9.\;\int\sqrt{\mathbf a^{\mathbf2}+\mathbf x^{\mathbf2}\;}\mathbf{dx}ආකාරයේ අනුකල
- මෙවැනි අනුකලන x=a\tan{\theta} ආදේශය භාවිතා කරමු.
උදා.\int\sqrt{9+x^2}dx සලකමු.
x=3\tan\theta ආදේශ කරමු.
x විෂයෙන් අවකලනයෙන්,
dx=3\sec^2\theta\;d\theta
\begin{array}{rcl}\tan\theta&=&\frac x3\\\sec\theta&=&\;\sqrt{1+\tan^2\theta}\\\sec\theta&=&\;\sqrt{1+\frac{x^2}9}=\frac13\sqrt{9+x^2}\\\sec\theta&=&\frac13\sqrt{9+x^2}\end{array}
\begin{array}{rcl}I&=&\int\sqrt{9+x^2}dx\\I&=&\int\sqrt{9+9\tan^2\theta}.3\sec^2\theta\;d\theta\\I&=&\int\sqrt{9(1+\tan^2\theta)}.3\sec^2\theta\;d\theta\\I&=&\int\sqrt{9\sec^2\theta}.3\sec^2\theta\;d\theta\\I&=&\int3\sec\theta.3\sec^2\theta\;d\theta\\I&=&\int9\sec^3\theta\;d\theta\end{array}
- කොටස් වශයෙන් අනුකලනය භාවිතයෙන් මෙතැන් සිට ගණන විසදමු.
\begin{array}{rcl}u&=&sec\theta\\\frac{du}{d\theta}&=&sec\theta tan\theta\\\frac{dv}{d\theta}&=&sec2\theta\\\int dv&=&\int sec^2\theta d\theta\\v&=&tan\theta\end{array}
\begin{array}{l}I=9\int\sec\theta.\sec^2\theta\;d\theta\\I=9\sec\theta\tan\theta-9\int\sec\theta.\tan^2\theta\;d\theta\\I=9\sec\theta\tan\theta-9\int\sec\theta.\left(\sec^2\theta-1\right)\;d\theta\\I=9\sec\theta\tan\theta+9\int\sec\theta d\theta-9\int\sec^3\theta\;d\theta;(I=9\int\sec^3\theta\;d\theta)\\2I=9\sec\theta\tan\theta+9\ln{\vert\sec\theta+\tan{\theta\vert}}+c\\I=\frac92.\;\frac13\sqrt{9+x^2}.\frac x3+\frac92.\ln{\vert\;}\frac13\sqrt{9+x^2}+\frac x3\vert+c\\I=\frac32.x\sqrt{9+x^2}.+\frac92.\ln{\vert\;}\frac13\sqrt{9+x^2}+\frac x3\vert+c\text{ : c යනු අභිමත නියතයකි.}\end{array}
10.\;\int\sqrt{\mathbf x^{\mathbf2}-\mathbf a^{\mathbf2}\;}\mathbf{dx}ආකාරයේ අනුකල
මෙවැනි අනුකලන x=a\sec\theta ආදේශය භාවිතා කරමු.
උදා.
\int\sqrt{x^2-7}dx සලකමු.
x=\sqrt{7\;}\sec\theta ආදේශ කරමු.
x විෂයෙන් අවකලනයෙන්,
dx=\sqrt{7\;}\sec\theta\;tan\theta\;d\theta
\begin{array}{rcl}\sec\theta&=&\frac x{\sqrt7}\\\tan\theta&=&\sqrt{\sec^2\theta-1}\\\tan\theta&=&\sqrt{\frac{x^2}7-1}J=\int\sqrt{x^2-7}dx\\J&=&\int\sqrt{7\sec^2\theta-7}.\sqrt7\sec\theta\;tan\theta\;d\theta\\J&=&\int\sqrt{{7(\sec}^2\theta-1)}.\sqrt7\sec\theta\;tan\theta\;d\theta\\J&=&\int\sqrt{7\tan^2\theta}.\sqrt7\sec\theta\;tan\theta\;d\theta\\J&=&\int\sqrt7\tan\theta.\sqrt7\sec\theta\;tan\theta\;d\theta\\J&=&\int7\sec\theta\;tan\;^2\theta\;d\theta\\J&=&7\int\sec\theta\;{(\sec}^2\theta-1)\;d\theta\\J&=&7\int\sec^3\theta\;d\theta-7\int\sec{\theta\;d\theta\rightarrow(1)}\end{array}
J_1=7\int\sec^3\theta\;d\theta ලෙස ගනිමු.J_1=7\int\sec\theta\sec^2\theta\;d\theta
කොටස් වශයෙන් අනුකලනය භාවිතයෙන් මෙතැන් සිට ගණන විසදමු.
