{2}}+1}{\dfrac{1}{2}+1}$

$\int u^{\frac{5}{2}} d u=\frac{2}{7} u^{\frac{7}{2}}$

$=\frac{2}{3}u^{\frac{3}{2}}+\frac{2}{7}u^{\frac{7}{2}}$

$\mathrm{Substitute\:back}\:u=\tan \left(x\right)$

$=\frac{2}{3}\tan ^{\frac{3}{2}}\left(x\right)+\frac{2}{7}\tan ^{\frac{7}{2}}\left(x\right)$

$\sf{Add\:a\:constant\:to\:the\:solution}$

$\sf{=\dfrac{2}{3}\tan ^{\dfrac{3}{2}}\left(x\right)+\dfrac{2}{7}\tan ^{\dfrac{7}{2}}\left(x\right)+C}$

$\boxed{\boxed{\bf{\int \sqrt{\tan \left(x\right)}\sec ^4\left(x\right)dx=\tfrac{2}{3}\tan ^{\tfrac{3}{2}}\left(x\right)+\tfrac{2}{7}\tan ^{\tfrac{7}{2}}\left(x\right)+C}}}$