Question

Evaluate: ${\int }_0^{\frac{\pi }{2}}\frac{si{n}^{2}x}{sinx+cosx}dx.$

OR

Evaluate: ${\int }_{-1}^2(e^{3x}+7x-5)dx$ as a limit of sums.

Answer 1

$LetI={\int }_0^{\frac{\pi }{2}}\frac{si{n}^{2}x}{sinx+cosx}dx......\left(i\right)$

$\Rightarrow I={\int }_0^{\frac{\pi }{2}}\frac{si{n}^2(\frac{\pi }{2}-x)}{sin(\frac{\pi }{2}-x)+cos(\frac{\pi }{2}-x)}dx..\left[\text{Using Property,}{\int }_0^{a}f\right(x)dx={\int }_0^{a}f(a-x\left)dx\right]$

$\Rightarrow I={\int }_0^{\frac{\pi }{2}}\frac{co{s}^{2}x}{sinx+cosx}dx.......\left(ii\right)Adding,\left(i\right)and\left(ii\right),\Rightarrow 2I={\int }_0^{\frac{\pi }{2}}\frac{si{n}^{2}x+co{s}^{2}x}{sinx+cosx}dx\Rightarrow 2I={\int }_0^{\frac{\pi }{2}}\frac{dx}{sinx+cosx}\Rightarrow 2I=\frac{1}{\sqrt{2}}{\int }_0^{\frac{\pi }{2}}\frac{dx}{sinx.\frac{1}{\sqrt{2}}+cosx\frac{1}{\sqrt{2}}}\Rightarrow 2I=\frac{1}{\sqrt{2}}{\int }_0^{\frac{\pi }{2}}\frac{dx}{sinx.cos\frac{\pi }{4}+cosx.sin\frac{\pi }{4}}$

$\Rightarrow 2I=\frac{1}{\sqrt{2}}{\int }_0^{\pi /2}cosec(\frac{\pi }{4}+x)dx\Rightarrow 2I=\frac{1}{\sqrt{2}}{[ln|cosec(\frac{\pi }{4}+x)-cotcosec(\frac{\pi }{4}+x)|]}_0^{\pi /2}\Rightarrow 2I=\frac{1}{\sqrt{2}}[ln|cosec(\frac{\pi }{4}+\frac{\pi }{2})-cot(\frac{\pi }{4}+\frac{\pi }{2})|]-[ln|cosec(\frac{\pi }{4}+0|)-cot(\frac{\pi }{4}+0)]\Rightarrow 2I=\frac{1}{\sqrt{2}}[ln|\sqrt{2}-(-1)|-ln|\sqrt{2}-1]\Rightarrow =\frac{1}{2\sqrt{2}}\left[ln|\frac{\sqrt{2}+1}{\sqrt{2}-1}|\right]$

OR

${\int }_{-1}^2(e^{3x}+7x-5)dx.Heref\left(x\right)=e^{3x}+7x-5a=-1,b=2,h=\frac{b-a}{n}=\frac{3}{n}\text{By definition}{\int }_{-1}^2(e^{3x}+7x-5)dx={lim}_{n\to \infty }{\sum }_{r=1}^{n}h.f(a+rh){lim}_{n\to \infty }{\sum }_{r=1}^{n}h.f(-1+rh)={lim}_{n\to \infty }{\sum }_{r=1}^{n}h.(e^{3(-1+rh)}+7(-1+rh)-5)$

${lim}_{n\to \infty }[h.e^{-3}.e^{3h}(1+e^{3h}+e^{6h}+.....+e^{3nh})+7h^2(1+2+3+.....+n)-12nh]$

${lim}_{n\to \infty }[\frac{h{e}^{3h}}{e^3}\times \frac{e^{3nh}-1}{e^{3h}-1}+7h^2\frac{n(n+1)}{2}-12nh]$
${lim}_{n\to \infty }[\left(\frac{h{e}^{3\times \frac{3}{n}}}{n{e}^3}\right)\times (e^{3\times \frac{3}{n}}-1)\times \frac{n}{3\times 3}+\frac{63}{n^2}\times \frac{n(n+1)}{2}-12\times 3]$

Now applying the limit we get

$=\frac{e^9-1}{3e^3}+\frac{63}{2}-36=\frac{e^9-1}{3e^3}-\frac{9}{2}$