Integracija trigonometrijskim smenama - zadaci sa rešenjima

Rešenje:
Uvedimo smenu [tex]x=4\sin \theta[/tex], pa je [tex]dx=4\cos \theta d\theta [/tex]

Tada je [tex]\sqrt{\left( 16-x^{2}\right) }=4\sqrt{1-\sin ^{2}\theta }=4\sqrt{\cos^{2}\theta }=4\cos \theta [/tex]

Dati integral je jednak
[tex]\int \frac{1}{\sqrt{\left( 16-x^{2}\right) }}dx=\int \frac{1}{4\cos \theta }4\cos \theta d\theta =\int d\theta [/tex]

[tex]\theta =\arcsin \frac{x}{4}[/tex] pa je [tex]\int \frac{1}{\sqrt{\left(16-x^{2}\right) }}dx=\arcsin \frac{x}{4}+C[/tex]

Rešenje:
Uvedimo smenu [tex]x=5\sec \theta[/tex], pa je [tex]dx=5\sec \theta \tan \theta\, d\theta[/tex].
Tada je [tex]x^{2}-25=25\left(\sec ^{2}\theta -1\right)=25\tan ^{2}\theta[/tex], pa je

[tex]\int \frac{1}{x^{2}-25}dx=\int \frac{5\sec \theta \tan \theta}{25\tan ^{2}\theta}d\theta =\frac{1}{5}\int \frac{\sec \theta}{\tan \theta}d\theta =\frac{1}{5}\int \csc \theta\, d\theta =-\frac{1}{5}\ln \left\vert \csc \theta +\cot \theta \right\vert +C[/tex]

Vraćamo se na promenljivu [tex]x[/tex]: [tex]\sec \theta =\frac{x}{5}[/tex], [tex]\cos \theta =\frac{5}{x}[/tex], [tex]\sin \theta =\frac{\sqrt{x^{2}-25}}{x}[/tex], pa je
[tex]\csc \theta =\frac{x}{\sqrt{x^{2}-25}}[/tex] i [tex]\cot \theta =\frac{5}{\sqrt{x^{2}-25}}[/tex]

[tex]\int \frac{1}{x^{2}-25}dx=-\frac{1}{5}\ln \left\vert \frac{x+5}{\sqrt{x^{2}-25}}\right\vert +C=-\frac{1}{5}\ln \sqrt{\left\vert \frac{x+5}{x-5}\right\vert }+C[/tex]

[tex]\int \frac{1}{x^{2}-25}dx=\frac{1}{10}\ln \left\vert \frac{x-5}{x+5}\right\vert +C[/tex]

Rešenje:
Integral možemo izračunati smenom [tex]x=\tan \theta [/tex], [tex]dx=\sec^{2}\theta d\theta [/tex] pa je
[tex]\int x\sqrt{1+x^{2}}dx=\int \tan \theta \sqrt{1+\tan^{2} \theta}\sec^{2}\theta d\theta =\int \tan \theta \sec^{3}\theta d\theta[/tex]

[tex]\int \sec \theta \tan \theta \sec^{2}\theta d\theta[/tex]
Ako je [tex]u=\sec \theta, du=\sec \theta \tan \theta d\theta[/tex]

Dakle, [tex]\int \sec \theta \tan \theta \sec^{2}\theta d\theta =\int u^{2}du=\frac{u^{3}}{3}+C=\frac{1}{3}\sec^{3}\theta +C[/tex]

Koristimo trougao.
Dakle, [tex]\int x\sqrt{1+x^{2}}dx=\frac{1}{3}\left( 1+x^{2}\right) ^{3/2}+C[/tex]

Rešenje:
Integral možemo izračunati smenom [tex]x=\tan \theta [/tex] [tex]\ [/tex]pa je [tex]dx=\sec ^{2}\theta d\theta [/tex],

Dakle, [tex]\int \frac{9x}{\sqrt{1+x^{2}}}dx=9\int \frac{\tan \theta }{\sqrt{1+\tan ^{2}\theta} }\sec ^{2}\theta d\theta =9\int \frac{\tan \theta }{\sec \theta }\sec ^{2}\theta d\theta =9\int \tan \theta \sec \theta d\theta =9\sec \theta +C[/tex]

Na osnovu trougla ispod dobijamo:
[tex]\int \frac{9x}{\sqrt{1+x^{2}}}dx=9\sqrt{1+x^{2}}+C[/tex]

