Answer
\[\frac{π}{4}-\frac{1}{2}\]
Work Step by Step
Let \[I=\int_{0}^{\frac{π}{4}}\frac{x\sin x}{\cos^3 x}dx\]
\[I=\int_{0}^{\frac{π}{4}} x\tan x(\sec^2 x)dx\;\;\;\ldots (1)\]
Consider indefinite integral
\[I_1=\int x\left(\tan x \:\sec^2 x\right)dx\]
Using integration by parts
\[I_1=x\int\left(\tan x \:\sec^2 x\right)dx-\int\left((x)'\int\left(\tan x \:\sec^2 x\right)dx\right)dx\]
\[I_1=\frac{x\tan^2 x}{2}-\int\frac{\tan^2 x}{2}dx\]
\[I_1=\frac{x\tan^2 x}{2}-\frac{1}{2}\int\left(\sec^2 x-1\right)dx\]
\[I_1=\frac{x\tan^2 x}{2}-\frac{\tan x}{2}+\frac{ x}{2}+C\;\;\;\;\ldots (2)\]
Where $C$ is constant of integration
Using (2) in (1)
\[I=\left[\frac{x\tan^2 x}{2}-\frac{\tan x}{2}+\frac{ x}{2}\right]_{0}^{\frac{π}{4}}\]
\[I=\frac{π}{8}-\frac{1}{2}+\frac{π}{8}=\frac{π}{4}-\frac{1}{2}\]
Hence \[I=\frac{π}{4}-\frac{1}{2}\;\;.\]
