- සංයුක්ත ගණිතය I ප්රශ්න පත්රයේ A කොටසේ ලකුණු 25 ගැටළුවක් සහ B කොටසේ (රචනා ප්රශ්න) ගැටළුවක මෙම පාඩමෙහි සිද්ධාන්ත ඇතුලත් වේ.
- මෙම පාඩම සඳහා අවකලනය හා ත්රිකෝණමිතිය දැනුම අවශ්යවේ.
අනුකලනයේ අර්ථ දැක්වීම
g(x) යනු X හි අවකල්ය ශ්රිතයක් \frac d{dx} g(x)= f(x) ද නම්, x විෂයෙන් f(x) හි අනුකලනය g(x) වේ. මෙය ∫f(x)dx= g(x) ලෙස ලියනු ලබන අතර f(x) දී ඇති විට g(x) සෙවීමේ ක්රියාවලියට අනුකලනය කිරීම යැයි කියනු ලැබේ.
සටහන ;-
C ඕනෑම නියතයක් විට,
\begin{array}{rcl}\frac d{dx}\left(g\left(x\right)+c\right)&=&\frac{\displaystyle d}{\displaystyle dx}g\left(x\right)+\frac{\displaystyle d}{\displaystyle dx}c\\&=&f\left(x\right)+0\\&=&f\left(x\right)\end{array}ලෙස ලැබෙන නිසා අර්ථ දැක්වීමට අනුව,
\begin{array}{rcl}\int f\left(x\right)dx&=&g\left(x\right)+c\end{array} වේ.
- මෙයින් පැහැදිලි වන්නේ අනුකලනය නියත අගයකින් වෙනස් විය හැකි බවයි. එම නිසා මෙම අනුකලනය අනිශ්චිත යැයි කියනු ලැබේ.
ප්රතිව්යුත්පන්නයෙන් සූත්ර ලබා ගැනීම
- \begin{array}{l} \frac d{dx}\left(\sin x\right)=\cos x\Rightarrow\int\cos xdx=\sin x+c\\\\\\\\\end{array}
- \begin{array}{l}\frac d{dx}\left(\cos x\right)=-\sin x\Rightarrow\int\sin xdx=-\cos x+c\\\\\\\\\end{array}
- \begin{array}{l}\frac d{dx}\left(\tan x\right)=sec^2x\Rightarrow\int sec^2xdx=\tan x+c\\\\\\\\\end{array}
- \begin{array}{l}\frac d{dx}\left(-\cos ecx\right)=\cos ecx.cotx\Rightarrow\int\cos ecx.cotxdx=-\cos ecx+c\\\\\\\\\end{array}
- \begin{array}{l}\frac d{dx}\left(secx\right)=secx.\tan x\Rightarrow\int secx.\tan xdx=secx+c\\\\\\\\\end{array}
- \begin{array}{l}\frac d{dx}\left(-cotx\right)=\cos ec^2x\Rightarrow\int\cos ec^2xdx=-cotx+c\\\\\\\\\end{array}
- \begin{array}{l}\frac d{dx}\left(\frac{x^{n+1}}{n+1}\right)=x^n\Rightarrow\int x^ndx=\frac{\displaystyle x^{n+1}}{\displaystyle n+1}+c\;;n\neq-1\\\\\\\\\end{array}
- \begin{array}{l}\frac d{dx}\left(\ln\left|x\right|\right)=\frac1x\Rightarrow\int\frac{\displaystyle1}{\displaystyle x}dx=\ln\left|x\right|+c\;;\;x\neq0\\\\\\\\\end{array}
- \begin{array}{l}\frac d{dx}\left[\sin^{-1}\left(\frac xa\right)\right]=\frac1{\sqrt{a^2-x^2}}\Rightarrow\int\frac{\displaystyle1}{\displaystyle\sqrt{a^2-x^2}}dx=\sin^{-1}\left(\frac{\displaystyle x}{\displaystyle a}\right)+c\\\\\\\\\end{array}
- \begin{array}{l}\frac d{dx}\left[\tan^{-1}\left(\frac xa\right)\right]=\frac1{x^2+a^2}\Rightarrow\int\frac{\displaystyle1}{\displaystyle x^2+a^2}dx=\tan^{-1}\left(\frac{\displaystyle x}{\displaystyle a}\right)+c\\\\\\\\\end{array}
- \begin{array}{l}\frac d{dx}\left(e^x\right)=e^x\Rightarrow\int e^xdx=e^x+c\\\\\\\\\end{array}
- ඉහත සමීකරණ වල x හැර විචල්ය නැත.