Rešenje:
Dopunjavamo do potpunog kvadrata izraz

[tex]4x-x^{2}=-\left( x-2\right)^{2}+4=4-\left( x-2\right)^{2}[/tex]
Dakle, [tex]\int \frac{1}{\sqrt{4x-x^{2}}}dx=\int \frac{1}{\sqrt{4-\left( x-2\right)^{2}}}dx[/tex].
Neka je [tex]u=x-2\Longrightarrow du=dx[/tex]

Tada je [tex]\int \frac{1}{\sqrt{4-\left( x-2\right)^{2}}}dx=\int \frac{1}{\sqrt{4-u^{2}}}du[/tex] Uvodimo trigonometrijsku smenu [tex]u=2\sin \theta[/tex]

[tex]du=2\cos \theta d\theta \Longrightarrow \int \frac{1}{\sqrt{4-u^{2}}}du =\int \frac{1}{\sqrt{4-4\sin ^{2}\theta }}du[/tex] [tex]=\int \frac{2\cos \theta d\theta }{2\sqrt{\cos ^{2}\theta }}=\int d\theta =\theta +C=\arcsin \frac{u}{2}+C[/tex]
Vraćamo se na promenljivu x.
[tex]u=x-2[/tex]

Dakle, [tex]\int \frac{1}{\sqrt{4x-x^{2}}}dx=\arcsin \frac{x-2}{2}+C[/tex]

Rešenje:
Neka je [tex]x=4\sin \theta[/tex], [tex]\theta \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right][/tex], pa je [tex]dx=4\cos \theta\, d\theta[/tex] i [tex]\sqrt{16-x^{2}}=\sqrt{16-16\sin ^{2}\theta }=4\cos \theta[/tex].

[tex]\int \frac{x^{2}}{\sqrt{16-x^{2}}}dx=\int \frac{16\sin ^{2}\theta }{4\cos \theta }4\cos \theta\, d\theta =16\int \sin ^{2}\theta\, d\theta[/tex]

Kako je [tex]\sin ^{2}\theta =\frac{1-\cos 2\theta }{2}[/tex],
[tex]16\int \sin ^{2}\theta\, d\theta =8\theta -4\sin 2\theta +C=8\theta -8\sin \theta \cos \theta +C[/tex]

Vraćamo se na promenljivu [tex]x[/tex]: [tex]\theta =\arcsin \frac{x}{4}[/tex], [tex]\sin \theta =\frac{x}{4}[/tex], [tex]\cos \theta =\frac{\sqrt{16-x^{2}}}{4}[/tex]
[tex]\int \frac{x^{2}}{\sqrt{16-x^{2}}}dx=8\arcsin \frac{x}{4}-\frac{x\sqrt{16-x^{2}}}{2}+C[/tex]

Rešenje:
Posle dopunjavanja do potpunog kvadrata dobijamo:
[tex]x^{2}+6x+12=\left( x+3\right)^{2}+3\Longrightarrow \int \frac{x}{\sqrt{ x^{2}+6x+12}}dx=\int \frac{x}{\sqrt{\left( x+3\right) ^{2}+3}}dx[/tex]

Uvedimo smenu
[tex]u=x+3\Longrightarrow du=dx,x=u-3[/tex] Dakle,
[tex]\int \frac{x}{\sqrt{\left( x+3\right) ^{2}+3}}dx=\int \frac{\left( u-3\right) }{\sqrt{u^{2}+3}}du[/tex]

Zatim uvodimo trigonometrijsku smenu [tex]u=\sqrt{3}\tan \theta [/tex]
[tex]du=\sqrt{3}\sec ^{2}\theta d\theta [/tex]

[tex]\int \frac{\left( u-3\right) }{\sqrt{u^{2}+3}}du=\int \frac{\left( \sqrt{3} \tan \theta -3\right) }{\sqrt{3\left( \tan ^{2}\theta +1\right) }}\sqrt{3} \sec ^{2}\theta d\theta =\int \frac{\left( \sqrt{3}\tan \theta -3\right) }{\sqrt{\sec ^{2}\theta }} \sec ^{2}\theta d\theta =\sqrt{3}\int \sec \theta \tan \theta d\theta -3\int \sec \theta d\theta =\sqrt{3}\sec \theta -3\ln \left\vert \sec \theta +\tan \theta \right\vert +C[/tex]

