নির্দিষ্ট সমাকলের তালিকা

গণিতে নির্দিষ্ট সমাকল

$\underset{a}{\overset{b}{\int }}f(x)dx$

xy-সমতলের গ্রাফ f, x-অক্ষ, তথা x = a ও x = b রেখা দ্বারা সীমাবদ্ধ ক্ষেত্রের ক্ষেত্রফলের সমান হয়। (x-অক্ষের উপরের ক্ষেত্রফল ধনাত্মক নেওয়া হয় যেখানে x-অক্ষের নীচের ক্ষেত্র ঋনাত্মক)

মূলদ বা অমূলদ অপেক্ষক সমন্বিত নির্দিষ্ট সমাকল

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{dx}{x^2+a^2}=\frac{\pi }{2a}$

$\underset{0}{\overset{a}{\int }}\frac{x^{m}dx}{x^n+a^n}=\frac{\pi a^{m-n+1}}{n\sin [(m+1)\pi /n)]}$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{x^{p-1}dx}{1+x}=\frac{\pi }{\sin p\pi }0<p<1$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{x^{m}dx}{1+2x\cos \beta +x^2}=\frac{\pi }{\sin (m\pi )}\frac{\sin (m\beta )}{\sin \beta }$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{dx}{\sqrt{a^2-x^2}}=\frac{\pi }{2}$

$\underset{0}{\overset{a}{\int }}\sqrt{a^2-x^2}dx=\frac{\pi a^2}{4}$

$\underset{0}{\overset{a}{\int }}{x}^m(a^n-x^n{)}^{p}dx=\frac{a^{m+1+np}\mathrm{\Gamma }[(m+1)/n]\mathrm{\Gamma }(p+1)}{n\mathrm{\Gamma }[((m+1)/n)+p+1]}$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{x^{m}dx}{({x^n+a^n)}^r}=\frac{(-1{)}^{r-1}\pi a^{m+1-nr}\mathrm{\Gamma }[(m+1)/n]}{n\sin [(m+1)\pi /n](r-1)!\mathrm{\Gamma }[(m+1)/n-r+1]}0<m+1<nr$

ত্রিকোণমিতিক অপেক্ষক সমন্বিত নির্দিষ্ট সমাকল

$\underset{0}{\overset{\pi }{\int }}\sin mx\sin nxdx=\{\begin{array}{cc}0& \text{if }m\ne n\\ \pi /2& \text{if }m=n\end{array}m,n\text{ integers}$

$\underset{0}{\overset{\pi }{\int }}\cos mx\cos nxdx=\{\begin{array}{cc}0& \text{if }m\ne n\\ \pi /2& \text{if }m=n\end{array}m,n\text{ integers}$

$\underset{0}{\overset{\pi }{\int }}\sin mx\cos nxdx=\{\begin{array}{cc}0& \text{if }m+n\text{ even}\\ \frac{2m}{m^2-n^2}& \text{if }m+n\text{ odd}\end{array}m,n\text{ integers}.$

$\underset{0}{\overset{\pi /2}{\int }}{\sin }^{2}xdx=\underset{0}{\overset{\pi /2}{\int }}{\cos }^{2}xdx=\pi /4$

$\underset{0}{\overset{\pi /2}{\int }}{\sin }^{2m}xdx=\underset{0}{\overset{\pi /2}{\int }}{\cos }^{2m}xdx=\frac{1\times 3\times 5\times \cdots \times (2m-1)}{2\times 4\times 6\times \cdots \times 2m}\frac{\pi }{2}m=1,2,3,\dots$

$\underset{0}{\overset{\pi /2}{\int }}{\sin }^{2m+1}xdx=\underset{0}{\overset{\pi /2}{\int }}{\cos }^{2m+1}xdx=\frac{2\times 4\times 6\times \cdots \times 2m}{1\times 3\times 5\times \cdots \times (2m-1)}m=1,2,3,\dots$

