1000 Solved Problems in Modern Physics

1.3 Solutions 53

1.45 (a)

∫

tan^6 xsec^4 xdx=

∫

tan^6 x(tan^2 x+1) sec^2 xdx

=

∫

(tanx)^8 sec^2 xdx+

∫

tan^6 xsec^2 xdx

=

∫

(tanx)^8 d(tanx)+

∫

(tanx)^6 d(tanx)

=

tan^9 x

9

+

tan^7 x

7

+C

(b)

∫

tan^5 xsec^3 xdx=

∫

tan^4 xsec^2 xsecxtanxdx

=

∫

(sec^2 x−1)^2 sec^2 xsecxtanxdx

=

∫

(sec^6 x−2sec^4 x+sec^2 x)d(secx)

=

sec^7 x

7

− 2

sec^5 x

5

+

sec^3 x

3

+C

1.46

∫ 4

2

2 x+ 4

x^2 − 4 x+ 8

dx=

∫ 4

2

2 x− 4 + 8

(x−2)^2 + 4

dx

=

∫ 4

2

2 x− 4

(x−2)^2 + 4

dx+ 8

∫ 4

2

dx

(x−2)^2 + 4

=ln [(x−2)^2 +4]^42 +(8/2) tan−^11

=ln 2+π

1.47 Let us first find the area OMP which is half of the required area OPP′.Forthe
upper branch of the curve,y=x^3 /^2 , and summing up all the strips between
the limitsx=0 andx=4, we get
Area OMP=

∫ 4

0 ydx=

∫ 4

0 x

3 / (^2) dx= 64
5.
Hence area OPP′= 2 x^645 = 25 .6 units.
Note: for the lower branchy=x^3 /^2 and the area will be− 64 /5. The area
will be negative simply because for the lower branch they-coordinates are
negative. The result for the area OPP′pertains to total area regardless of sign.
Fig. 1.12Semi-cubical
parabola