সৃজনশীলঃ ‘দৃশ্যকল্প-১: f(x)=tan⁡x.দৃশ্যকল্প-২: 9×2+25y2=225.…’ ঢাকা বোর্ড ২০২১ – যোগজীকরণ, উচ্চতর গণিত, উচ্চতর গণিত ১ম পত্র (এইচএসসি)

দেওয়া আছে, $f\left(x\right)=\tan x$

প্রদত্ত রাশি, $=\int \frac{\mathrm{dx}}{5+f(\frac{\pi }{2}-x)}$

$=\int \frac{dx}{5+\tan (\frac{\pi }{2}-x)}$

$=\int \frac{dx}{5+\cot x}$

$=\int \frac{1}{5+\frac{\cos x}{\sin x}}dx=\int \frac{\sin x}{5\sin x+\cos x}dx$

ধরি, $\sin x=m(5\sin x+\cos x)+n\frac{d}{dx}(5\sin x+\cos x)$

$=5m\sin x+m\cos x+n(5\cos x-\sin x)$

$=5m\sin x+m\cos x+5n\cos x-n\sin x$

$\therefore \sin x=(5m-n)\sin x+(m+5n)\cos x$

উভয়পক্ষ হতে $\sin x$ ও cosx এর সহগ সমীকৃত করে পাই,

$5m-n=1......\left(i\right)$

$m+5n=0.......\left(ii\right)$

এখন, (i) $\times 5\Rightarrow 25m-5n=5$

$\underset{―}{\left(ii\right)\times 1\Rightarrow m+5n=0}$

$$ (+) করে, $26m=5$

বা, $m=\frac{5}{26}$

(ii)নং এ $m$ এর মান বসিয়ে পাই,

$\frac{5}{26}+5n=0$

বা, $5n=-\frac{5}{26}$

$\therefore n=-\frac{1}{26}$

$\therefore \int \frac{\sin x}{5\sin x+\cos x}dx=\int \frac{\frac{5}{26}(5\sin x+\cos x)-\frac{1}{26}(5\cos x-\sin x)}{5\sin x+\cos x}dx$

$=\frac{5}{26}\int dx-\frac{1}{26}\int \frac{5\cos x-\sin x}{5\sin x+\cos x}dx$

$=\frac{5}{26}x-\frac{1}{26}\ln |5\sin x+\cos x|+c$

নির্ণেয় যোগজ : $\frac{5}{26}x-\frac{1}{26}\ln |5\sin x+\cos x|+c$.

দৃশ্যকল্প-২: $9x^2+25y^2=225$

বা, $\frac{x^2}{25}+\frac{y^2}{9}=1$

বা, $\frac{{\dot{y}}^2}{9}=1-\frac{x^2}{25}$

বা, $y^2=\frac{9}{25}(25-x^2)$

বা, $y=\frac{3}{5}\sqrt{25-x^2}$

এবং $x-3=0$

বা, $x=3$

$\therefore$ আবদ্ধ ক্ষেত্রের ক্ষেত্রফল

$\begin{array}{cc}& ={\int }_3^5\mathrm{dx}\\ & ={\int }_3^5\frac{3}{5}\sqrt{25-x^2}\mathrm{dx}\\ & =\frac{3}{5}{\int }_{{\sin }^{-1}\frac{3}{5}}\sqrt{25-25{\sin }^2\theta \cdot 5}\cos \theta d\theta \\ & =3{\int }_{{\sin }^{-1}\frac{3}{5}}^{5}5\sqrt{1-{\sin }^2\theta \cdot \cos \theta d\theta }\\ & =15{\int }_{{\sin }^{-1}\frac{3}{5}}^{\frac{\pi }{2}}{\cos }^2\theta d\theta {\int }_{{\sin }^{-1}\frac{3}{5}}^{\frac{\pi }{2}}2{\cos }^2\theta d\theta \end{array}$

$\begin{array}{cc}& =\frac{15}{2}{\int }_{{\sin }^{-1}\frac{3}{5}}^{\frac{\pi }{2}}(1+\cos 2\theta )d\theta \\ & =\frac{15}{2}{[\theta +\frac{\sin 2\theta }{2}]}_{{\sin }^{-1}\frac{3}{5}}^{\frac{\pi }{2}}\end{array}$

$\begin{array}{cc}& \therefore \sqrt{5^2-3^2}=4\\ & {\sin }^{-1}\theta =\frac{3}{5}\\ & \therefore {\cos }^{-1}\theta =\frac{4}{5}\end{array}$

$=\frac{15}{2}{[\theta +\frac{2\sin \theta \cdot \cos \theta }{2}]}_{{\sin }^{-1\frac{3}{5}}}^{\frac{\pi }{2}}$

$=\frac{15}{2}[(\frac{\pi }{2}+\sin \frac{\pi }{2}\cdot \cos \frac{\pi }{2})-({\sin }^{-1}\frac{3}{5}+\frac{3}{5}\cdot \frac{4}{5})]$

$=\frac{15}{2}(\frac{\pi }{2}+0-{\sin }^{-1}\frac{3}{5}-\frac{12}{25})$

$=\frac{15}{2}(\frac{\pi }{2}-{\sin }^{-1}\frac{3}{5}-\frac{12}{25})$বর্গ একক

নির্ণেয় ক্ষেত্রফল $\frac{15}{2}(\frac{\pi }{2}-{\sin }^{-1}\frac{3}{5}-\frac{12}{25})$বর্গ একক।়