সৃজনশীলঃ ‘f(x) = sinx…’ কুমিল্লা ভিক্টোরিয়া সরকারি কলেজ ২০২৪ – যোগজীকরণ, উচ্চতর গণিত, উচ্চতর গণিত ১ম পত্র (এইচএসসি)

দেওয়া আছে,

$f\left(x\right)=\sin x$

$y=\frac{1}{\sqrt{2}}\sqrt{1+f(3\pi /2+2x)}$

$=\frac{1}{\sqrt{2}}\sqrt{1+\sin (3\pi /2+2x)}$

$=\frac{1}{\sqrt{2}}\sqrt{1-\cos 2x}$

$=\frac{1}{\sqrt{2}}\sqrt{2{\sin }^{2}x}$

$=\frac{1}{\sqrt{2}}\cdot \sqrt{2}\sin x$

$y=\sin x$

$\therefore$ ছায়ারেখা অংশের ক্ষেত্রফল $={\int }_0^{\pi }\sin xdx$

$=[-\cos x{]}_0^{\pi }$

$=[-\cos \pi +\cos 0]$

 = 1+1

 = 2 (Ans)

দেওয়া আছে,

$f\left(x\right)=\sin x$

$\int \frac{dx}{f\{{\sin }^{-1}(x-1\left)\right\}\sqrt{{\left[f\{{\sin }^{-1}(x-1\left)\right\}\right]}^2+2x}}$

$=\int \frac{dx}{\sin \{{\sin }^{-1}(x-1\left)\right\}\sqrt{{[\sin \{{\sin }^{-1}(x-1\left)\right\}]}^2+2x}}$

$=\int \frac{dx}{(x-1)\sqrt{(x-1{)}^2+2x}}$

$=\int \frac{dx}{(x-1)\sqrt{x^2-2x+1+2x}}$

$=\int \frac{dx}{(x-1)\sqrt{x^2+1}}$

$=\int \frac{dz}{z^2\cdot \frac{1}{z}\sqrt{{(1+\frac{1}{z})}^2+1}}$

ধরি, $x-1=\frac{1}{z}$

$\Rightarrow \theta 1=-\frac{1}{z^2}\frac{dz}{dx}$

$\Rightarrow dx=\frac{-dz}{z^2}$

$=\int \frac{dz}{z\sqrt{1+\frac{2}{z}+\frac{1}{z^2}+1}}$

$=\int \frac{dz}{z\sqrt{2z^2+2z+1}}$

$=\int \frac{dz}{\sqrt{2z^2+2z+1}}$

$=\int \frac{dz}{\sqrt{2(z^2+z)+1}}$

$=\int \frac{dz}{\sqrt{2{(z+\frac{1}{2})}^2-\frac{1}{4})+1}}$

$=\int \frac{dz}{\sqrt{2{(z+\frac{1}{2})}^2-\frac{1}{2}+1}}$

$=\int \frac{dz}{\sqrt{2(z+\frac{1}{2})+\frac{1}{2}}}$

$=\frac{1}{\sqrt{2}}\int \frac{dz}{\sqrt{{\left(\frac{1}{2}\right)}^2+{(z+\frac{1}{2})}^2}}$

$=\frac{1}{\sqrt{2}}\ln |z+\frac{1}{2}+\sqrt{{(z+\frac{1}{2})}^2+{\left(\frac{1}{2}\right)}^2}+c$

$[\because \int \frac{dx}{\sqrt{x^2+a^2}}=\ln |x+\sqrt{x^2+a^2}]$

$=\frac{1}{\sqrt{2}}\ln |z+\frac{1}{2}+\sqrt{z^2+2\cdot z\frac{1}{2}+\frac{1}{4}+\frac{1}{4}}+c$

$=\frac{1}{\Omega }\ln |z+\frac{1}{2}+\sqrt{z^2+z+\frac{1}{2}}+c$

$=\frac{1}{\sqrt{2}}\ln |(\frac{1}{x-1}+\frac{1}{2})+\sqrt{{\left(\frac{1}{x-1}\right)}^2+\left(\frac{1}{x-1}\right)+\frac{1}{2}}$ + C

$\left[\begin{array}{c}\because \frac{1}{z}=x-1\\ \therefore z=\frac{1}{x-1}\end{array}\right]$