Iz trougla ispod, gde je [tex]a=\sqrt{3},u=x[/tex]

Dakle, [tex]=\sqrt{3}\frac{\sqrt{3+u^{2}}}{\sqrt{3}}-3\ln \left\vert \frac{\sqrt{ 3+u^{2}}}{\sqrt{3}}+\frac{u}{\sqrt{3}}\right\vert +C[/tex]

Vraćamo se na promenljivu x
[tex]u=x+3[/tex]

[tex]\int \frac{x}{\sqrt{x^{2}+6x+12}}dx=\sqrt{3}\frac{\sqrt{3+\left( x+3\right)^{2}}}{\sqrt{3}}-3\ln \left\vert \frac{\sqrt{3+\left( x+3\right) ^{2}}}{\sqrt{3}}+\frac{\left( x+3\right) }{\sqrt{3}}\right\vert +C[/tex]

Rešenje:
Dopunjavamo do potpunog kvadrata: [tex]2x-x^{2}=1-(x-1)^{2}[/tex], pa je
[tex]\int \frac{x^{2}}{\sqrt{2x-x^{2}}}dx=\int \frac{x^{2}}{\sqrt{1-(x-1)^{2}}}dx[/tex]

Neka je [tex]u=x-1[/tex], [tex]du=dx[/tex], [tex]x=u+1[/tex]:
[tex]\int \frac{(u+1)^{2}}{\sqrt{1-u^{2}}}du[/tex]

Sada uvodimo trigonometrijsku smenu [tex]u=\sin \theta[/tex], [tex]du=\cos \theta\, d\theta[/tex], [tex]\sqrt{1-u^{2}}=\cos \theta[/tex]:
[tex]\int \frac{(\sin \theta +1)^{2}}{\cos \theta }\cos \theta\, d\theta =\int \left( \sin ^{2}\theta +2\sin \theta +1\right) d\theta[/tex]

[tex]=\frac{\theta }{2}-\frac{\sin 2\theta }{4}-2\cos \theta +\theta +C=\frac{3\theta }{2}-\frac{1}{2}\sin \theta \cos \theta -2\cos \theta +C[/tex]

Vraćamo se na promenljivu [tex]x[/tex]: [tex]\theta =\arcsin (x-1)[/tex], [tex]\sin \theta =x-1[/tex], [tex]\cos \theta =\sqrt{1-(x-1)^{2}}=\sqrt{2x-x^{2}}[/tex]
[tex]\int \frac{x^{2}}{\sqrt{2x-x^{2}}}dx=\frac{3}{2}\arcsin (x-1)-\frac{x-1}{2}\sqrt{2x-x^{2}}-2\sqrt{2x-x^{2}}+C=\frac{3}{2}\arcsin (x-1)-\frac{x+3}{2}\sqrt{2x-x^{2}}+C[/tex]

Rešenje:
Odgovor: samo (i) i (iii).

[tex]\int \frac{x-9}{x^{2}-6x}dx\Longrightarrow [/tex](i) [tex]\int \frac{x-9}{x\left( x-6\right) }dx[/tex]

(iii) [tex]\frac{x-9}{x\left( x-6\right) }=\frac{A}{x}+\frac{B}{x-6}=\frac{ A\left( x-6\right) +Bx}{x\left( x-6\right) }[/tex]
Dobijamo [tex]x-9=A\left( x-6\right) +Bx[/tex]

Tada je [tex]x-9=\left( A+B\right) x-6A\Longrightarrow \left\{ \begin{array}{c} A+B=1 \\ -6A=-9 \end{array} \right. \Longrightarrow A=\frac{3}{2},B=-\frac{1}{2}[/tex]

Tada je [tex]\int \frac{x-9}{x\left( x-6\right) }dx=\int \left( \frac{\frac{3}{2}}{x }-\frac{\frac{1}{2}}{x-6}\right) dx=\frac{3}{2}\int \frac{1}{x}dx-\frac{1}{2} \int \frac{1}{x-6}dx=\frac{3}{2}\ln x-\frac{1}{2}\ln \left\vert x-6\right\vert +C=\ln \sqrt{ x^{3}}-\ln \sqrt{x-6}+\ln C=\ln \left\vert C\sqrt{\frac{x^{3}}{x-6}}\right\vert [/tex]

