# 演習問題

$\int \frac{1}{sinx+cosx+1}dx$ 　を $tan\frac{x}{2}=t$ 　と置換して解きなさい．

$y=\sqrt{2x+3}$

$y={e}^{-x}$

$y=\mathrm{sin6}x$

$y=\mathrm{cos2}x$

$y=\sqrt[3]{\frac{x+1}{x-1}}$

$y=\mathrm{tan}2x$

$y=log\left(log\left(log\left(log5x\right)\right)\right)$

$y=\sqrt[5]{5x-7}$

$y={sin}^{3}4x{cos}^{4}3x$

$y=log\left(sinx\right)$

$y=\frac{sin6x}{cos2x}$

$y=\frac{{e}^{6}cos6x}{log3x}$

$y=log|\frac{1-2x}{1+2x}|$

$y=log\frac{{e}^{3x}-3}{{e}^{3x}+1}$

$y=\frac{{e}^{x}-{e}^{-x}}{{e}^{x}+{e}^{-x}}$

$y=\frac{1}{n}log\frac{{e}^{-2nx}+{e}^{2nx}}{4}$ 　　（$n$ は自然数）

$y={\mathrm{sin}}^{-1}\frac{x}{\sqrt{7}}$

$y={\mathrm{cos}}^{-1}\sqrt{2x}$

$y={\mathrm{cos}}^{-1}\frac{2}{x}$

$y={\mathrm{tan}}^{-1}6x$

$y={\mathrm{tan}}^{-1}\frac{13}{x}$

$y=\mathrm{log}\left(x+\sqrt{{x}^{2}+4}\right)$

$y={7}^{{x}^{3}-x}$

$y=3\cdot 5{}^{{x}^{4}-3x}$

$y=log\left(sin3x\right)$

$y={x}^{x}$　 ($x>0$ )

$y={e}^{3x}cos6x$

$y=\mathrm{log}\left(\mathrm{sin}5x\right)$

$y={e}^{ax}cosbx$

$y={e}^{ax}\mathrm{cos}bx$

$z=\sqrt{3x-4y}$

$z=log\left(x-y\right)$

$z=\sqrt{5{x}^{2}+7{y}^{3}}$

$z=\mathrm{log}\left(7{x}^{4}+5{y}^{3}\right)$

$z=\mathrm{sin}4{x}^{2}+\mathrm{cos}5{y}^{3}$

$z=3\mathrm{sin}\left(4{x}^{3}-7{y}^{4}\right)$

$z=5\mathrm{cos}x{y}^{2}$

$z={\mathrm{sin}}^{-1}3{x}^{2}{y}^{3}$

$z={\mathrm{cos}}^{-1}\frac{y}{x}$

$f\left(x,y\right)=\sqrt{{x}^{2}-xy}$

$f\left(x,y\right)={sin}^{-1}\frac{x}{y}$

$z=log\sqrt{{x}^{2}+{y}^{2}}\text{\hspace{0.17em}}$ ならば ${\left(\frac{\partial z}{\partial x}\right)}^{2}+{\left(\frac{\partial z}{\partial y}\right)}^{2}=\frac{1}{{e}^{2z}}\text{\hspace{0.17em}}$ である．