\begin{array}{rcl}u&=&sec\theta\\\frac{du}{d\theta}&=&sec\theta tan\theta\\\frac{dv}{d\theta}&=&sec^2\theta\\\frac{dv}{d\theta}&=&sec^2\theta d\theta\\v&=&tan\theta\end{array}
\begin{array}{rcl}J_1&=&7\int\sec\theta.\sec^2\theta\;d\theta\\J_1&=&7\sec\theta\tan\theta-7\int\sec\theta.\tan^2\theta\;d\theta\\J_1&=&7\sec\theta\tan\theta-7\int\sec\theta.\left(\sec^2\theta-1\right)\;d\theta\\J_1&=&7\sec\theta\tan\theta+7\int\sec\theta d\theta-7\int\sec^3\theta\;d\theta;(J_1=7\int\sec^3\theta\;d\theta)\\2J_1&=&7\sec\theta\tan\theta+7\int\sec\theta d\theta\\J_1&=&\frac72\sec\theta\tan\theta+\frac72\int\sec\theta d\theta\end{array}
දැන් J_1=7\int\sec^3\theta\;d\theta ,(1) හි ආදේශ කරමු.
\begin{array}{rcl}J&=&\frac72\sec\theta\tan\theta+\frac72\int\sec\theta d\theta-7\int\sec\theta\;d\theta\\J&=&\frac72\sec\theta\tan\theta-\frac72\int\sec\theta d\theta\\J&=&\frac72\sec\theta\tan\theta-\frac72\ln{\vert\sec\theta+\tan{\theta\vert}}+c\\J&=&\frac72.\;\frac x{\sqrt7}.\;\sqrt{\frac{x^2}7-1}-\frac72.\ln\left|\frac x{\sqrt7}+\sqrt{\frac{x^2}7-1}\right|+c\\J&=&\;\frac x2.\;\sqrt{x^2-7}-\frac72.\ln\left|\frac x{\sqrt7}+\sqrt{\frac{x^2}7-1}\right|+c\end{array}:c යනු අභිමත නියතයකි.
11.\;\int\left(\mathbf a^{\mathbf2}-\mathbf x^{\mathbf2}\right)^{\mathbf3/\mathbf2}ආකාරයේ අනුකල
- මෙවැනි අනුකලනx=a\sin\theta හෝ x=a\cos\theta ආදේශ භාවිතා කරමු.
උදා.
\int\left(4-x^2\right)^{3/2}dxසලකමු.
x=2\sin\theta ආදේශ කරමු.
x විෂයෙන් අවකලනයෙන්,
\begin{array}{rcl}dx&=&2\;cos\theta\;d\theta\\\cos\theta&=&\;\sqrt{1-\sin^2\theta}=\;\sqrt{1-\frac{x^2}4}=\;\frac{\sqrt{4-x^2}}2\end{array}\begin{array}{l}I=\int\left(4-x^2\right)^{3/2}dx\\I=\int\sqrt{4-x^2}\left(4-x^2\right)dx\\I=\int\sqrt{4-4\sin^2\theta}.\left(4-4\sin^2\theta\right).2\cos\theta d\theta\\I=\int\sqrt{4\cos^2\theta}.\left(4\cos^2\theta\right).2\cos\theta d\theta=16\int\cos^4\theta\;d\theta\\I=16\int\cos^3\theta.\cos\theta\;d\theta\end{array}
කොටස් වශයෙන් අනුකලනය භාවිතයෙන් මෙතැන් සිට ගණන විසදමු.
