Question

Evaluate the integral of (√tan x )sec⁴ x dx

Answer 1

$Solution: \\Let ~I=\int(\sqrt {tan~ x}) sec^4 x ~dx \\~~~~~~~I=\int(\sqrt {tan~ x}) sec^2 x ~sec^2 x ~dx \\~~~~~~~I=\int(\sqrt {tan ~x}) (1+tan^2 x) ~sec^2 x ~dx \\~~~~~~~I=\int[(\sqrt {tan~ x})sec^2 x (1+tan^2 x) ] ~dx \\~~~~~~~I=\int[(\sqrt {tan ~x} )~sec^2 x +(\sqrt {tan~ x} )~sec^2 x~tan^2 x] ~dx \\~~~~~~~I=\int[(\sqrt {tan~ x} )~sec^2 x +(\sqrt {tan ~x} )~tan^2 x~sec^2 x] ~dx \\~~~~~~~I=\int(\sqrt {tan~ x} )~sec^2 x~dx +\int(\sqrt {tan ~x} )~~tan^2 x~sec^2 x ~dx \\Let ~u= tan ~x \Rightarrow du=sec^2 x~dx \\~~~~~~~~I=\int(\sqrt {u} )~du +\int(\sqrt {u} )~~u^2~du \\~~~~~~~~~~~=\int u^{\frac{1}{2}} ~du +\int u^{\frac{1}{2}} ~u^2 ~du \\~~~~~~~~~~~=\int u^{\frac{1}{2}} ~du +\int u^{\frac{5}{2}}~du \\ \therefore I= \frac{u^{\frac{1}{2}+1}}{\frac{1}{2}+1}+\frac{u^{\frac{5}{2}+1}}{\frac{5}{2}+1}+C \\ ~~~~I= \frac{u^{\frac{3}{2}}}{\frac{3}{2}}+\frac{u^{\frac{7}{2}}}{\frac{7}{2}}+C \\~~~~I=\frac{2}{3}u^{\frac{3}{2}}+\frac{2}{7}u^{\frac{7}{2}}+C \\ Now ~ back ~substitute~u=tan ~x \\ \therefore \int(\sqrt {tan~ x}) sec^4 x ~dx=\frac{2}{3}~tan^{\frac{3}{2}} x+\frac{2}{7}~tan^{\frac{7}{2}} x+C$