$\underset{0}{\overset{\pi /2}{\int }}{\sin }^{2p-1}{\cos }^{2q-1}xdx=\frac{\mathrm{\Gamma }(p)\mathrm{\Gamma }(q)}{2\mathrm{\Gamma }(p+q)}$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{\sin px}{x}dx=\{\begin{array}{cc}\pi /2& \text{if }p>0\\ 0& \text{if }p=0\\ -\pi /2& \text{if }p<0\end{array}$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{\sin px\cos qx}{x}dx=\{\begin{array}{cc}0& \text{ if }p>q>0\\ \pi /2& \text{ if }0<p<q\\ \pi /4& \text{ if }p=q>0\end{array}$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{\sin px\sin qx}{x^2}dx=\{\begin{array}{cc}\pi p/2& \text{ if }0<p\le q\\ \pi q/2& \text{ if }0<q\le p\end{array}$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{{\sin }^{2}px}{x^2}dx=\frac{\pi p}{2}$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{1-\cos px}{x^2}dx=\frac{\pi p}{2}$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{\cos px-\cos qx}{x}dx=\ln \frac{q}{p}$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{\cos px-\cos qx}{x^2}dx=\frac{\pi (q-p)}{2}$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{\cos mx}{x^2+a^2}dx=\frac{\pi }{2a}{e}^{-ma}$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{x\sin mx}{x^2+a^2}dx=\frac{\pi }{2}{e}^{-ma}$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{\sin mx}{x(x^2+a^2)}dx=\frac{\pi }{2a^2}(1-e^{-ma})$

$\underset{0}{\overset{2\pi }{\int }}\frac{dx}{a+b\sin x}=\frac{2\pi }{\sqrt{a^2-b^2}}$

$\underset{0}{\overset{2\pi }{\int }}\frac{dx}{a+b\cos x}=\frac{2\pi }{\sqrt{a^2-b^2}}$

$\underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{dx}{a+b\cos x}=\frac{{\cos }^{-1}(b/a)}{\sqrt{a^2-b^2}}$

$\underset{0}{\overset{2\pi }{\int }}\frac{dx}{(a+b\sin x{)}^2}=\underset{0}{\overset{2\pi }{\int }}\frac{dx}{(a+b\cos x{)}^2}=\frac{2\pi a}{(a^2-b^2{)}^{3/2}}$

$\underset{0}{\overset{2\pi }{\int }}\frac{dx}{1-2a\cos x+a^2}=\frac{2\pi }{1-a^2}0<a<1$

$\underset{0}{\overset{\pi }{\int }}\frac{x\sin xdx}{1-2a\cos x+a^2}=\{\begin{array}{cc}\frac{\pi }{a}\ln (1+a)& \text{if }|a|<1\\ \pi \ln (1+1/a)& \text{if }|a|>1\end{array}$

$\underset{0}{\overset{\pi }{\int }}\frac{\cos mxdx}{1-2a\cos x+a^2}=\frac{\pi a^m}{1-a^2},a^2<1,m=0,1,2,\dots$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\sin a{x}^{2}dx=\underset{0}{\overset{\mathrm{\infty }}{\int }}\cos a{x}^2=\frac{1}{2}\sqrt{\frac{\pi }{2a}}$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\sin a{x}^n=\frac{1}{n{a}^{1/n}}\mathrm{\Gamma }(1/n)\sin \frac{\pi }{2n},n>1$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\cos a{x}^n=\frac{1}{n{a}^{1/n}}\mathrm{\Gamma }(1/n)\cos \frac{\pi }{2n},n>1$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{\sin x}{\sqrt{x}}dx=\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{\cos x}{\sqrt{x}}dx=\sqrt{\frac{\pi }{2}}$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{\sin x}{x^p}dx=\frac{\pi }{2\mathrm{\Gamma }(p)\sin (p\pi /2)},0<p<1$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{\cos x}{x^p}dx=\frac{\pi }{2\mathrm{\Gamma }(p)\cos (p\pi /2)},0<p<1$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\sin a{x}^2\cos 2bxdx=\frac{1}{2}\sqrt{\frac{\pi }{2a}}(\cos \frac{b^2}{a}-\sin \frac{b^2}{a})$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\cos a{x}^2\cos 2bxdx=\frac{1}{2}\sqrt{\frac{\pi }{2a}}(\cos \frac{b^2}{a}+\sin \frac{b^2}{a})$

চলসূচকীয় অপেক্ষক সমন্বিত নির্দিষ্ট সমাকল

$\underset{0}{\overset{\mathrm{\infty }}{\int }}{e}^{-ax}\cos bxdx=\frac{a}{a^2+b^2}$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}{e}^{-ax}\sin bxdx=\frac{b}{a^2+b^2}$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{e^{-ax}\sin bx}{x}dx={\tan }^{-1}\frac{b}{a}$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{e^{-ax}-e^{-bx}}{x}dx=\ln \frac{b}{a}$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}{e}^{-a{x}^2}dx=\frac{1}{2}\sqrt{\frac{\pi }{a}}$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}{e}^{-a{x}^2}\cos bxdx=\frac{1}{2}\sqrt{\frac{\pi }{a}}{e}^{-b^2/4a}$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}{e}^{-(a{x}^2+bx+c)}dx=\frac{1}{2}\sqrt{\frac{\pi }{a}}{e}^{(b^2-4ac)/4a}\mathrm{erfc}\frac{b}{2\sqrt{a}},\text{ where }\mathrm{erfc}(p)=\frac{2}{\sqrt{\pi }}\underset{p}{\overset{\mathrm{\infty }}{\int }}{e}^{-x^2}dx$

