Интегралы, помогите )))

Решите задачу:

$1)\ \int_{ \pi /4}^{ \pi /3} \dfrac{dx}{cos^2x} = tg\ x|_{ \pi /4}^{ \pi /3}=tg \frac{ \pi }{3}- tg \frac{ \pi }{4} = \sqrt{3} -1.$

$2)\ f(x)= \begin{cases} \frac{1}{ \sqrt{x}}, \ x \leq 1 \\ x^3,\ x\ \textgreater \ 1 \end{cases}\\ \\ \int_{1/4}^{2} f(x)dx = \int_{1/4}^{1} \frac{1}{ \sqrt{x}} dx+ \int_{1}^{2} x^3 dx=2 \sqrt{x} |_{1/4}^{1}+ \frac{x^4}{4}|{1}^{2} =\\ \\ =2-1+-0,25=4,75.$