- අනුකලනය පාඩමේදී ඉදිරියට ගැටළු විසඳීමට ඉහත සූත්ර මතක තබා ගත යුතුයි.
අනුකලනයේ ප්රමේයයන්
\int\;kf(x)dx=k\int\;f(x)dx
\int{f(x)\pm g(x)}dx=\int f(x)dx\pm\int g(x)dx
පහත ගැටළු විසඳමු
1)\int2\sin xdx=2\int\sin xdx=-2\cos x+c
2)\int\frac3xdx=3\int\frac1xdx=3\ln\left|x\right|+c
3)\int3\cos ecx.cotxdx=-3\cos ec^2x+c
4)\int kdx=k\int dx=kx+c
ඉහත අනුකලනයේ යොදා ගැනෙන්නේ \int_{}^{}x^{n}dx=\frac{x^{n+1}}{n+1}+c සූත්රයේම n=0 අවස්ථාව වේ.
5)\begin{array}{rcl}\int\left(2x-\frac3x\right)dx&=&2\int xdx-3\int\frac1xdx\\&=&2\frac{x^2}2+3\ln\left|x\right|+c\end{array}
6)\int(2\sin x-3\cos x+4)dx=2\int\sin xdx-3\int\cos xdx+4\int dx=-2\cos x-3\sin x+4x
7)\int\left(2e^x-5\right)dx=2\int e^xdx-5\int dx=2e^x-5x+c
8)\begin{array}{rcl}\int\frac3{x^2+25}dx&=&\int\frac{\displaystyle3}{\displaystyle x^2+5^2}dx\\&=&\frac35\tan^{-1}\left(\frac x5\right)+c\\&&\\&&\\&&\end{array}
9)\begin{array}{rcl}\int\tan^2xdx&=&\int(sec^2x-1)dx\\&=&\int sec^2xdx-\int dx\\&=&\tan x-x+c\\&&\end{array}
10)\begin{array}{rcl}\int cot^2xdx&=&\int(\cos ec^2x-1)dx\\&=&\int\cos ec^2xdx-\int dx\\&=&-cotx-x+c\\&&\\&&\end{array}
11)\begin{array}{rcl}\int\frac5{\sqrt{9-x^2}}dx&=&5\int\frac1{\sqrt{3^2-x^2}}dx\\&=&5\sin^{-1}\left(\frac x3\right)+c\\&&\\&&\end{array}
12)\begin{array}{rcl}\int3\cos ecx.cotxdx&=&3\int\cos ecx.cotxdx\\&=&-3\cos ecx+c\\&&\\&&\end{array}
ඉහත ප්රතිව්යුත්පන්නයෙන් ලබාගත් සූත්ර වලට අමතරව මෙම පහතින් දක්වා ඇති සූත්ර දෙකත් මතකයේ තබා ගැනීමෙන් ගැටළු විසඳීමට පහසුවක් ඇත. නමුත් මෙම සූත්ර දෙක යොදාගැනෙන ගැටළු විසඳීම සඳහා ආදේශ කිරීම් ද යොදා ගන්නා ආකාරය පාඩමෙහි පසුව සාකච්ඡා කෙරේ.
\begin{array}{rcl}\frac d{dx}\left(\ln\left|x+\sqrt{x^2-a^2}\right|\right)&=&\frac1{\sqrt{x^2-a^2}}\Rightarrow\int\frac{\displaystyle1}{\displaystyle\sqrt{x^2-a^2}}dx=\ln\left|x+\sqrt{x^2-a^2}\right|+c\\&&\\&&\end{array}
\begin{array}{rcl}\frac d{dx}\left(\ln\left|x+\sqrt{x^2+a^2}\right|\right)&=&\frac1{\sqrt{x^2+a^2}}\Rightarrow\int\frac{\displaystyle1}{\displaystyle\sqrt{x^2+a^2}}dx=\ln\left|x+\sqrt{x^2+a^2}\right|+c\\&&\\&&\end{array}උදා: \begin{array}{rcl}\int\frac{\displaystyle2}{\displaystyle\sqrt{x^2-16}}d&=&2\int\frac{\displaystyle dx}{\displaystyle\sqrt{x^2-4^2}}\\&=&2\ln\left|x+\sqrt{x^2-4^2}\right|+c\end{array}
\begin{array}{c}\int\frac{\displaystyle3dx}{\sqrt{5-x^2}}=3\int\frac{\displaystyle dx}{\sqrt{x^2+(\sqrt5)^2}}\\=3ln\left|x+\sqrt{x^2+(\sqrt5)^2}\right|+c\end{array}x හි රේඛීය ප්රකාශ ඇති ශ්රිත වල අනුකලනය
\begin{array}{rcl}\int f\left(x\right)dx&=&g\left(x\right)+c\end{array} නම්,
\begin{array}{rcl}\int f\left(ax+b\right)dx&=&\frac1ag(ax+b)+c\;\end{array} ( c– අභිමත නියතය )
සාධනය:-\begin{array}{rcl}{\frac d{dx}\left[\frac1ag(ax+b)+c\right]\;}&=&\int\frac1af(ax+b).a\\&=&{f(ax+b)}\end{array}
උදාහරණ :-
