সৃজনশীলঃ ‘f(x)=3−2x,g(x)=3+4x−4×2,h(x)=a2−x2…’ সিলেট বোর্ড ২০২১ – যোগজীকরণ, উচ্চতর গণিত, উচ্চতর গণিত ১ম পত্র (এইচএসসি)

দেওয়া আছে, $f\left(x\right)=3-2x$

$g\left(x\right)=3+4x-4x^2$

$\therefore \int \frac{f\left(x\right)}{g\left(x\right)}dx=\int \frac{3-2x}{\sqrt{3+4x-4x^2}}dx$

$=\int \frac{\frac{1}{4}(4-8x)+2}{\sqrt{3+4x-4x^2}}dx$

$=\frac{1}{4}\int \frac{4-8x}{\sqrt{3+4x-4x^2}}dx+\int \frac{2}{\sqrt{3+4x-4x^2}}dx$

ধরি,

$z=3+4x-4x^2$

$\to \frac{dz}{dx}=(4-8x)$

$\to dz=(4-8x)dx$

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

$=\frac{1}{4}\cdot 2\sqrt{z}+\int \frac{2dx}{\sqrt{4-1+4x-4x^2}}$

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

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

$=\frac{1}{2}\sqrt{z}+2({\sin }^{-1}\frac{1-2x}{2})-\frac{1}{2}+c$

$=\frac{1}{2}\left(\sqrt{3+4x-4x^2}\right)-{\sin }^{-1}\frac{1-2x}{2}+C$

ধরি,

$x=a\sin \theta$

$\Rightarrow \frac{dx}{d\theta }=a\cos \theta$

$\Rightarrow dx=a\cos \theta \cdot d\theta$

$\int h\left(x\right)dx=\int \sqrt{a^2-x^{2}dx}$

$=\int \sqrt{a^2-a^2{\sin }^2\theta }\cdot a\cos \theta d\theta$

$=\int \sqrt{a^2(1-{\sin }^2\theta )}\cdot a\cos \theta d\theta$

$=\int \sqrt{a^2\cdot {\cos }^2\theta }\cdot a\cos \theta d\theta$

$=a^2\int {\cos }^2\theta d\theta$

$=a^2\int {\cos }^2\theta d\theta$

$=a^2\int 2\cdot \frac{1}{2}{\cos }^2\theta d\theta$

$=\frac{a^2}{2}\int (1+\cos 2\theta )d\theta$

$=\frac{a^2}{2}[\theta +\frac{1}{2}{\sin }^2\theta ]$

$=\frac{a^2}{2}\theta +\frac{a^2}{2}\sin \theta \cdot \cos \theta +c$

$=\frac{a^2}{2}\theta +\frac{a^2}{2}\sin \theta \sqrt{1-{\sin }^2\theta }+c$

$=\frac{a^2}{2}{\sin }^{-1}\frac{x}{a}+\frac{a^2}{2}\cdot \frac{x}{a}\sqrt{1-\frac{x^2}{a^2}}+c$

$=\frac{x\sqrt{a^2-x^2}}{2}+\frac{a^2}{2}{\sin }^{-1}\frac{x}{a}+c$

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