Aptarimas:Matematika/Integravimas keičiant kintamąjį

• $\int {\sqrt {x^{2}+a}}\;dx=\int {\sqrt {\sin ^{2}(t)+a}}\cdot \cos(t)dt={\sqrt {\sin ^{2}(t)+a}}\cdot \sin(t)-\int ({\sqrt {\sin ^{2}(t)+a}})'\cdot \sin(t)dt=$

$={\sqrt {\sin ^{2}(t)+a}}\cdot \sin(t)-\int {\frac {1}{2}}\cdot {\frac {2\sin(t)\cdot \cos(t)}{\sqrt {\sin ^{2}(t)+a}}}\cdot \sin(t)dt={\sqrt {\sin ^{2}(t)+a}}\cdot \sin(t)-\int {\frac {\cos(t)}{\sqrt {\sin ^{2}(t)+a}}}dt=$

$={\sqrt {\sin ^{2}(t)+a}}\cdot \sin(t)-\int {\frac {\cos(t)}{\sqrt {{\frac {1-\cos(2t)}{2}}+a}}}dt={\sqrt {\sin ^{2}(t)+a}}\cdot \sin(t)-\int {\frac {\cos(t)}{\sqrt {\frac {1+2a-\cos(2t)}{2}}}}dt={\sqrt {\sin ^{2}(t)+a}}\cdot \sin(t)-\int {\frac {{\sqrt {2}}\cos(t)}{\sqrt {1+2a-\cos(2t)}}}dt=$

kur $x=\sin(t)$, $t=\arcsin(x)$, $dx=\cos(t)dt$; $\sin ^{2}(t)={\frac {1-\cos(2t)}{2}},$

• $\int {\sqrt {x^{2}+a}}\;dx=-\sin(t)\int {\sqrt {\cos ^{2}(t)+a}}{\mathsf {d}}t=-[{\sqrt {\cos ^{2}(t)+a}}\cdot (-\cos(t))-\int ({\sqrt {\cos ^{2}(t)+a}})'\cdot (-\cos(t))dt]=$

$={\sqrt {\cos ^{2}(t)+a}}\cdot \cos(t)-\int ({\sqrt {\cos ^{2}(t)+a}})'\cdot \cos(t)dt=$

$={\sqrt {\cos ^{2}(t)+a}}\cdot \cos(t)-\int {\frac {1}{2}}\cdot {\frac {2\cos(t)\cdot (-\sin(t))}{\sqrt {\cos ^{2}(t)+a}}}\cdot \cos(t)dt={\sqrt {\cos ^{2}(t)+a}}\cdot \cos(t)+\int {\frac {\sin(t)}{\sqrt {\cos ^{2}(t)+a}}}dt=$

$={\sqrt {\cos ^{2}(t)+a}}\cdot \cos(t)+\int {\frac {\sin(t)}{\sqrt {{\frac {\cos(2t)+1}{2}}+a}}}dt={\sqrt {\cos ^{2}(t)+a}}\cdot \cos(t)+{\sqrt {2}}\int {\frac {\sin(t)}{\sqrt {\cos(2t)+1+2a}}}dt=$

$={\sqrt {\cos ^{2}(t)+a}}\cdot \cos(t)+{\sqrt {2}}\int {\frac {\sin(t)}{\sqrt {\cos(2t)+1+2a}}}{\frac {{\mathsf {d}}(\cos(2t)+1+2a)}{-2\sin(2t)}}=$

kur $x=\cos(t)$, $t=\arccos(x)$, $dx=-\sin(t)dt$; $\cos ^{2}(t)={\frac {\cos(2t)+1}{2}};$

${\mathsf {d}}(\cos(2t)+1+2a)=-2\sin(2t){\mathsf {d}}t,$

${\frac {{\mathsf {d}}(\cos(2t)+1+2a)}{-2\sin(2t)}}={\mathsf {d}}t,$