BASIC CALCULUS
Quarter 4
Week 3&4
(I)Find the antiderivative of the following functions using substitution.
1. ∫ [ 2 e sin (x+ 1) cos ( x+1 ) ] dx
u=sin (x +1)
du=cos ( x +1 ) dx
∫ [ 2 e sin (x+ 1) cos ( x+1 ) ] dx=2 ∫ e u du
u
¿2e +c
substitute u=sin ( x +1 ) back intothe expression
∴∫ [ 2e cos ( x+1 ) dx ]=2e
sin ( x +1) sin ( x +1)
+c
2.∫ ( x −5) 2 xdx
2 5
2
u=x −5
du=2 xdx
∫ (x 2−5)5 2 xdx=∫ u5 du
1 6
¿ u +c
6
2
substitute u=x −5 back into the expression
1 2
∴∫ ( x −5 ) 2 xdx= ( x −5) +c
2 6
6
x2− x
3.∫ dx
1+3 x2 −2 x 3
2 3
u=1+ 3 x −2 x
2
du=6 x−6 x dx
¿−6 ( x −x ) dx
2
2
∫ 1+3xx−2 −2
x
x
3
−1 1
dx = ∫ du
6 u
, 1
¿− ln |u|+ c
6
2 3
substitute u=1+3 x −2 x
x2 −x −1
∴∫ dx= ln |1+3 x2 −2 x 3|+c
2
1+3 x −2 x 3
6
cosx
4.∫ 1+2 sinx dx
u=1+ 2 sinx
du=2 cosxdx
cosx 1 1
∫ 1+2 sinx dx= 2 ∫ u du
1
¿ ln |u|+ c
2
substitute u=1+2 sinx back intothe expression
cosx 1
∴∫ dx= ln |1+2 sinx|+c
1+2 sinx 2
5 3 2
3 x −2 x +5 x −2
5.∫ dx
x3 +1
5 3 2
3 x −2 x +5 x −2
Lets first simplify ∨break downthe expression 3
x +1
3 x 5−2 x 3 +5 x2−2 ( 3 x +5 x ) −(2 x + 2)
5 2 3
3
= 3
x +1 x +1
2 3 3
x (3 x +5) 2( x +1)
¿ −
3
x +1 (x¿¿ 3+ 1) ¿
2 3
x (3 x +5)
¿ 3
−2
x +1
5 3 2 2 3
3 x −2 x +5 x −2 x (3 x +5)
∫ 3
x +1
dx=∫ 3
x +1
dx −2∫ dx
|
2 3
x (3 x +5)
Lets ∫ 3
dx cause−2 ∫ dx=−2 x +c
x +1
3 3
u=x +1 ⟹ x =u−1
2
du=3 x dx
Quarter 4
Week 3&4
(I)Find the antiderivative of the following functions using substitution.
1. ∫ [ 2 e sin (x+ 1) cos ( x+1 ) ] dx
u=sin (x +1)
du=cos ( x +1 ) dx
∫ [ 2 e sin (x+ 1) cos ( x+1 ) ] dx=2 ∫ e u du
u
¿2e +c
substitute u=sin ( x +1 ) back intothe expression
∴∫ [ 2e cos ( x+1 ) dx ]=2e
sin ( x +1) sin ( x +1)
+c
2.∫ ( x −5) 2 xdx
2 5
2
u=x −5
du=2 xdx
∫ (x 2−5)5 2 xdx=∫ u5 du
1 6
¿ u +c
6
2
substitute u=x −5 back into the expression
1 2
∴∫ ( x −5 ) 2 xdx= ( x −5) +c
2 6
6
x2− x
3.∫ dx
1+3 x2 −2 x 3
2 3
u=1+ 3 x −2 x
2
du=6 x−6 x dx
¿−6 ( x −x ) dx
2
2
∫ 1+3xx−2 −2
x
x
3
−1 1
dx = ∫ du
6 u
, 1
¿− ln |u|+ c
6
2 3
substitute u=1+3 x −2 x
x2 −x −1
∴∫ dx= ln |1+3 x2 −2 x 3|+c
2
1+3 x −2 x 3
6
cosx
4.∫ 1+2 sinx dx
u=1+ 2 sinx
du=2 cosxdx
cosx 1 1
∫ 1+2 sinx dx= 2 ∫ u du
1
¿ ln |u|+ c
2
substitute u=1+2 sinx back intothe expression
cosx 1
∴∫ dx= ln |1+2 sinx|+c
1+2 sinx 2
5 3 2
3 x −2 x +5 x −2
5.∫ dx
x3 +1
5 3 2
3 x −2 x +5 x −2
Lets first simplify ∨break downthe expression 3
x +1
3 x 5−2 x 3 +5 x2−2 ( 3 x +5 x ) −(2 x + 2)
5 2 3
3
= 3
x +1 x +1
2 3 3
x (3 x +5) 2( x +1)
¿ −
3
x +1 (x¿¿ 3+ 1) ¿
2 3
x (3 x +5)
¿ 3
−2
x +1
5 3 2 2 3
3 x −2 x +5 x −2 x (3 x +5)
∫ 3
x +1
dx=∫ 3
x +1
dx −2∫ dx
|
2 3
x (3 x +5)
Lets ∫ 3
dx cause−2 ∫ dx=−2 x +c
x +1
3 3
u=x +1 ⟹ x =u−1
2
du=3 x dx