Rešenje:
[tex]\int \frac{1}{x^{2}-9}dx=\int \frac{1}{\left( x+3\right) \left( x-3\right) } dx[/tex]
Rastavljamo na parcijalne razlomke

[tex]\frac{1}{\left( x+3\right) \left( x-3\right) }=\frac{A}{x+3}+\frac{B}{x-3}= \frac{A\left( x-3\right) +B\left( x+3\right) }{\left( x+3\right) \left(x-3\right) }\Longrightarrow [/tex]

[tex]1=\left( A+B\right) x+3\left( B-A\right)[/tex]

Dobijamo [tex]\left\{ \begin{array}{c} A+B=0 \\ 3\left( B-A\right) =1 \end{array} \right.[/tex]

Kako je [tex]A=-B\Longrightarrow 3\left( B-A\right) =3\left( B+B\right)=1\Longrightarrow B=\frac{1}{6}[/tex]

Tada je [tex]A=-\frac{1}{6}[/tex]
Dakle, [tex]\int \frac{1}{x^{2}-9}dx=\int \left( \frac{-\frac{1}{6}}{x+3}+\frac{\frac{1}{6}}{x-3}\right) dx=-\frac{1}{6}\int \frac{1}{x+3}dx+\frac{1}{6}\int \frac{1}{x-3}dx=\frac{1}{6}(\ln \left( x-3\right) -\ln (x+3))+C[/tex]

[tex]\int \frac{1}{x^{2}-9}dx=\ln (C\sqrt[6]{\frac{x-3}{x+3}})[/tex]

Rešenje:
Rastavljamo imenilac na činioce: [tex]x^{2}+3x-4=(x+4)(x-1)[/tex]

[tex]\frac{5}{x^{2}+3x-4}=\frac{A}{x+4}+\frac{B}{x-1}=\frac{A(x-1)+B(x+4)}{(x+4)(x-1)}[/tex]
[tex](A+B)x+(4B-A)=5\Longrightarrow \left\{ \begin{array}{l} A+B=0 \\ 4B-A=5 \end{array}\right.[/tex]
Iz prve jednačine [tex]A=-B[/tex], pa je [tex]5B=5[/tex], [tex]B=1[/tex], [tex]A=-1[/tex].

[tex]\int \frac{5}{x^{2}+3x-4}dx=\int \left( \frac{-1}{x+4}+\frac{1}{x-1}\right) dx=\ln \left\vert x-1\right\vert -\ln \left\vert x+4\right\vert +C=\ln \left\vert \frac{x-1}{x+4}\right\vert +C[/tex]

Rešenje:
Rastavljamo imenilac na činioce: [tex]x^{2}+11x+18=(x+9)(x+2)[/tex]

[tex]\frac{x+2}{x^{2}+11x+18}=\frac{A}{x+9}+\frac{B}{x+2}=\frac{A(x+2)+B(x+9)}{(x+9)(x+2)}[/tex]
[tex](A+B)x+(2A+9B)=x+2\Longrightarrow \left\{ \begin{array}{l} A+B=1 \\ 2A+9B=2 \end{array}\right.[/tex]
Iz prve jednačine [tex]A=1-B[/tex], pa je [tex]2-2B+9B=2[/tex], tj. [tex]B=0[/tex], [tex]A=1[/tex].
Zaista, razlomak se skraćuje: [tex]\frac{x+2}{(x+9)(x+2)}=\frac{1}{x+9}[/tex] za [tex]x\neq -2[/tex].

[tex]\int \frac{x+2}{x^{2}+11x+18}dx=\int \frac{1}{x+9}dx=\ln \left\vert x+9\right\vert +C[/tex]

Rešenje:
Odgovor: samo (i) i (iii).

