Question

integrate the integral​
step by step

Answer 1

Taking the indefinite integral,

$\displaystyle\longrightarrow\sf{\int\sqrt{40^2+(30-10t)^2}\ dt=\int\sqrt{(10\times4)^2+[10(3-t)]^2}\ dt}$

$\displaystyle\longrightarrow\sf{\int\sqrt{40^2+(30-10t)^2}\ dt=10\int\sqrt{16+(t-3)^2}\ dt}$

Let,

$\displaystyle\longrightarrow\sf{u=t-3}$

$\displaystyle\longrightarrow\sf{du=dt}$

We see $\sf{u=-3}$ for $\sf{t=0}$ and $\sf{u=0}$ for $\sf{t=3.}$

Then,

$\displaystyle\longrightarrow\sf{\int\sqrt{40^2+(30-10t)^2}\ dt=10\int\sqrt{16+u^2}\ du}$

Let,

$\displaystyle\longrightarrow\sf{u=4\cot v\quad\iff\quad v=\cot^{-1}\left(\dfrac{u}{4}\right)}$

$\displaystyle\longrightarrow\sf{du=-4\csc^2v\ dv}$

Then,

$\displaystyle\longrightarrow\sf{\int\sqrt{40^2+(30-10t)^2}\ dt=-40\int\csc^2v\sqrt{16+(4\cot v)^2}\ dv}$

$\displaystyle\longrightarrow\sf{\int\sqrt{40^2+(30-10t)^2}\ dt=-40\int\csc^2v\sqrt{16+16\cot^2v}\ dv}$

$\displaystyle\longrightarrow\sf{\int\sqrt{40^2+(30-10t)^2}\ dt=-160\int\csc^2v\sqrt{1+\cot^2v}\ dv}$

$\displaystyle\longrightarrow\sf{\int\sqrt{40^2+(30-10t)^2}\ dt=-160\int\csc^3v\ dv\quad\quad\dots(1)}$

But,

$\displaystyle\longrightarrow\sf{\int\csc^3v\ dv=\int\csc v\cdot\csc^2v\ dv}$

Performing integration by parts,

$\displaystyle\longrightarrow\sf{\int\csc^3v\ dv=\csc v\cdot\int\csc^2v\ dv-\int\left(\dfrac{d}{dv}\,\csc v\cdot\int\csc^2v\ dv\right)\ dv}$

$\displaystyle\longrightarrow\sf{\int\csc^3v\ dv=-\csc v\cot v-\int-\csc v\cot v\cdot-\cot v\ dv}$

$\displaystyle\longrightarrow\sf{\int\csc^3v\ dv=-\csc v\cot v-\int\csc v\cot^2v\ dv}$

$\displaystyle\longrightarrow\sf{\int\csc^3v\ dv=-\csc v\cot v-\int\csc v(\csc^2v-1)\ dv}$

$\displaystyle\longrightarrow\sf{\int\csc^3v\ dv=-\csc v\cot v-\int\csc^3v\ dv+\int\csc v\ dv}$

$\displaystyle\longrightarrow\sf{2\int\csc^3v\ dv=-\csc v\cot v+\int\csc v\ dv}$

$\displaystyle\longrightarrow\sf{\int\csc^3v\ dv=-\dfrac{\csc v\cot v\displaystyle+\int\dfrac{\csc v(-\cot v-\csc v)}{(\csc v+\cot v)}\ dv}{2}}$

$\displaystyle\longrightarrow\sf{\int\csc^3v\ dv=-\dfrac{\csc v\cot v\displaystyle+\int\dfrac{-\csc v\cot v-\csc^2v}{\csc v+\cot v}\ dv}{2}}$

$\displaystyle\longrightarrow\sf{\int\csc^3v\ dv=-\dfrac{\csc v\cot v\displaystyle+\ln|\csc v+\cot v|}{2}}$

Then (1) becomes,

$\displaystyle\longrightarrow\sf{\int\sqrt{40^2+(30-10t)^2}\ dt=160\cdot \dfrac{\csc v\cot v\displaystyle+\ln|\csc v+\cot v|}{2}}$

