3 Fill the blanks (1) int _(0)^(pi )/(2)sin^6xdx= ( int (2)/(sqrt (x^2)-4x+3)dx=

(1) $\int_{0}^{\frac{\pi}{2}} \sin^6 x \, dx = \frac{8}{15}$<br />(2) $\int \frac{2}{\sqrt{x^2 - 4x + 3}} \, dx = 2 \sqrt{x - 3} + C$<br /><br />Explanation:<br />(1) To evaluate the integral $\int_{0}^{\frac{\pi}{2}} \sin^6 x \, dx$, we can use the power-reducing formula for $\sin^2 x$ and then integrate term by term. The result is $\frac{8}{15}$.<br /><br />(2) For the integral $\int \frac{2}{\sqrt{x^2 - 4x + 3}} \, dx$, we can complete the square in the denominator to get $\sqrt{(x - 3)^2 - 4}$. Then, we can use the substitution $u = x - 3$ to simplify the integral. The result is $2 \sqrt{x - 3} + C$, where $C$ is the constant of integration.