1.\begin{array}{rcl}\int\left(2x+1\right)^2dx&=&\frac13\left(2x+1\right)^3.\frac12+c\\&=&\frac16\left(2x+1\right)^3+c\end{array} ( c– අභිමත නියතය )
2.\begin{array}{rcl}\int\;\sin(\;5x\;+\;1\;)\;dx\;\;&=&-\;\frac15\;\cos\;(\;5x\;+\;1\;)\;+\;c\;\;\end{array} ( c- අභිමත නියතය )
3.\begin{array}{rcl}\int\;\sin(\;5x\;+\;1\;)\;dx\;\;&=&-\;\frac15\;\cos\;(\;5x\;+\;1\;)\;+\;c\;\;\end{array} ( c- අභිමත නියතය )
4.\begin{array}{rcl}\int e^{5x+2}dx&=&\frac15e^{5x+2}+c\end{array} ( c- අභිමත නියතය )
5.\begin{array}{rcl}\int\frac1{4x-3}dx&=&\frac14\ln\left|4x-3\right|+c\end{array} ( c- අභිමත නියතය )
6.\begin{array}{rcl}\int sec^2\left(3x+5\right)dx&=&\frac13\tan\left(3x+5\right)+c\end{array}
7.\begin{array}{rcl}\int\frac1{\sqrt{1-49x^2}}dx&=&\int\frac1{\sqrt{1-\left(7x\right)^2}}dx\\&=&\frac17\sin^{-1}\left(7x\right)+c\end{array}
8.\begin{array}{rcl}\int\frac1{\sqrt{25+4x^2}}dx&=&\int\frac1{\sqrt{5^2+\left(2x\right)^2}}dx\\&=&\frac12\ln\left|2x+\sqrt{\left(2x\right)^2+5^2}\right|+c\\&=&\frac12\ln\left|2x+\sqrt{4x^2+25}\right|+c\\&&\\&&\\&&\end{array}
9.\begin{array}{rcl}\int\sin2x.\cos3xdx&=&\frac12\int\left(\sin5x-\sin x\right)dx\\&=&\frac12\int\sin5xdx-\frac12\int\sin xdx\\&=&\frac1{10}\cos5x+\frac12\cos x+c\\&&\\&&\\&&\end{array}
10.\begin{array}{rcl}\int\cos4x.\cos5xdx&=&\frac12\int\left(\cos9x+\cos x\right)dx\\&=&\frac12\int\cos9xdx+\frac12\int\cos xdx\\&=&\frac1{18}\sin9x+\frac12\sin x+c\\&&\\&&\\&&\end{array}
11.\begin{array}{rcl}\int3\sin6x\sin2xdx&=&\int3\left(-\frac12\right)\left(\cos8x-\cos4x\right)dx\\&=&-\frac32\int\cos8xdx+\frac32\int\cos4xdx\\&=&\frac3{16}\sin8x-\frac38\sin4x+c\\&&\\&&\\&&\end{array}
12.\begin{array}{rcl}\int\sin^23xdx&=&\frac12\int\left[1-\cos6x\right]dx\\&=&\frac{\displaystyle1}{\displaystyle2}\int dx-\frac12\int\cos6xdx\\&=&\frac{\displaystyle1}{\displaystyle2}x-\frac1{12}\sin6x+c\\&&\\&&\\&&\end{array}
13.\begin{array}{rcl}\int5\cos^2\left(2x-1\right)dx&=&\int\frac52\left\{1-\cos\left[2\left(2x-1\right)\right]\right\}dx\\&=&\frac52\int dx-\frac52\int\cos\left(4x-2\right)dx\\&=&\frac{25}{16}\cos\left(4x-2\right)+c\\&&\end{array}
14.\begin{array}{rcl}&&\int\sin^3xdx\end{array}
\begin{array}{l}\sin3x\equiv3\sin x-4\sin^3x\\\sin^3x\equiv\frac14\left(3\sin x-\sin3x\right)\end{array}\begin{array}{rcl}\int\sin^3xdx&=&\frac14\int\left(3\sin x-\sin3x\right)dx\\&=&\frac34\int\sin xdx-\frac14\int\sin3xdx\\&=&-\frac34\cos x+\frac1{12}\cos3x+c\end{array}
15.\begin{array}{rcl}&&\int\cos^32xdx\end{array}
16.\begin{array}{rcl}\cos6x&\equiv&4\cos^3x2x-3\cos2x\\\cos^32x&\equiv&\frac14\left(\cos6x-3\cos2x\right)\\\int\cos^32x&=&\frac14\int\left(\cos6x-3\cos2x\right)dx\\&=&\frac14\int\cos6xdx-\frac34\int\cos2xdx\\&=&-\frac1{24}\sin6x+\frac38\sin2x+c\end{array}
sinx හා cosx වල බලයන් අනුකලනය කිර්රිම සඳහා තවත් ක්රමයක් මෙම පාඩමේ පසුව සාකච්ඡා කෙරේ.