\begin{array}{rcl}u&=&\cos^3\theta\\\frac{du}{d\theta}&=&-3\cos^3\theta\sin\theta\\\frac{dv}{d\theta}&=&\cos\theta\\\int dv&=&\int\cos\theta\;d\theta\\v&=&\sin\theta\end{array}
\begin{array}{l}I=16\int\cos^3\theta.\cos\theta\;d\theta\\I=16\sin\theta\cos^3\theta-16\int\sin\theta.{(-3\cos}^2\theta.\sin{\theta)}\;d\theta\\I=16\sin\theta\cos^3\theta+48\int\sin^2\theta.\cos^2\theta.\;d\theta\\I=16\sin\theta\cos^3\theta+48\int{{(1-\cos}^2\theta).\cos^2\theta.\;d\theta}\\I=16\sin\theta\cos^3\theta+48\int\cos^2\theta.\;d\theta-48\int\cos^4\theta d\theta\\I=16\sin\theta\cos^3\theta+48\int{\cos^2\theta.\;d\theta-3I(\;;}\;I=16cos4\theta d\theta\text{බැවින්})\\4I=16\sin\theta\cos^3\theta+48\int\frac{\left(1+\cos2\theta\right)}2d\theta\\I=4\sin\theta\cos^3\theta+12\int d\theta+12\int\cos2\theta\;d\theta\\I=4\sin\theta\cos^3\theta+12\;\theta+12.\frac{\sin2\theta}2+c\\I=4.\frac x2.\left(1-\frac{x^2}4\right)^{3/2}+12\;\sin^{-1}\left(\frac x2\right)+6\sin\left[2\;\left(\sin^{-1}\frac x2\right)\right]+c\\I=2x\left(1-\frac{x^2}4\right)^{3/2}+12\;\sin^{-1}\left(\frac x2\right)+6\sin\left[2\;\left(\sin^{-1}\frac x2\right)\right]+c\text{ : c යනු අභිමත නියතයකි.}\end{array}
12.\int\left(\mathbf a^{\mathbf2}+\mathbf x^{\mathbf2}\right)^{\mathbf3/\mathbf2}dxආකාරයේ අනුකල
- මෙවැනි අනුකලන x=a\tan\thetaආදේශය භාවිතා කරමු.
උදා.
\int\left(1+x^2\right)^{3/2}dxසලකමු.
x=\tan\theta ආදේශ කරමු.
\sec\theta=\sqrt{1+x^2}
x විෂයෙන් අවකලනයෙන්,
dx=\sec^2\theta\;d\theta
\begin{array}{rcl}I&=&\int\left(1+x^2\right)^{3/2}dx\\I&=&\int\sqrt{1+x^2}\left(1+x^2\right)dx\\I&=&\int\sqrt{1+\tan^2\theta}.\left(1+\tan^2\theta\right).\sec^2\theta d\theta\\I&=&\int\sqrt{\sec^2\theta}.\left(\sec^2\theta\right).\sec^2\theta d\theta\\I&=&\int\sec^5\theta\;d\theta\\I&=&\int\sec^3\theta.\sec^2\theta\;d\theta\end{array}
කොටස් වශයෙන් අනුකලනය භාවිතයෙන් මෙතැන් සිට ගණන විසදමු.
\begin{array}{rcl}u&=&\sin^3\theta\\\frac{du}{d\theta}&=&3sec^2\theta sec\theta\tan\theta\\\frac{du}{d\theta}&=&3\tan\theta sec^3d\theta\\\frac{dv}{d\theta}&=&sec^2\theta\\\int dv&=&\int sec^2\theta d\theta\\v&=&\tan\theta\\&&\\&&\\&&\end{array}
\begin{array}{rcl}I&=&\int\sec^3\theta.\sec^2\theta\;d\theta\\I&=&tan\;\theta\sec^3\theta-\int\tan\theta.{(3\sec}^3\theta.\tan{\theta)}\;d\theta\\I&=&tan\;\theta\sec^3\theta-3\int\sec^3\theta.\tan^2\theta.\;d\theta\\I&=&tan\;\theta\sec^3\theta-3\int{\sec^3\theta.{(\sec}^2\theta-1).\;d\theta}\\I&=&tan\;\theta\sec^3\theta-3\int\sec^5\theta.\;d\theta+3\int\sec^3\theta d\theta\\I&=&tan\;\theta\sec^3\theta-3I+3\int\sec^3\theta d\theta\;(;I=sec5\theta d\theta\text{බැවින්)}\\4I&=&tan\;\theta\sec^3\theta+3\int\sec^3\theta d\theta\;\rightarrow(1)\end{array}
J_1=3\int\sec^3\theta\;d\theta ලෙස ගනිමු.J_1=3\int\sec\theta\sec^2\theta\;d\theta
කොටස් වශයෙන් අනුකලනය භාවිතයෙන් මෙතැන් සිට ගණන විසදමු.
