całka Educal:

	sin x		1
Wykąż że ∫0π/2		dx=		arctg √2
	√1+2sin2x		√2

Mariusz:

	sin(x)
∫0π/2		dx
	√1+4sin(x)cos(x)

	sin(x)
∫0π/2		dx
	cos(x)√1+tg2(x)+4tg(x)

	sin(x)   cos(x)
∫0π/2		dx
	√1+tg2(x)+4tg(x)

	tg(x)
∫0π/2		dx
	√1+tg2(x)+4tg(x)

t = tg(x)

	t	1
∫0∞			dt
	1+t2	√1+4t+t2

√1+4t+t² = u − t 1+4t+t² = u²−2ut+t² 1+4t = u²−2ut 2ut+4t = u²−1 (2u+4)t = u²−1

	u2−1
t =
	2u+4

	(u2−1)2
1+t2 = 1+
	(2u+4)2

	u4−2u2+1+4u2+16u+16
1+t2 =
	(2u+4)2

	u4+2u2+16u+17
1+t2 =
	(2u+4)2

	u2−1
√1+4t+t2 =u −
	2u+4

	u2+4u+1
√1+4t+t2 =
	2u+4

	2u(2u+4)−2(u2−1)
dt =		du
	(2u+4)2

	4u2+8u−2u2+2
dt =		du
	(2u+4)2

	2u2+8u+2
dt =		du
	(2u+4)2

	u2+4u+1
dt = 2		du
	(2u+4)2

u = t+√1+4t+t² t = 0 , u = 1 t → ∞ , u →∞

	u2−1		(2u+4)2		2u+4
2∫1∞		*		*
	2u+4		u4+2u2+16u+17		u2+4u+1

	u2+4u+1
*		du
	(2u+4)2

	u2−1
2∫1∞		du
	u4+2u2+16u+17

u⁴+2u²+16u+17 = (u²−au+b)(u²+au+c) u⁴+2u²+16u+17 = (u⁴+au³+cu²−au³−a²u²−acu+bu²+abu+bc) u⁴+2u²+16u+17 = u⁴ + (b+c−a²)u²+(ab−ac)u+bc u⁴+2u²+16u+17 = u⁴ + (b+c−a²)u²+a(b−c)u+bc b+c−a² = 2 a(b−c) = 16 bc = 17 b+c = 2 + a²

	16
b−c =
	a

4bc = 68

	16
2b = 2 + a2 +
	a

	16
2c = 2 + a2 −
	a

4bc = 68

	16		16
(2 + a2 +		)(2 + a2 −		)=68
	a		a

	16		16
(2 + a2 +		)(2 + a2 −		)−68=0
	a		a

	256
(a4+4a2+4)−		−68 = 0
	a2

	256
a4+4a2−64 −		= 0
	a2

a⁶+4a⁴ − 64a² − 256 = 0 a⁴(a²+4)−64(a²+4) = 0 (a⁴−64)(a²+4) = 0 a = 2√2

	1		16
b =		( 2 + a2 +		)
	2		a

	1		16
c =		( 2 + a2 −		)
	2		a

	1		16
b =		(2 + 8 +		)
	2		2√2

	1		16
c =		(2 + 8 −		)
	2		2√2

	1		16√2
b =		(10 +		)
	2		4

	1		16√2
c =		(10 −		)
	2		4

	1
b =		(10 + 4√2)
	2

	1
c =		(10 − 4√2)
	2

b = 5+2√2 c = 5−2√2 u⁴+2u²+16u+17 = (u² − 2√2u + 5+2√2)(u² + 2√2u + 5−2√2)

	u2−1
2∫1∞		du
	u4+2u2+16u+17

	u2−1
2∫1∞		du
	(u2 − 2√2u + 5+2√2)(u2 + 2√2u + 5−2√2)

(u−√2)²+3+2√2=u²−2√2u+2+3+2√2 (u+√2)²+3−2√2=u²+2√2u+2+3−2√2 (u−√2)²+3+2√2=u²−2√2u+5+2√2 (u+√2)²+3−2√2=u²+2√2u+5−2√2

	u2−1
2		=
	(u2 − 2√2u + 5+2√2)(u2 + 2√2u + 5−2√2)

Au+B		Cu+D
	+
u2 − 2√2u + 5+2√2		u2 + 2√2u + 5−2√2

(Au+B)(u² + 2√2u + 5−2√2) + (Cu+D)(u² − 2√2u + 5+2√2) = 2u²−2 A(u³ + 2√2u² + (5−2√2)u)+B(u² + 2√2u + (5−2√2)) +C(u³ − 2√2u² + (5+2√2)u) + D(u² − 2√2u + (5+2√2)) = 2u²−2 (A+C)u³ +(2√2A + B −2√2C + D)u²+((5−2√2)A+2√2B+(5+2√2)C −2√2D) +((5−2√2)B+(5+2√2)D) = 2u²−2 A+C = 0 2√2A + B −2√2C + D = 2 (5−2√2)A+2√2B+(5+2√2)C −2√2D=0 ((5−2√2)B+(5+2√2)D) = −2 C = −A 4√2A+B+D = 2 −4√2A+2√2B−2√2D=0 ((5−2√2)B+(5+2√2)D) = −2 C = −A 4√2A+B+D = 2 (1+2√2)B+(1−2√2)D = 2 (5−2√2)B+(5+2√2)D = −2 6B+6D = 0 B+D = 0 D = −B C = −A

	√2
A =
	4

D = −B (5−2√2)B−(5+2√2)B = −2 −4√2B=−2

	√2
B =
	4

√2		u+1		√2		u+1
	∫		du−		∫		du
4		u2 − 2√2u + 5+2√2		4		u2 + 2√2u + 5−2√2

√2		2u+2		√2		2u+2
	∫		−		∫		du
8		u2 − 2√2u + 5+2√2		8		u2 + 2√2u + 5−2√2

√2		2u−2√2+2√2+2
	∫		du
8		u2 − 2√2u + 5+2√2

	√2		2u+2√2+2−2√2
−		∫		du
	8		u2 + 2√2u + 5−2√2

√2		2u−2√2		√2
	∫		du+		∫U{√2+1}{(u − √2)2 +
8		u2 − 2√2u + 5+2√2		4

(√2+1)²}du

	√2		2u+2√2		√2
−		∫		du+		∫U{√2−1}{(u + √2)2 +
	8		u2 + 2√2u + 5−2√2		4

(√2−1)²}du

√2		√2
	ln(u2 − 2√2u + 5+2√2)−		ln(u2 + 2√2u + 5−2√2)
8		8

	√2		u − √2		√2		u+√2
+		arctg(		)+		arctan(		)
	4		√2+1		4		√2−1

√2		u2 − 2√2u + 5+2√2
	ln(		)
8		u2 + 2√2u + 5−2√2

	√2		u − √2		√2		u+√2
+		arctg(		)+		arctan(		)
	4		√2+1		4		√2−1

	√2		u2 − 2√2u + 5+2√2
=		ln(limu→∞		})
	8		u2 + 2√2u + 5−2√2

√2

6

√2

π

1−√2

π

−

ln(

)+

(

−arctg(

)+

−a

8

6

4

2

1+√2

2

	1+√2
rctg(		))
	√2−1

√2		1+√2		1−√2
	(arctg(		)−arctg(		))
4		1−√2		1+√2

√2		π		π
	((		+arctg(√2))−(		−arctg(√2)))
4		4		4

	√2
=		(2arctg(√2))
	4

	√2
=		arctg(√2)
	2

	1
=		arctg(√2)
	√2