Integralai

Sveiki , prasau pagalbos, reikia apskaiciuoti integralus , taikant integravimo dalimis formule. Dekui is anksto.

1. ∫arctg(√x)dx

2. ∫xsin(πx)dx

1.
[tex]\int{\arctan{\sqrt{x}}}dx=x\arctan{\sqrt{x}}-\int{x \cdot \frac{1}{1+x} \cdot \frac{1}{2\sqrt{x}}dx}=x\arctan{\sqrt{x}}-\int{\frac{\sqrt{x}dx}{2(1+x)}}[/tex]

[tex]\int{\frac{\sqrt{x}dx}{2(1+x)}}=[x=t^2, dx=2tdt]=\int{\frac{t \cdot 2tdt}{2(1+t^2)}}=\int{\frac{(t^2+1-1)dt}{1+t^2}}=\int{dt}-\int{\frac{dt}{1+t^2}}=[/tex]
[tex]=t-\arctan{t}=\sqrt{x}-\arctan{\sqrt{x}}[/tex]

2.
[tex]\int{x\sin{\pi x}dx}=-\frac{1}{\pi}\int{xd\cos{\pi x}}=-\frac{1}{\pi}(x\cos{\pi x}-\int{\cos{\pi x}}dx)=[/tex]
[tex]=-\frac{1}{\pi}(x\cos{\pi x}-\frac{1}{\pi}\sin{\pi x}})[/tex]

brainstorm1.
[tex]\int{\arctan{\sqrt{x}}}dx=x\arctan{\sqrt{x}}-\int{x \cdot \frac{1}{1+x} \cdot \frac{1}{2\sqrt{x}}dx}=x\arctan{\sqrt{x}}-\int{\frac{\sqrt{x}dx}{2(1+x)}}[/tex]

[tex]\int{\frac{\sqrt{x}dx}{2(1+x)}}=[x=t^2, dx=2tdt]=\int{\frac{t \cdot 2tdt}{2(1+t^2)}}=\int{\frac{(t^2+1-1)dt}{1+t^2}}=\int{dt}-\int{\frac{dt}{1+t^2}}=[/tex]
[tex]=t-\arctan{t}=\sqrt{x}-\arctan{\sqrt{x}}[/tex]

2.
[tex]\int{x\sin{\pi x}dx}=-\frac{1}{\pi}\int{xd\cos{\pi x}}=-\frac{1}{\pi}(x\cos{\pi x}-\int{\cos{\pi x}}dx)=[/tex]
[tex]=-\frac{1}{\pi}(x\cos{\pi x}-\frac{1}{\pi}\sin{\pi x}})[/tex]