Question

Solve:
${\int }_0^{10}{\sec }^2(3x+6)dx$.

Answer 1

$I={\int }_0^{10}{\sec }^2(3x+6)dx$
$=\frac{\tan (3x+6)}{3}{|}_0^{10}=\frac{1}{3}\left[\tan \right({36}^{\circ }-\tan \left(6^{\circ }\right)]$
$[\because {\int }_a^{b}f(x)dx=I(x){|}_a^b=I(b)-I(a\left)\right]$.