416 INTEGRAL CALCULUS
from 12, Table 40.1, page 398. Hence
∫
dx
7 −3 sinx+6 cosx
=tan−^1⎛⎜
⎝tanx
2− 32⎞⎟
⎠+cProblem 7. Determine:∫
dθ
4 cosθ+3 sinθFrom equations (1) to (3),
∫
dθ
4 cosθ+3 sinθ
=∫2dt
1 +t^24(
1 −t^2
1 +t^2)
+ 3(
2 t
1 +t^2)=∫
2dt
4 − 4 t^2 + 6 t=∫
dt
2 + 3 t− 2 t^2=−1
2∫
dtt^2 −3
2t− 1=−1
2∫
dt
(
t−3
4) 2
−25
16=1
2∫
dt
(
5
4) 2
−(
t−3
4) 2=1
2⎡⎢
⎢
⎣12(
5
4)ln⎧
⎪⎪
⎨⎪⎪
⎩5
4+(
t−3
4)5
4−(
t−3
4)⎫
⎪⎪
⎬⎪⎪
⎭⎤⎥
⎥
⎦+cfrom problem 11, Chapter 41, page 411=1
5ln⎧
⎪⎨⎪⎩1
2+t2 −t⎫
⎪⎬⎪⎭+cHence
∫
dθ
4 cosθ+3 sinθ=1
5ln⎧
⎪⎨⎪⎩1
2+tanθ
22 −tanθ
2⎫
⎪⎬⎪⎭+cor1
5ln⎧
⎪⎨⎪⎩1 +2 tanθ
24 −2 tanθ
2⎫
⎪⎬⎪⎭+cNow try the following exercise.Exercise 167 Further problems on the
t=tanθ/2 substitutionIn Problems 1 to 4, integrate with respect to the
variable.1.∫
dθ
5 +4 sinθ
⎡⎢
⎣2
3tan−^1⎛⎜
⎝5 tanθ
2+ 43⎞⎟
⎠+c⎤⎥
⎦2.∫
dx
1 +2 sinx
⎡⎢
⎣1
√
3ln⎧
⎪⎨⎪⎩tanx
2+ 2 −√
3tanx
2+ 2 +√
3⎫
⎪⎬⎪⎭+c⎤⎥
⎦3.∫
dp
3 −4 sinp+2 cosp
⎡⎢
⎣1
√
11ln⎧
⎪⎨⎪⎩tanp
2− 4 −√
11tanp
2− 4 +√
11⎫
⎪⎬⎪⎭+c⎤⎥
⎦4.∫
dθ
3 −4 sinθ
⎡⎢
⎣1
√
7ln⎧
⎪⎨⎪⎩3 tanθ
2− 4 −√
73 tanθ
2− 4 +√
7⎫
⎪⎬⎪⎭+c⎤⎥
⎦- Show that
∫
dt
1 +3 cost
=1
2√
2ln⎧
⎪⎨⎪⎩√
2 +tant
2
√
2 −tant
2⎫
⎪⎬⎪⎭+c- Show that
∫π/ 303dθ
cosθ= 3 .95, correct to 3significant figures.- Show that
∫π/ 20dθ
2 +cosθ=π
3√
3