Question

Solve: $\stackrel{4}{\underset{0}{\int }}(1-{\sec }^2\theta {)}^2$

Answer 1

solve : $\int (1-{\sec }^2\theta {)}^{2}d\theta =I$

$I={\int }_0^4({\sec }^2\theta -1){\tan }^2\theta d\theta$

$={\int }_0^4{\sec }^2\theta {\tan }^2\theta d\theta -{\int }_0^4{\tan }^2\theta d\theta$

$={\int }_0^4{\sec }^2\theta {\tan }^2\theta d\theta -{\int }_0^4({\sec }^2\theta -1)d\theta$

1) for ${\int }_0^4{\sec }^2\theta {\tan }^2\theta d\theta \to \tan \theta =\mu$

${\sec }^2\theta d\theta =d\mu$

$\therefore {\int }_0^{\tan 4}{\mu }^2.du$

$={\left[\frac{{\mu }^3}{3}\right]}_0^{\tan 4}=\frac{{\tan }^{3}4}{3}$

2) ${\int }_0^4({\sec }^2\theta -1)d\theta ={\int }_0^4{\sec }^2\theta d\theta -{\int }_0^{4}d\theta$

$=[\tan \theta {]}_0^4-[\theta {]}_0^4$

$={\tan }^4-4$

$\therefore I=\frac{{\tan }^{3}4}{3}-\tan 4+4$