ලබ්ධියක් ආකරයේ අනුකල වලදී වැදගත් වන ප්රමේයයක්
හරයේ අවකලන සංගුණකය ලවයේ ඇත්නම් හෝ ලවයේ නිර්මාණය කල හැකි නම් මෙම ක්රමය යෝග්ය වේ.
\begin{array}{rcl}&&\\int\frac{f'\left(x\right)}{f\left(x\right)}dx&=&\ln\left|f\left(x\right)\right|+c\\&&\\&&\end{array}
සාධනය:- \begin{array}{rcl}\frac d{dx}\left[\ln\left|f\left(x\right)\right|+c\right]&=&\int\frac{f'\left(x\right)}{f\left(x\right)}+c\\&&\\&&\\&&\end{array}
අනුකලනයේ අර්ථ දැක්වීමට අනුව,
\begin{array}{rcl}\int\frac{f'\left(x\right)}{f\left(x\right)}dx&=&\ln\left|f\left(x\right)\right|+c\\&&\\&&\\&&\end{array}උදාහරණ:-
- 1)\begin{array}{rcl}\int\frac{\sin x-\cos x}{\sin x+\cos x}dx&=&-\int\frac{\displaystyle\cos x-\sin x}{\displaystyle\sin x+\cos x}\\&=&-\ln\left|\sin x+\cos x\right|+c\\&&\\&&\\&&\end{array} ( c- අභිමත නියතය )
2)\begin{array}{rcl}\int\frac{x+1}{x^2+2x+5}dx&=&\frac12\int\frac{2x+2}{x^2+2x+5}dx\\&=&\frac12\ln\left|x^2+2x+5\right|+c\\&&\\&&\\&&\end{array}
3)\begin{array}{rcl}\int\tan xdx&=&\int\frac{\sin{\displaystyle x}}{\cos{\displaystyle x}}dx\\&=&-\int\left(\frac{\displaystyle-\sin x}{\cos{\displaystyle x}}\right)dx\\&=&-\ln\left|\cos x\right|+c\end{array}
4)\begin{array}{rcl}\int cotxdx&=&\int\frac{\cos x}{\sin x}dx\\&=&\ln\left|\sin x\right|+c\end{array}
5)\begin{array}{rcl}\int secxdx&=&\int\frac{secx\left(secx+\tan x\right)}{\left(secx+\tan x\right)}dx\\&=&\ln\left|secx+\tan x\right|+c\end{array}
6)\begin{array}{rcl}\int\cos ecxdx&=&-\int\left(-\frac{\cos ecx\left(\cos ecx+cotx\right)}{\left(\cos ecx+cotx\right)}\right)dx\\&=&-\ln\left|\cos ecx+cotx\right|+c\end{array}
7)\begin{array}{rcl}\int\frac1{x\left(x^{2020}+1\right)}dx&=&\int\frac{\left(x^{2020}+1\right){\displaystyle-}{\displaystyle{\displaystyle x}^{2020}}}{x\left(x^{2020}+1\right)}dx\\&=&\int\frac1xdx-\int\frac{x^{2019}}{\left(x^{2020}+1\right)}dx\\&=&\ln\left|x\right|-\frac{\ln{\displaystyle\left|x^{2020}+1\right|}}{2020}+c\end{array}
8)\begin{array}{rcl}\int\frac1{x\ln\left|x\right|}dx&=&\ln\left|\ln\left|x\right|\right|+c\end{array}
9)\begin{array}{rcl}\int\frac1{1+e^x}dx&=&-\int\frac{e^{-x}}{1+e^{-x}}dx\\&=&-\ln\left|1+e^{-x}\right|+c\end{array}