সৃজনশীলঃ ‘f(x) = sin…’ বরিশাল বোর্ড ২০২১ – যোগজীকরণ, উচ্চতর গণিত, উচ্চতর গণিত ১ম পত্র (এইচএসসি)

f(x) = sinx

$\begin{array}{cc}& \therefore f(\frac{\pi }{2}-\theta )=\sin (\frac{\pi }{2}-\theta )\\ & =\cos \theta \end{array}$

$\therefore \int \frac{d\theta }{1+3{\left\{f(\frac{\pi }{2}-\theta )\right\}}^2}$

$=\int \frac{d\theta }{1+3(\cos \theta {)}^2}$

$=\int \frac{d\theta }{1+3{\cos }^2\theta }$

$=\int \frac{\frac{d\theta }{{\cos }^2\theta }}{\frac{1+3{\cos }^2\theta }{{\cos }^2\theta }}$

$=\int \frac{\frac{1}{{\cos }^2\theta }d\theta }{\frac{1}{{\cos }^2\theta }+\frac{3{\cos }^2}{{\cos }^2\theta }}$

$=\int \frac{{\sec }^2\theta d\theta }{{\sec }^2\theta +3}$

$=\int \frac{{\sec }^2\theta d\theta }{1+{\tan }^2\theta +3}$ [যেহেতু, ${\sec }^2\theta =1+{\tan }^2\theta$ ]

ধরি

$\tan \theta =z$

বা, $\frac{d}{d\theta }(\tan \theta )=\frac{d}{d\theta }\left(z\right)$

বা, ${\sec }^2\theta =\frac{\mathrm{dz}}{d}$

$\therefore dz={\sec }^2\theta d\theta$

$\begin{array}{cc}& =\int \frac{{\sec }^2\theta d\theta }{{\tan }^2\theta +4}\\ & \therefore \int \frac{{\sec }^2\theta d\theta }{{\tan }^2\theta +4}=\int \frac{dz}{z^2+4}\\ & =\int \frac{dz}{z^2+2^2}\\ & =\int \frac{dz}{2^2+z^2}\\ & =\frac{1}{2}{\tan }^{-1}\left(\frac{z}{2}\right)+c\\ & =\frac{1}{2}{\tan }^{-1}\left(\frac{\tan \theta }{2}\right)+c\text{ [Ans.] }\end{array}$

এখানে, $f(\frac{\pi }{2}-x)=\sin (\frac{\pi }{2}-x)$

$=\cos X$

$\therefore {\{1+f(\frac{\pi }{2}-x)\}}^{2}f\left(x\right)dx$

$=(1+\cos x{)}^2\sin xdx$

ধরি, $\cos x=z$

বা, $\frac{d}{dx}(\cos x)=\frac{d}{dx}\left(z\right)$

বা, $-\sin x=\frac{\mathrm{dz}}{\mathrm{dx}}$

বা, $-\sin xdx=dz$

$\therefore \sin xdx=-dz$

$={\int }_1^0(1+z{)}^2(-dz)$

$=-{\int }_1^0(1+z{)}^{2}dz$

$={\int }_0^1(1+z{)}^{2}dz$

$={\int }_0^1(1+2z+z^2)dz$

$={\int }_0^{1}1\cdot dz+{\int }_0^{1}2zdz+{\int }_0^1{z}^{2}dz$

$=[z{]}_0^1+2\left[\frac{z^2}{2}\right]1+\left[\frac{z^3}{3}\right]\begin{array}{cc}& 1\\ & 0\end{array}$

$=(1-0)+2\times (\frac{1^2}{2}-\frac{0^2}{2})+(\frac{1^3}{3}-\frac{0^3}{3})$

$=1+2\times \frac{1}{2}+\frac{1}{3}$

$=1+1+\frac{1}{3}$

$=\frac{3+3+1}{3}=\frac{7}{3}$

$\therefore {\{1+f(\frac{\pi }{2}-x)\}}^{2}f\left(x\right)dx=\frac{7}{3}$ [Showed]