সৃজনশীলঃ ‘f(x)=11+sin⁡x−cox,g(x)=2+x1+x2…’ দিনাজপুর বোর্ড ২০২১ – যোগজীকরণ, উচ্চতর গণিত, উচ্চতর গণিত ১ম পত্র (এইচএসসি)

দেওয়া আছে,

$f\left(x\right)=\frac{1}{1+\sin x-\cos x}$

$\therefore \int f\left(x\right)dx=\int \frac{1}{1+\sin x-\cos x}dx$

$=\int \frac{1}{1+\frac{2\tan \frac{x}{2}}{1+{\tan }^2\frac{x}{2}}-\frac{1-{\tan }^2\frac{x}{2}}{1+{\tan }^2\frac{x}{2}}dx}$

$=\int \frac{1+{\tan }^2\frac{x}{2}}{1+2\tan \frac{x}{2}-1+{\tan }^2\frac{x}{2}}dx$

$=\int \frac{{\sec }^2\frac{x}{2}}{2{\tan }^2\frac{x}{2}+2\tan \frac{x}{2}}dx$

ধরি,

$\tan \frac{x}{2}=y$

$\Rightarrow \frac{1}{2}{\sec }^2\frac{x}{2}dx=dy$

$\Rightarrow {\sec }^2\frac{x}{2}dx=2dy$

$\therefore \int f\left(x\right)dx=\int \frac{2dy}{2y^2+2y}$

$=\int \frac{1}{y(y+1)}dy$

$=\int \{\frac{1}{y(0+1)}+\frac{1}{(y+1)(-1)}\}dy$

$=\int \frac{1}{y}dy-\int \frac{1}{(y+1)}dy$

$=\int \frac{1}{y}dy-\int \frac{1}{y+1}dy$

$=\ln |y|-\ln |y+1|+c$

$=\ln \left|\frac{y}{y+1}\right|+c$

$\therefore \int f\left(x\right)dx=\ln \left|\frac{\tan \frac{x}{2}}{\tan \frac{x}{2}+1}\right|+c$

দেওয়া আছে,

$\begin{array}{cc}& g\left(x\right)=\frac{2+x}{1+x^2}\\ & \therefore {\int }_0^{1}g\left(x\right)dx={\int }_0^1\frac{2+x}{1+x^2}dx\\ & ={\int }_0^1\frac{2}{1+x^2}dx+{\int }_0^1\frac{x}{1+x^2}dx\\ & =2{\int }_0^1\frac{1}{1+x^2}dx+\frac{1}{2}{\int }_0^1\frac{2x}{1+x^2}dx\\ & =2{[{\tan }^{-1}x]}_0^1+\frac{1}{2}{[\ln (1+x^2)]}_0^1\\ & [\because \int \frac{f^{\prime }\left(x\right)}{f\left(x\right)}dx=\ln f(x)+c]\\ & =2({\tan }^{-1}1-{\tan }^{-1}0)+\frac{1}{2}[\ln 2-\ln 1]\\ & =2(\frac{\pi }{4}-0)+\frac{1}{2}(\ln 2-0)\\ & =\frac{\pi }{2}+\frac{1}{2}\ln 2\end{array}$

$\therefore$ নির্ণেয় মানঃ

$\frac{\pi }{2}+\frac{1}{2}\ln 2$