91Ó°ÊÓ

Find the values of \( p \) for which the integral converges and evaluate the integral for those values of \( p \). \( \displaystyle \int_0^1 \frac{1}{x^p}\ dx \)

Find the area of the region bounded by the given curves. \( y = \sin^2 x \) , \( y = \sin^3 x \) , \( 0 \le x \le \pi \)

Evaluate the integral. \( \displaystyle \int x^3 \sqrt{x + c}\ dx \)

Find the area of the region bounded by the given curves. \( y = \tan x \) , \( y = \tan^2 x \) , \( 0 \le x \le \frac{\pi}{4} \)

Evaluate the integral. \( \displaystyle \int \frac{x \ln x}{\sqrt{x^2 - 1}}\ dx \)

Find the values of \( p \) for which the integral converges and evaluate the integral for those values of \( p \). \( \displaystyle \int_e^\infty \frac{1}{x (\ln x)^p}\ dx \)

Evaluate the integral. \( \displaystyle \int \frac{dx}{x^4 - 16} \)

Use a graph to find approximate x-coordinates of the points of intersection of the given curves. Then find (approximately) the area of the region bounded by the curves. \( y = \arcsin \left(\frac{1}{2} x \right) \), \( y = 2 - x^2 \)

Find the values of \( p \) for which the integral converges and evaluate the integral for those values of \( p \). \( \displaystyle \int_0^1 x^p \ln x\ dx \)

The German Mathematician Karl Weierstrass (1815-1897) noticed that the substitution \( t = \tan (\frac{x}{2}) \) will convert any rational function of \( \sin x \) and \( \cos x \) into an ordinary rational function of \( t \). (a) If \( t = \tan (\frac{x}{2}) \) , \( -\pi < x < \pi \) , sketch a right triangle or use trigonometric identities to show that \( \cos \left (\dfrac{x}{2} \right) = \dfrac{1}{\sqrt{1 + t^2}} \) and \( \sin \left (\dfrac{x}{2} \right) = \dfrac{t}{\sqrt{1 + t^2}} \) (b) Show that \( \cos x = \dfrac{1 - t^2}{1 + t^2} \) and \( \sin x = \dfrac{2t}{1 + t^2} \) (c) Show that $$ dx = \frac{2}{1 + t^2}\ dt $$