log
10
x
;
∫
log
10
x
d
x
=
0.4343
x
(
log
ϵ
x
−
1
)
+
C
.
a
x
(p.149)
;
∫
a
x
d
x
=
a
x
log
ϵ
a
+
C
.
cos
a
x
;
∫
cos
a
x
d
x
=
1
a
sin
a
x
+
C
{\displaystyle {\begin{alignedat}{4}&\log _{10}x;&&&&\int \log _{10}x\,dx&&=0.4343x(\log _{\epsilon }x-1)+C.\\&a^{x}&&{\text{(p.149)}};\quad \quad &&\int a^{x}\,dx&&={\dfrac {a^{x}}{\log _{\epsilon }a}}+C.\\&\cos ax;&&&&\int \cos ax\,dx&&={\frac {1}{a}}\sin ax+C\\\end{alignedat}}}
(for if
y
=
sin
a
x
{\displaystyle y=\sin ax}
d
y
d
x
=
a
cos
a
x
{\displaystyle {\dfrac {dy}{dx}}=a\cos ax}
cos
a
x
{\displaystyle \cos ax}
y
=
1
a
sin
a
x
{\displaystyle y={\dfrac {1}{a}}\sin ax}
sin
a
x
;
∫
sin
a
x
d
x
=
−
1
a
cos
a
x
+
C
.
{\displaystyle \sin ax;\;\quad \quad \quad \int \sin ax\,dx=-{\frac {1}{a}}\cos ax+C.}
Try also
cos
2
θ
{\displaystyle \cos ^{2}\theta }
cos
2
θ
=
cos
2
θ
−
sin
2
θ
=
2
cos
2
θ
−
1
{\displaystyle \cos 2\theta =\cos ^{2}\theta -\sin ^{2}\theta =2\cos ^{2}\theta -1}
hence
cos
2
θ
=
1
2
(
cos
2
θ
+
1
{\displaystyle \cos ^{2}\theta ={\tfrac {1}{2}}(\cos 2\theta +1}
and
∫
cos
2
θ
d
θ
=
1
2
∫
(
cos
2
θ
+
1
)
d
θ
=
1
2
∫
cos
2
θ
d
θ
+
1
2
∫
d
θ
.
=
sin
2
θ
4
+
θ
2
+
C
.
(See also p. 227.)
{\displaystyle {\begin{aligned}\int \cos ^{2}\theta \,d\theta &={\tfrac {1}{2}}\int (\cos 2\theta +1)\,d\theta \\&={\tfrac {1}{2}}\int \cos 2\theta \,d\theta +{\tfrac {1}{2}}\int d\theta .\\&={\frac {\sin 2\theta }{4}}+{\frac {\theta }{2}}+C.\quad {\text{(See also p. 227.)}}\end{aligned}}}
See also the Table of Standard Forms on pp. 252, 253. You should make such a table for yourself, putting in it only the general functions which you have successfully differentiated and integrated. See to it that it grows steadily!
