Questão 05 a) ∫ x
2
3x
e dx
u = x2 du = 2 x dx 3x dv = e e3x v = 3 = x2
3x
e 2 − ∫ e 3x x dx 3 3
u = x du = dx dv = e3x e3x v = 3 3x
=
3x
3x
e 2 3x 2 e 2 3x e 2 3x 2 x − x e + ∫ e dx = x − x e x +2 3 9 9 3 9 27 =
9 e 3x x 2−6 e 3x x +2 e 3x e3x = (9x2−6x+ 2)+ c 27 27
b)
∫ x cos 5x dx
u = x du = dx dv = cos 5x dx sin 5x v = 5
1 1 x sin 5x− ∫ sin 5x dx = 5 5 u = 5x du = 5 dx
−1 1 sin u du = cos u ∫ 25 25
1 1 sin 5x cos 5x 1 sin 5x + (cos5x ) = + = ( 5 sin 5x+cos 5x ) +c 5x 25 5x 25 25
c)
∫ √ x ln x dx
u = ln x dx du = x dv = √ x dx 3 2 v = x2 3 3
3
3
3
3
3
2 2 2 x2 2 2 2 2 x2 −1 x ln x− ∫ dx = x 2 ln x− ∫ x 2∗x dx = x 2∗x − ∫ dx∗ = 3 3 x 3 3 3 3 (3∗2−1 ) 3
3
2 2 4 2 x ln x − ∫ x 2 dx = ( √ x 3 3 ln x−2)+c 3 9 9
d)
∫ x cos x dx
u = x du = dx dv = cos x dx v = sin x
x sin x−∫ sin x dx = x sin x +cos x +c
e)
∫ arctan x dx u = tan −1 x 1 du = 2 dx x +1 dv = dx v = x x tan−1 x−∫
x = x +1 2
2
u = x +1 du = 2x dx du = x dx 2 1 du 1 1 x tan −1 x− ∫ = x tan−1 x − ln u = x tan−1 x − ln (x 2 +1)+ c 2 u 2 2
f)
∫ x 2 sin x dx u = x2 du = 2x dx dv = sin x dx v = −cos x −x 2 cos x +2 ∫ x cos x dx
u = x du = dx dv = cos x dx v = sin x −x 2 cos x +2 ( x sin x −∫ sin x dx ) = −x 2 cos x+ 2x sin x + 2cos x +c = 2x sin x−(x 2−2)cos x +c
g)
∫ x e x dx
x
u = x du = dx dv = e x dx v = ex x
x
x
x
= x e −∫ e dx = x e −e = e ( x−1)+ c
h)
∫ x sin x dx
−x cos x +∫ x cos dx
u = x du = dx dv = sin x dx v = −cos x u = x
du = dx
−x cos x +∫ cos u du = − x cos x + sen u = −x cos x +sin x dx
i)
∫ x 2 e x dx 2
u = x du = 2x dx dv = e x dx v = ex 2
x
x
= x e −2∫ x e dx
x
x
u = x du = dx dv = e x dx v = ex x
x
x
= x e −∫ e dx = x e −e = e ( x−1)+ c = x 2 e x −2( x e x −∫ e x dx) = x 2 e x −2 x e x + 2 e x = 2 e x (
j)
∫ x ln x dx
u = ln x dx du = x dv = x dx x2 v = 2
x2 1 x2 x2 x2 1 2 ln x− ∫ dx = ln x− = x (2 ln x−1)+c 2 2 x 2 4 4
k)
∫ x 2 ln x dx
u = ln x dx du = x 2 dv = x dx x3 v = 3
x2 − x+1)+c 2
3
3
3
2
x 1 x x x 1 ln x− ∫ dx = ln x− = x 3 (3 ln x−1)+ c 3 3 x 3 9 9
l)
∫ x sec2 x dx
u = x du = dx dv = sec 2 x dx v = tan x
x tan x −∫ tan x dx = x tan x +ln (cos x )+ c
m)
∫ e x cos x dx
∫ e x cos x dx
u = cos x du = −sin x dx dv = e x dx v = ex x
x
= e cos x+∫ e sin x dx u = sin x du = cos x dx dv = e x dx v = ex
∫ e x cos x dx I = e x cos x +e x sin x−I
= e x cos x +e x sin x−∫ e x cos x dx = ex x x ⇒ 2 I = e cos x +e sin x = (cos x +sin x )+c 2
n)
∫ e−2x sin x dx
∫e
−2x
u = sin x du = cos x dx dv = e−2x dx −e −2x v = 2
e 2x −2x sin x dx = − sin x +1 over2 ∫ e cos x dx 2
u = cos x du = −sin x dx dv = e −2x dx e−2x v = 2 2x
∫ e−2x sin x dx
= −
−2x
e 1 −e sin x + ( 2 2 2
⇒I = −
cos x−
1 e−2x sin x dx ) = ∫ 2
e 2x 1 −2x I sin x− e cos x− = 2 4 4 2x
⇒I +
⇒
I e 1 = − sin x− e−2x cos x = 4 2 4
5 e 2x 1 I = − sin x− e −2x cos x = 4 2 4
⇒I = −
(
)
4 e 2x 1 sin x− e −2x cos x = 5 2 4
1 ⇒ I = − e −2x ( 2 sin x +cos x ) 5
o)
∫ arcsin x dx
u = arcsin x 1 du = dx dx √1−x 2 dv = dx v = x x dx √ 1−x 2
u = 1− x 2
∫ arcsin x dx
= x arcsin x−∫
∫ arcsin x dx
= x arcsin x +
1 du dx 2 ∫ √u
∫ arcsin x dx
= x arcsin x +
2 √ u dx 2
∫ arcsin x dx
= x arcsin x + √ u = x arcsin x+ √1−x 2 +c
p) é idêntico o exercício da letra e
du = −2 dx
q)
π
∫0 sin t dx
π
∫0 sin t dx
u = t du = dt dv = sin t dx v = −cos t π
= [−t cos t +∫ cos t dt ]0 = [t cos t+sin t ]π0 =
−π cos π+sin π = −π∗(−1)+0 = π
r)
∫ x sin 3x dx
−x 1 cos x + ∫ cos 3x dx 3 3
u = x du = dx dv = sin 3x dx −1 v = cos x 3
u = x
du = dx
−x 1 x 1 1 cos 3x+ ∫ cos u du = − cos 3x+ sen u = (−3x cos 3x+sin 3x dx ) 3 9 3 9 9
