problem set 3

1.

${\int }_{-\pi /2}^0{e}^q\sin (q)dq$

$f(q)=\sin (q)$
$g^{\prime }(q)=e^q$

$\int f(q)g^{\prime }(q)dq=f(q)g(q)-\int g(q)f^{\prime }(q)dq$
${\int }_{-\pi /2}^0{e}^q\sin (q)dq=\sin (q)e^q-{\int }_{-\pi /2}^0{e}^q\cos (q)dq$

${\int }_{-\pi /2}^0{e}^q\cos (q)dq$
$f(q)=\cos (q)$
$g^{\prime }(q)=e^q$
${\int }_{-\pi /2}^0{e}^q\cos (q)dq=\cos (q)e^q+{\int }_{-\pi /2}^0{e}^q\sin (q)dq$

${\int }_{-\pi /2}^0{e}^q\sin (q)dq=[\sin (q)e^q{]}_{-\pi /2}^0-[\cos (q)e^q{]}_{\pi /2}^0-{\int }_{-\pi /2}^0{e}^q\sin (q)dq$
$2{\int }_{-\pi /2}^0{e}^q\sin (q)dq=[\sin (q)e^q{]}_{-\pi /2}^0-[\cos (q)e^q{]}_{\pi /2}^0$

$[\sin (q)e^q{]}_{-\pi /2}^0=\sin (0)e^0-\sin (-\frac{\pi }{2})e^{-\pi /2}=e^{-\pi /2}$
$[\cos (q)e^q{]}_{-\pi /2}^0=\cos (0)e^0-\cos (-\frac{\pi }{2})e^{-\pi /2}=1$

$2{\int }_{-\pi /2}^0{e}^q\sin (q)dq=e^{-\pi /2}-1$
${\int }_{-\pi /2}^0{e}^q\sin (q)dq=\frac{e^{-\pi /2}-1}{2}$

4.

${\int }_0^{\pi /4}\sin (2\theta )cos(2\theta )d\theta$

$u=2\theta$
$\frac{d}{d\theta }[u]=\frac{d}{d\theta }[2\theta ]$
$\frac{du}{d\theta }=2$
$d\theta =\frac{du}{2}$

${\int }_0^{\pi /4}\sin (2\theta )cos(2\theta )d\theta =2{\int }_0^{\pi /2}\sin (u)\cos (u)du$

$f(u)=\sin (u)$
$g^{\prime }(u)=\cos (u)$

$2{\int }_0^{\pi /2}\sin (u)\cos (u)du=[\sin (u)\sin (u){]}_0^{\pi /2}-{\int }_0^{\pi /2}\sin (u)\cos (u)du$

$3{\int }_0^{\pi /2}\sin (u)\cos (u)du=[{\sin }^2(u){]}_0^{\pi /2}$
$={\sin }^2(\pi /2)-{\sin }^2(0)$
$=1$

${\int }_0^{\pi /2}\sin (u)\cos (u)du=\frac{1}{3}$
${\int }_0^{\pi /4}\sin (2\theta )\cos (2\theta )d\theta =\frac{1}{3}$

23.

$\int \frac{1}{\sqrt{1+e^x}}dx$

$u=\sqrt{1+e^x}$
$\frac{du}{dx}=\frac{d}{dx}[\sqrt{1+e^x}]$
$\frac{du}{dx}=\frac{1}{2}(1+e^x{)}^{-\frac{1}{2}}\cdot (1\cdot e^x)$
$du=\frac{e^x}{2\sqrt{1+e^x}}dx$

$du=\frac{1}{\sqrt{1+e^x}}dx\cdot \frac{e^x}{2}$
$du\cdot \frac{2}{e^x}=\frac{1}{\sqrt{1+e^x}}dx$
$\int \frac{2}{e^x}du$

$u^2=1+e^x$
$u^2-1=e^x$
$\int \frac{2}{u^2-1}du$
$2\int \frac{1}{u^2-1}du$

$\frac{1}{u^2-1}=\frac{1}{(u+1)(u-1)}$
$\frac{A}{u+1}+\frac{B}{u-1}$
$\frac{A(u-1)+B(u+1)}{(u+1)(u-1)}$

$A(u-1)+B(u+1)=1$
$Au-A+Bu+B=1$
$(A+B)u+(-A+B)$

$A+B=0$
$-A+B=1$
$A=-0.5,B=0.5$

$\frac{1}{u^2-1}=\frac{1}{2}(\frac{1}{u-1}-\frac{1}{u+1})$

$\int [\frac{1}{u-1}-\frac{1}{u+1}]du$
$\int \frac{1}{u-1}du-\int \frac{1}{u+1}du$
$\ln (|u-1|)-\ln (|u+1|)$
$\ln (|\sqrt{1+e^x}-1|)-\ln (|\sqrt{1+e^x}+1|)$