Advanced book on Mathematics Olympiad

536 Real Analysis

455.Denote the value of the integral byI. With the substitutiont=abx we have

I=

∫b

a

e

bt

−e

ta

ab

t

·

−ab

t^2

dt=−

∫b

a

e

ta

−e

bt

t

dt=−I.

HenceI=0.

456.The substitutiont= 1 −xyields

I=

∫ 1

0

√ (^32) ( 1 −t) (^3) − 3 ( 1 −t) (^2) −( 1 −t)+ 1 dt=−

∫ 1

0

√ (^32) t (^3) − 3 t (^2) −t+ 1 dt=−I.
HenceI=0.
(Mathematical Reflections, proposed by T. Andreescu)
457.Using the substitutionsx=asint, respectively,x=acost, we find the integral to
be equal to both the integral

L 1 =

∫π/ 2

0

sint

sint+cost

dt

and the integral

L 2 =

∫π/ 2

0

cost

sint+cost

dt.

Hence the desired integral is equal to

1

2

(L 1 +L 2 )=

1

2

∫ π 2

0

1 dt=

π

4

.

458.Denote the integral byI. With the substitutiont=π 4 −xthe integral becomes

I=

∫ 0

π 4 ln

(

1 +tan

(π

4

−t

))

(− 1 )dt=

∫ π 4

0

ln

(

1 +

1 −tant

1 +tant

)

dt

=

∫ π 4

0

ln

2

1 +tant

dt=

π

4

ln 2−I.

Solving forI, we obtainI=π 8 ln 2.

459.With the substitution arctanx=tthe integral takes the form

I=

∫ π 4

0

ln( 1 +tant)dt.