Higher Engineering Mathematics, Sixth Edition

Reduction formulae 433

=

∫

tann−^2 xsec^2 xdx−In− 2

i.e.In=

tann−^1 x

n− 1

−In− 2

Whenn=7,

I 7 =

∫

tan^7 xdx=

tan^6 x

6

−I 5

I 5 =

tan^4 x

4

−I 3 and I 3 =

tan^2 x

2

−I 1

I 1 =

∫

tanxdx=ln(secx)

from Problem 9, Chapter 39, page 394

Thus

∫

tan^7 xdx=

tan^6 x

6

−

[

tan^4 x

4

−

(

tan^2 x

2

−ln(secx)

)]

Hence

∫

tan^7 xdx

=

1

6

tan^6 x−

1

4

tan^4 x+

1

2

tan^2 x

−ln(secx)+c

Problem 14. Evaluate, using a reduction formula,

∫ π

2

0

sin^2 tcos^6 tdt.

∫ π

2

0

sin^2 tcos^6 tdt=

∫ π

2

0

( 1 −cos^2 t)cos^6 tdt

=

∫ π

2

0

cos^6 tdt−

∫ π

2

0

cos^8 tdt

If In=

∫ π

2

0

cosntdt

then
∫ π 2

0

sin^2 tcos^6 tdt=I 6 −I 8

and from equation (6),

I 6 =

5

6

I 4 =

5

6

[

3

4

I 2

]

=

5

6

[

3

4

(

1

2

I 0

)]

and I 0 =

∫ π 2

0

cos^0 tdt

=

∫ π 2

0

1dt=[x]

π

2

0 =

π

2

Hence I 6 =

5

6

·

3

4

·

1

2

·

π

2

=

15 π

96

or

5 π

32

Similarly,I 8 =

7

8

I 6 =

7

8

·

5 π

32

Thus

∫ π

2

0

sin^2 tcos^6 tdt=I 6 −I 8

=

5 π

32

−

7

8

·

5 π

32

=

1

8

·

5 π

32

=

5 π

256

Problem 15. Use integration by parts to

determine a reduction formula for

∫

(lnx)ndx.

Hence determine

∫

(lnx)^3 dx.

LetIn=

∫

(lnx)ndx.

Using integration by parts, letu=(lnx)n, from which,

du

dx

=n(lnx)n−^1

(

1

x

)

and du=n(lnx)n−^1

(

1

x

)

dx

and let dv=dx, from which,v=

∫

dx=x

Then In=

∫

(lnx)ndx

=(lnx)n(x)−

∫

(x)n(lnx)n−^1

(

1

x

)

dx