[tex]\int \frac{5x^{2}-12x-12}{x^{3}-4x}dx\Longrightarrow \frac{5x^{2}-12x-12}{x^{3}-4x}=\frac{5(x^{2}-\frac{12}{5}x-\frac{12}{5})}{x\left( x+2\right)\left( x-2\right) }=\frac{5(x-\frac{6}{5}-\frac{4}{5}\sqrt{6})(x-\frac{6}{5}+\frac{4}{5}\sqrt{6})}{x\left( x+2\right) \left( x-2\right) }[/tex] (i) sada [tex]\frac{5(x-\frac{6}{5}- \frac{4}{5}\sqrt{6})(x-\frac{6}{5}+\frac{4}{5}\sqrt{6})}{x\left( x+2\right)\left( x-2\right) }=\frac{A}{x}+\frac{B}{x+2}+\frac{C}{x-2}=\frac{A(x+2)(x-2)+Bx(x-2)+Cx(x+2)}{ x\left( x+2\right) \left( x-2\right) }[/tex]

Tada je [tex]A(x+2)(x-2)+Bx(x-2)+Cx(x+2)=5(x-\frac{6}{5}-\frac{4}{5}\sqrt{6})(x-\frac{6}{5}+\frac{4}{5}\sqrt{6})[/tex]

Ako je [tex]x=2[/tex] pa je [tex]8C=5(2-\frac{6}{5}-\frac{4}{5}\sqrt{6})(2-\frac{6}{5}+\frac{4}{5}\sqrt{6})= -16[/tex]

Dakle, [tex]C=-2[/tex] ako je [tex]x=-2[/tex] pa je [tex]8B=5(-2-\frac{6}{5}-\frac{4}{5}\sqrt{6})(-2-\frac{6}{5}+\frac{4}{5}\sqrt{6})[/tex]
[tex]8B=32[/tex] pa je [tex]B=4[/tex] i ako je [tex]x=0[/tex] pa je
[tex]-4A=5(-\frac{6}{5}-\frac{4}{5}\sqrt{6})(-\frac{6}{5}+\frac{4}{5}\sqrt{6} )= -12[/tex] pa je [tex]A=3[/tex]

[tex]\int \frac{5x^{2}-12x-12}{x^{3}-4x}dx=\int \left( \frac{3}{x}+\frac{4}{x+2}-\frac{2}{x-2}\right) dx=3\ln x+4\ln \left\vert x+2\right\vert -2\ln \left\vert x-2\right\vert +\ln C=\ln \left\vert C\frac{x^{3}\left( x+2\right)^{4}}{\left( x-2\right) ^{2}}\right\vert [/tex]

Rešenje:
Odgovor: samo (i) i (iii).
[tex]\frac{x+2}{x^{2}-4x}=\frac{x+2}{x(x-4)}=\frac{A}{x}+\frac{B}{x-4}=\frac{ A(x-4)+Bx}{x(x-4)}[/tex] pa je [tex]A(x-4)+Bx=x+2[/tex] pa je [tex]\left( A+B\right) x-4A=x+2\Longrightarrow \left\{ \begin{array}{c} A+B=1 \\ -4A=2 \end{array} \right. [/tex] pa je [tex]A=-\frac{1}{2}[/tex] [tex]B=\frac{3}{2}[/tex] pa je [tex]\int \frac{x+2}{x^{2}-4x}dx=\int \left( \frac{-\frac{1 }{2}}{x}+\frac{\frac{3}{2}}{x-4}\right) dx=-\frac{1}{2}\int \frac{1}{x}dx+ \frac{3}{2}\int \frac{1}{x-4}dx[/tex] [tex]=-\frac{1}{2}\ln x+\frac{3}{2}\ln \left\vert x-4\right\vert +\ln C=\ln \left\vert C\sqrt{\frac{\left( x-4\right) ^{3}}{x}}\right\vert [/tex]

Rešenje:
Odgovor: (i), (ii) i (iii).

[tex]\frac{x^{2}+2x+1}{x(x-1)(x+1)}=\frac{\left( x+1\right)^{2}}{x(x-1)(x+1)}= \frac{x+1}{x(x-1)}=\frac{A}{x}+\frac{B}{x-1}=\frac{A(x-1)+Bx}{x(x-1)}[/tex]

Tada je [tex]x+1=\left( A+B\right) x-A\Longrightarrow \left\{ \begin{array}{c} \left( A+B\right) =1 \\ -A=1 \end{array} \right. \Longrightarrow A=-1[/tex]

[tex]B=2[/tex]
Tada je [tex]\int \frac{x^{2}+2x+1}{x(x-1)(x+1)}dx=\int \left( \frac{-1}{x}+\frac{2}{x-1}\right) dx=-\ln x+2\ln \left\vert x-1\right\vert +\ln C=\ln \left\vert C\frac{\left( x-1\right) ^{2}}{x}\right\vert [/tex]