evaluate the integrals\n59. $int_{ln(pi/6)}^{ln(pi/2)}2e^{v}cos e^{v}dv$\n60. $int_{0}^{sqrt{lnpi}}2x…

Explanation:

Step1: Solve integral 59

Let $u = e^v$. Then $du=e^v dv$. When $v = \ln(\frac{\pi}{6})$, $u=\frac{\pi}{6}$; when $v=\ln(\frac{\pi}{2})$, $u = \frac{\pi}{2}$. The integral $\int_{\ln(\frac{\pi}{6})}^{\ln(\frac{\pi}{2})}2e^v\cos(e^v)dv=2\int_{\frac{\pi}{6}}^{\frac{\pi}{2}}\cos(u)du$. $2\int_{\frac{\pi}{6}}^{\frac{\pi}{2}}\cos(u)du=2[\sin(u)]_{\frac{\pi}{6}}^{\frac{\pi}{2}}$.

Step2: Calculate the value of $2[\sin(u)]_{\frac{\pi}{6}}^{\frac{\pi}{2}}$

$2(\sin(\frac{\pi}{2})-\sin(\frac{\pi}{6}))=2(1 - \frac{1}{2})=1$.

Step3: Solve integral 60

Let $t = e^{x^2}$. Then $dt = 2x e^{x^2}dx$. When $x = 0$, $t = 1$; when $x=\sqrt{\ln\pi}$, $t=\pi$. The integral $\int_{0}^{\sqrt{\ln\pi}}2x e^{x^2}\cos(e^{x^2})dx=\int_{1}^{\pi}\cos(t)dt$. $\int_{1}^{\pi}\cos(t)dt=[\sin(t)]_{1}^{\pi}$.

Step4: Calculate the value of $[\sin(t)]_{1}^{\pi}$

$\sin(\pi)-\sin(1)=0 - \sin(1)=-\sin(1)$.

Step5: Solve integral 61

Let $u = 1 + e^r$. Then $du=e^r dr$. The integral $\int\frac{e^r}{1 + e^r}dr=\int\frac{du}{u}$. $\int\frac{du}{u}=\ln|u|+C=\ln(1 + e^r)+C$.

Step6: Solve integral 62

Multiply the numerator and denominator of $\int\frac{dx}{1 + e^x}$ by $e^{-x}$: $\int\frac{e^{-x}}{1 + e^{-x}}dx$. Let $w=1 + e^{-x}$, then $dw=-e^{-x}dx$. So $\int\frac{e^{-x}}{1 + e^{-x}}dx=-\int\frac{dw}{w}=-\ln|w|+C=-\ln(1 + e^{-x})+C$.

Answer:

59. $1$
60. $-\sin(1)$
61. $\ln(1 + e^r)+C$
62. $-\ln(1 + e^{-x})+C$