Question

Evaluate : ${\int }_0^{\frac{\pi }{4}}\frac{sinx+cosx}{16+9sin2x}dx.$
OR
Evaluate ${\int }_1^3(x^2+3x+e^x)dx$, as the limit of the sum.

Answer 1

Let I = ${\int }_0^{\frac{\pi }{4}}\frac{sinx+cosx}{16+9sin2x}dx.$

Put $sinx-cosx=t\Rightarrow (cosx+sinx)dx=dt$ and , $(sinx-cosx{)}^2=t^2\Rightarrow sin2x=1-t^2$.

When x = 0 $\Rightarrow t=-1$, when $x=\frac{\pi }{4}\Rightarrow t=0$

So, I = ${\int }_{-1}^0\frac{1}{25-9t^2}dt={\int }_{-1}^0\frac{1}{5^2-(3t{)}^2}dt=\frac{1}{2\times 5}\times \frac{1}{3}[log|\frac{5+3t}{5-3t}|{]}_{-1}^0$

$\Rightarrow I=\frac{1}{30}[log|1|-log\left|\frac{2}{8}\right|]=\frac{1}{30}log4=\frac{1}{15}log2.$

OR Let I =${\int }_1^3(x^2+3x+e^x)dx$

We know ${\int }_a^{b}f\left(x\right)dx={lim}_{n\to \infty }h\left[f\right(a)+f(a+h)+f(a+2h)+...+f(a+(n-1)h\left)\right],\text{As}n\to \infty ,h\to 0\Rightarrow nh=b-a\therefore {\int }_a^{b}f\left(x\right)dx={lim}_{n\to \infty }h{\sum }_{r=0}^{n-1}f(a+rh)...\left(i\right)$

Here $f\left(x\right)=x^2+3x+e^x,a=1,b=3\therefore f(a+rh)=(a+rh{)}^2+3(a+rh)+e^{a+rh}\Rightarrow f(1+rh)=(1+rh{)}^2+3(1+rh)+e^{1+rh}=4+5rh+r^2{h}^2+e{e}^{rh}$

By using (i) ${\int }_3^1(x^2+3x+e^x)dx={lim}_{n\to \infty }h{\sum }_{r=0}^{n-1}[4+5rh+r^2{h}^2+e{e}^{rh}]\Rightarrow I={lim}_{n\to \infty }h\{h^2{\sum }_{r=0}^{n-1}{r}^2+5h{\sum }_{r=0}^{n-1}r+{\sum }_{r=0}^{n-1}4+e{\sum }_{r=0}^{n-1}{e}^{rh}\}$

$\Rightarrow i={lim}_{n\to \infty }\{h^2\frac{n(n-1)(2n-1)}{6}+5h\frac{n(n-1)}{2}+4n+e(1+e^h+e^{2h}+...+e^{(n-1)h})\}$

$\Rightarrow i={lim}_{n\to \infty }\{\frac{nh(nh-h)(2nh-h)}{6}+5\times \frac{nh(nh-h)}{2}+4nh+e\times (e^{nh-1}\left)\left(\frac{h}{e^h-1}\right)\right\}$

Since $n\to \infty ,h\Rightarrow 0\Rightarrow nh=3-1=2$

So, I = $\frac{2(2-0)(4-0)}{6}+\frac{5}{2}\times \left(2\right(2-0\left)\right)+4\times 2+e\times (e^2-1)\left(1\right)=\frac{62}{3}+(e^3-e)$

Hence I = ${\int }_3^1(x^2+3x+e^x)dx=\frac{62}{3}+e^3-e$