Question

help guyyyyss ........... ​

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,

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

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Answer 2

Answer:

$AnswEr : -</p><p></p><p>Given Integrand,</p><p></p><p>\displaystyle \sf \int \sqrt{tan \: x} {sec}^{4} x .dx∫tanxsec4x.dx</p><p></p><p>Let t = tan x</p><p></p><p>\begin{gathered} \longrightarrow \sf \: \dfrac{dt}{dx} = {sec}^{2} x \\ \\ \longrightarrow \sf \: dx = \dfrac{dt}{ {sec}^{2}x } \end{gathered}⟶dxdt=sec2x⟶dx=sec2xdt</p><p></p><p>Now,</p><p></p><p>\begin{gathered} \implies \displaystyle \sf \int \sqrt{t} \: {sec}^{4} x . \dfrac{dt}{ {sec}^{2}x } \\ \\ \implies \displaystyle \sf \int \sqrt{t} \: {sec}^{2} x .dt \\ \\ \implies \displaystyle \sf \int \sqrt{t} ( {t}^{2} + 1)dt \\ \\ \implies \displaystyle \sf \int \sqrt{t} dt + \int t^2 \sqrt{t} dt \\ \\ \implies \sf \: \dfrac{2 \sqrt{t {}^{3} } }{3} + \dfrac{2 \sqrt{t} {}^{7} }{7} + c \\ \\ \implies \boxed{ \boxed{ \sf \: \dfrac{2 \sqrt{(tan \: x) {}^{3} } }{3} + \dfrac{2 \sqrt{(tan \: x) {}^{7} } }{7} + c}}\end{gathered}⟹∫tsec4x.sec2xdt⟹∫tsec2x.dt⟹∫t(t2+1)dt⟹∫tdt+∫t2tdt⟹32t3+72t7+c⟹32(tanx)3+72(tanx)7+c</p><p></p><p></p><p>$

Step-by-step explanation:

thanks bro