$\displaystyle\longrightarrow\sf{\int\sqrt{40^2+(30-10t)^2}\ dt=80(\csc v\cot v+\ln|\csc v+\cot v|)}$

$\displaystyle\longrightarrow\sf{\int\sqrt{40^2+(30-10t)^2}\ dt=80\left(\csc\left(\cot^{-1}\left(\dfrac{u}{4}\right)\right)\cdot\dfrac{u}{4}+\ln\left|\csc\left(\cot^{-1}\left(\dfrac{u}{4}\right)\right)+\dfrac{u}{4}\right|\right)}$

$\displaystyle\longrightarrow\sf{\int\sqrt{40^2+(30-10t)^2}\ dt=80\left(\dfrac{u\sqrt{u^2+16}}{16}+\ln\left|\dfrac{u+\sqrt{u^2+16}}{4}\right|\right)}$

$\displaystyle\longrightarrow\sf{\int\sqrt{40^2+(30-10t)^2}\ dt=5u\sqrt{u^2+16}+80\ln\left|u+\sqrt{u^2+16}\right|-80\ln4}$

We do not consider the constant left in the integral.

$\displaystyle\longrightarrow\sf{\int\sqrt{40^2+(30-10t)^2}\ dt=5u\sqrt{u^2+16}+80\ln\left|u+\sqrt{u^2+16}\right|}$

Because it can be considered or added with the integral constant.

Now calculating the definite integral as mentioned to be found,

$\displaystyle\longrightarrow\sf{\int\limits_0^3\sqrt{40^2+(30-10t)^2}\ dt=80\left[\dfrac{u\sqrt{u^2+16}}{16}+\ln\left|\dfrac{u+\sqrt{u^2+16}}{4}\right|\right]_{-3}^0}$

RHS is in terms of u so the limits wrt u should be taken there.

$\displaystyle\longrightarrow\sf{\int\limits_0^3\sqrt{40^2+(30-10t)^2}\ dt=\left[5u\sqrt{u^2+16}+80\ln\left|u+\sqrt{u^2+16}\right|\right]_{-3}^0}$

$\displaystyle\longrightarrow\sf{\int\limits_0^3\sqrt{40^2+(30-10t)^2}\ dt=80\ln4-(-75+80\ln|-3+5|)}$

$\displaystyle\longrightarrow\sf{\int\limits_0^3\sqrt{40^2+(30-10t)^2}\ dt=80\ln4+75-80\ln2}$

$\displaystyle\longrightarrow\sf{\int\limits_0^3\sqrt{40^2+(30-10t)^2}\ dt=80\ln\left(\dfrac{4}{2}\right)+75}$

$\displaystyle\longrightarrow\sf{\int\limits_0^3\sqrt{40^2+(30-10t)^2}\ dt=80\ln2+75}$

$\displaystyle\longrightarrow\sf{\underline{\underline{\int\limits_0^3\sqrt{40^2+(30-10t)^2}\ dt=130.452}}}$

Answer 2

♣ Qᴜᴇꜱᴛɪᴏɴ :

$\sf{\int _0^3\:\sqrt{\left(40\right)^2+\left(30-10t\right)^2}dt}$

♣ ᴀɴꜱᴡᴇʀ :

$\boxed{\sf{\int _0^3\sqrt{40^2+\left(30-10t\right)^2}dt=160\left(\frac{15}{32}+\frac{1}{2}\ln \left(2\right)\right)\quad \left(\mathrm{Decimal:\quad }\:130.45177\dots \right)}}$

♣ ᴄᴀʟᴄᴜʟᴀᴛɪᴏɴꜱ :

$\text { Apply u - substitution: } u=30-10 t$

$\sf{=\int _{30}^0-\dfrac{\sqrt{1600+u^2}}{10}du}$

$\sf{\int _a^bf\left(x\right)dx=-\int _b^af\left(x\right)dx,\:a<b}$

$\sf{=-\int _0^{30}-\dfrac{\sqrt{1600+u^2}}{10}du}$

$\mathrm{Take\:the\:constant\:out}:\quad \int a\cdot f\left(x\right)dx=a\cdot \int f\left(x\right)dx$