$\underset{-\mathrm{\infty }}{\overset{+\mathrm{\infty }}{\int }}{e}^{-(a{x}^2+bx+c)}dx=\sqrt{\frac{\pi }{a}}{e}^{(b^2-4ac)/4a}$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}{x}^n{e}^{-ax}dx=\frac{\mathrm{\Gamma }(n+1)}{a^{n+1}}$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}{x}^m{e}^{-a{x}^2}dx=\frac{\mathrm{\Gamma }[(m+1)/2]}{2a^{(m+1)/2}}$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}{e}^{-a{x}^2-b/x^2}dx=\frac{1}{2}\sqrt{\frac{\pi }{a}}{e}^{-2\sqrt{ab}}$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{x}{e^x-1}dx=\zeta (2)=\frac{{\pi }^2}{6}$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{x^{n-1}}{e^x-1}dx=\mathrm{\Gamma }(n)\zeta (n)$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{x}{e^x+1}dx=\frac{1}{1^2}-\frac{1}{2^2}+\frac{1}{3^2}-\frac{1}{4^2}+\dots =\frac{{\pi }^2}{12}$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{\sin mx}{e^{2\pi x}-1}dx=\frac{1}{4}\coth \frac{m}{2}-\frac{1}{2m}$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}(\frac{1}{1+x}-e^{-x})\frac{dx}{x}=\gamma$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{e^{-x^2}-e^{-x}}{x}dx=\frac{\gamma }{2}$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}(\frac{1}{e^x-1}-\frac{e^{-x}}{x})dx=\gamma$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{e^{-ax}-e^{-bx}}{x\sec px}dx=\frac{1}{2}\ln \frac{b^2+p^2}{a^2+p^2}$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{e^{-ax}-e^{-bx}}{x\csc px}dx={\tan }^{-1}\frac{b}{p}-{\tan }^{-1}\frac{a}{p}$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{e^{-ax}(1-\cos x)}{x^2}dx={\cot }^{-1}a-\frac{a}{2}\ln (a^2+1)$

$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{e}^{-x^2}dx=\sqrt{\pi }$

$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{x}^{2(n+1)}{e}^{-x^2/2}dx=\frac{(2n+1)!}{2^{n}n!}\sqrt{2\pi }n=0,1,2,\dots$

লঘুগণকীয় অপেক্ষক সমন্বিত নির্দিষ্ট সমাকল

$\underset{0}{\overset{1}{\int }}{x}^m(\ln x{)}^{n}dx=\frac{(-1{)}^{n}n!}{(m+1{)}^{n+1}}m>-1,n=0,1,2,\dots$

$\underset{0}{\overset{1}{\int }}\frac{\ln x}{1+x}dx=-\frac{{\pi }^2}{12}$

$\underset{0}{\overset{1}{\int }}\frac{\ln x}{1-x}dx=-\frac{{\pi }^2}{6}$

$\underset{0}{\overset{1}{\int }}\frac{\ln (1+x)}{x}dx=\frac{{\pi }^2}{12}$

$\underset{0}{\overset{1}{\int }}\frac{\ln (1-x)}{x}dx=-\frac{{\pi }^2}{6}$

হাইপারবোলিক অপেক্ষক সমন্বিত নির্দিষ্ট সমাকল

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{\sin ax}{\sinh bx}dx=\frac{\pi }{2b}\tanh \frac{a\pi }{2b}$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{\cos ax}{\cosh bx}dx=\frac{\pi }{2b}\frac{1}{\cosh \frac{a\pi }{2b}}$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{x}{\sinh ax}dx=\frac{{\pi }^2}{4a^2}$

বিবিধ

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{f(ax)-f(bx)}{x}dx=[f(0)-f(\mathrm{\infty })]\ln \frac{b}{a}$

$\underset{-a}{\overset{a}{\int }}(a+x{)}^{m-1}(a-x{)}^{n-1}dx=(2a{)}^{m+n-1}\frac{\mathrm{\Gamma }(m)\mathrm{\Gamma }(n)}{\mathrm{\Gamma }(m+n)}$

আরও দেখুন

• সমাকলনের তালিকাসমূহ (List of integrals)
• গামা অপেক্ষক (Gamma function)
• সীমার তালিকা (List of limits)