\begin{array}{rcl}u&=&sec\theta\\\frac{du}{d\theta}&=&sec\theta.\tan\theta\\\frac{dv}{d\theta}&=&sec^2\theta\\\int dv&=&\int sec^2\theta d\theta\\v&=&\tan\theta\\&&\\&&\end{array}
\begin{array}{rcl}J_1&=&3\int\sec\theta.\sec^2\theta\;d\theta\\J_1&=&3\sec\theta\tan\theta-3\int\sec\theta.\tan^2\theta\;d\theta\\J_1&=&3\sec\theta\tan\theta-3\int\sec\theta.\left(\sec^2\theta-1\right)\;d\theta\\J_1&=&3\sec\theta\tan\theta+3\int\sec\theta d\theta-3\int\sec^3\theta\;d\theta;(J_1=3sec3\theta d\theta\text{බැවින්})\\2J_1&=&3\sec\theta\tan\theta+3\int\sec\theta d\theta\\J_1&=&\frac32\sec\theta\tan\theta+\frac32\int\sec\theta d\theta\\J_1&=&3\int\sec^3\theta\;d\theta,(1)\text{හිආදේශකරමු.}\end{array}
\begin{array}{l}J_1=3\int\sec\theta.\sec^2\theta\;d\theta\\J_1=3\sec\theta\tan\theta-3\int\sec\theta.\tan^2\theta\;d\theta\\J_1=3\sec\theta\tan\theta-3\int\sec\theta.\left(\sec^2\theta-1\right)\;d\theta\\J_1=3\sec\theta\tan\theta+3\int\sec\theta d\theta-3\int\sec^3\theta\;d\theta;(J_1=3sec3\theta d\theta\text{බැවින්})\\2J_1=3\sec\theta\tan\theta+3\int\sec\theta d\theta\\J_1=\frac32\sec\theta\tan\theta+\frac32\int\sec\theta d\theta\text{ : c යනු අභිමත නියතයකි.}\end{array}
13.\;\int\left(\mathbf x^{\mathbf2}-\mathbf a^{\mathbf2}\right)^{\mathbf3/\mathbf2}dx ආකාරයේ අනුකල
- මෙවැනි අනුකලන x=asec\;\theta ආදේශය භාවිතා කරමු.
උදා.
\int\left(x^2-1\right)^{3/2}dx\;සලකමු.
x=\sec\thetaආදේශ කරමු.
\tan\theta=\sqrt{x^2-1}
x විෂයෙන් අවකලනයෙන්,
dx=sec\theta\tan\theta\;d\theta
\begin{array}{rcl}I&=&\int\left(x^2-1\right)^{3/2}dx\\I&=&\int\sqrt{x^2-1}\left(x^2-1\right)dx\\I&=&\int\sqrt{\sec^2\theta-1}.\left(\sec^2\theta-1\right).\sec\theta\tan\theta d\theta\\I&=&\int\sqrt{\tan^2\theta}.\left(\tan^2\theta\right).sec\theta tan\theta d\theta\\I&=&\int sec\;\theta\tan^4\theta\;d\theta I=\int sec\;\theta\;\left(\sec^2\theta-1\right)^2\;d\theta\\I&=&\int sec\theta\;\left(\sec^4\theta-2\sec^2\theta+1\right)d\theta\;\\I&=&\int\sec^5\theta d\theta-2\int\sec^3\theta d\theta+\int{sec\theta d\theta\rightarrow(1)}\end{array}
\begin{array}{rcl}J&=&\int\sec^5\theta\;d\theta\text{ලෙසගනිමු}.\\J&=&\int\sec^3\theta.\sec^2\theta\;d\theta\end{array}
කොටස් වශයෙන් අනුකලනය භාවිතයෙන් මෙතැන් සිට ගණන විසදමු.
