Answer:
[tex]I=\frac{x^3}{3(25x^2+1)^{3/2}}[/tex]
Step-by-step explanation:
(a)
[tex]5x=tan{\theta}\\x = \frac{\tan{\theta}}{5}\\dx=\frac{1}{5}\sec^2{\theta}d\theta\\[/tex]
[tex]I=\frac{1}{125}\int\frac{tan^2{\theta}sec^2{\theta}}{(tan^2{\theta}+1)^{5/2}}d\theta[/tex]
[tex]I=\frac{1}{125}\int\frac{tan^2{\theta}sec^2{\theta}}{sec^5{\theta}}d\theta[/tex]
[tex]I=\frac{1}{125}\int\frac{tan^2{\theta}}{sec^3{\theta}}d\theta[/tex]
[tex]I=\frac{1}{125}\int\sin^2{\theta}{cos{\theta}}d\theta[/tex]
[tex]I=\frac{1}{125}\int\sin^2{\theta}d\sin{\theta}[/tex]
[tex]I=\frac{1}{125}\frac{1}{3}sin^3{\theta}}[/tex]
[tex]I=\frac{1}{125}\frac{1}{3}(\frac{5x}{\sqrt{25x^2+1})})^3[/tex]