$=-\left(-\dfrac{1}{10}\cdot \int _0^{30}\sqrt{1600+u^2}du\right)$

_______________________________

$\text { Apply Trig Substitution: } u=40 \tan (v)$

$=-\left(-\dfrac{1}{10}\cdot \int _0^{\arctan \left(\tfrac{3}{4}\right)}1600\sec ^3\left(v\right)dv\right)$

$\mathrm{Take\:the\:constant\:out}:\quad \int a\cdot f\left(x\right)dx=a\cdot \int f\left(x\right)dx$

$=-\left(-\dfrac{1}{10}\cdot \:1600\cdot \int _0^{\arctan \left(\tfrac{3}{4}\right)}\sec ^3\left(v\right)dv\right)$

$\sf{\mathrm{Apply\:Integral\:Reduction}:\quad \int \sec ^n\left(x\right)dx=\dfrac{\sec ^{n-1}\left(x\right)\sin \left(x\right)}{n-1}+\dfrac{n-2}{n-1}\int \sec ^{n-2}\left(x\right)dx}$

$\sf{\int \sec ^3\left(v\right)dv=\dfrac{\sec ^2\left(v\right)\sin \left(v\right)}{2}+\frac{1}{2}\int \sec \left(v\right)dv}$

$\sf{=-\left(-\dfrac{1}{10}\cdot \:1600\left(\left[\dfrac{\sec ^2\left(v\right)\sin \left(v\right)}{2}\right]^{\arctan \left(\dfrac{3}{4}\right)}_0+\dfrac{1}{2}\cdot \int _0^{\arctan \left(\dfrac{3}{4}\right)}\sec \left(v\right)dv\right)\right)}$

$\mathrm{Use\:the\:common\:integral}:\quad \int \sec \left(v\right)dv=\ln \left|\tan \left(v\right)+\sec \left(v\right)\right|$

$\sf{=-\left(-\dfrac{1}{10}\cdot \:1600\left(\left[\dfrac{\sec ^2\left(v\right)\sin \left(v\right)}{2}\right]^{\arctan \left(\dfrac{3}{4}\right)}_0+\dfrac{1}{2}\left[\ln \left|\tan \left(v\right)+\sec \left(v\right)\right|\right]^{\arctan \left(\dfrac{3}{4}\right)}_0\right)\right)}$

_______________________________

$\sf{Simplify\::-\left(-\dfrac{1}{10}\cdot \:1600\left(\left[\dfrac{\sec ^2\left(v\right)\sin \left(v\right)}{2}\right]^{\arctan \left(\dfrac{3}{4}\right)}_0+\dfrac{1}{2}\left[\ln \left|\tan \left(v\right)+\sec \left(v\right)\right|\right]^{\arctan \left(\dfrac{3}{4}\right)}_0\right)\right)}$

$\sf{=160\left(\left[\dfrac{1}{2}\sec \left(v\right)\tan \left(v\right)\right]^{\arctan \left(\dfrac{3}{4}\right)}_0+\dfrac{1}{2}\left[\ln \left|\tan \left(v\right)+\sec \left(v\right)\right|\right]^{\arctan \left(\dfrac{3}{4}\right)}_0\right)}$

_______________________________

$\begin{aligned}&\text { Compute the boundaries: }\left[\frac{1}{2} \sec (v) \tan (v)\right]_{0}^{\arctan \left(\frac{3}{4}\right)}=\frac{15}{32}\\\\&\text { Compute the boundaries: }[\ln |\tan (v)+\sec (v)|]_{0}^{\arctan \left(\frac{3}{4}\right)}=\ln (2)\end{aligned}$

$\boxed{\bf{=160\left(\dfrac{15}{32}+\dfrac{1}{2}\ln \left(2\right)\right)}}$

$\large{\boxed{\bf{=130.45177\:\:or\:\: 130.452\:\:(approx)}}}$

Refer to attachment for graph !!!!