\begin{array}{rcl}u&=&\sin^3\theta\\\frac{du}{d\theta}&=&3sec^2\theta sec\theta\tan\theta\\\frac{du}{d\theta}&=&3\tan\theta sec^3d\theta\\\frac{dv}{d\theta}&=&sec^2\theta\\\int dv&=&\int sec^2\theta d\theta\\v&=&\tan\theta\\&&\\&&\\&&\end{array}
\begin{array}{rcl}J&=&\int\sec^3\theta.\sec^2\theta\;d\theta\\J&=&tan\;\theta\sec^3\theta-\int\tan\theta.{(3\sec}^3\theta.\tan{\theta)}\;d\theta\\J&=&tan\;\theta\sec^3\theta-3\int\sec^3\theta.\tan^2\theta.\;d\theta\\J&=&tan\;\theta\sec^3\theta-3\int{\sec^3\theta.{(\sec}^2\theta-1).\;d\theta}\\J&=&tan\;\theta\sec^3\theta-3\int\sec^5\theta.\;d\theta+3\int\sec^3\theta d\theta\\J&=&tan\;\theta\sec^3\theta-3I+3\int\sec^3\theta d\theta(;J=sec5\theta d\theta\text{බැවින්})\\4J&=&tan\;\theta\sec^3\theta+3\int\sec^3\theta d\theta\;\\J&=&\frac14tan\;\theta\sec^3\theta+\frac34\int\sec^3\theta d\theta\rightarrow(2)\end{array}
\begin{array}{rcl}J_1&=&\int\sec^3\theta\;d\theta\text{ ලෙසගනිමු.}\\{\text{J}}_1&=&\int sec\theta sec^2\theta\;d\theta\end{array}
කොටස් වශයෙන් අනුකලනය භාවිතයෙන් මෙතැන් සිට ගණන විසදමු.
\begin{array}{rcl}u&=&sec\theta\\\frac{du}{d\theta}&=&sec\theta.\tan\theta\\\frac{dv}{d\theta}&=&sec^2\theta\\\int dv&=&\int sec^2\theta d\theta\\v&=&\tan\theta\\&&\\&&\end{array}
\begin{array}{rcl}J_1&=&\int\sec\theta.\sec^2\theta\;d\theta J_1\\&=&\sec\theta\tan\theta-\int\sec\theta.\tan^2\theta\;d\theta\\J_1&=&\sec\theta\tan\theta-\int\sec\theta.\left(\sec^2\theta-1\right)\;d\theta\\J_1&=&\sec\theta\tan\theta+\int\sec\theta d\theta-\int\sec^3\theta\;d\theta;(J_1=sec3\theta d\theta\text{බැවින්})\\2J_1&=&\sec\theta\tan\theta+\int\sec\theta d\theta\\J_1&=&\frac12\sec\theta\tan\theta+\frac12\int\sec\theta d\theta\rightarrow\left(3\right)\\&&\text{(2)න්(1)ටආදේශකරමු.}\end{array}
\begin{array}{rcl}I&=&\frac14tan\;\theta\sec^3\theta+\frac34\int\sec^3\theta d\theta-2\int\sec^3\theta d\theta+\int\sec\theta d\theta\\I&=&\frac14tan\;\theta\sec^3\theta-\frac54\int\sec^3\theta d\theta+\int{sec\theta d\theta\;\rightarrow(4)}\\&&\text{(3)න්(4)ටආදේශකරමු.}\end{array}
\begin{array}{l}I=\frac14sec^3\theta\;-\;\frac54\left\{\frac12sec\theta\;\tan\theta\;+\frac12\int sec\theta\;d\theta\right\}+\int sec\theta\;d\theta\\I=\frac14sec^3\theta\;-\;\frac58sec\theta\;\tan\theta\;-\frac58\int sec\theta\;d\theta+\int sec\theta\;d\theta\\I=\frac14sec^3\theta\;-\;\frac58sec\theta\;\tan\theta\;+\frac38\int sec\theta\;d\theta\\I=\frac14sec^3\theta\;-\;\frac58sec\theta\;\tan\theta\;+\frac38\ln\left|sec\theta\;+\tan\theta\;\right|+c\\I=\frac14.\sqrt{x^2-1}.x^3\;-\frac58.x.\sqrt{x^2-1}\;+\frac38\ln\left|x+\sqrt{x^2-1}\right|\;+c\text{ : c යනු අභිමත නියතයකි.}\end{array}
