Higher Engineering Mathematics

430 INTEGRAL CALCULUS

Hence

∫

cos^4 xdx

=

1

4

cos^3 xsinx+

3

4

(

1

2

cosxsinx+

1

2

x

)

=

1

4

cos^3 xsinx+

3

8

cosxsinx+

3

8

x+c

Problem 12. Determine a reduction formula

for

∫π

2

0

cosnxdxand hence evaluate

∫ π

2

0

cos^5 xdx.

From equation (5),

∫

cosnxdx=

1

n

cosn−^1 xsinx+

n− 1

n

In− 2

and hence

∫ π

2

0

cosnxdx=

[

1

n

cosn−^1 xsinx

]π 2

0

+

n− 1

n

In− 2

=[0−0]+

n− 1

n

In− 2

i.e.

∫ π

2

0

cosnxdx=In=

n− 1

n

In− 2 (6)

(Note that this is the same reduction formula as

for

∫π 2

0

sinnxdx(in Problem 10) and the result is

usually known asWallis’s formula).
Thus, from equation (6),
∫π
2

0

cos^5 xdx=

4

5

I 3 , I 3 =

2

3

I 1

and I 1 =

∫π

2

0

cos^1 xdx

=[sinx]

π

2

0 =(1−0)=^1

Hence

∫π

2

0

cos^5 xdx=

4

5

I 3 =

4

5

[

2

3

I 1

]

=

4

5

[

2

3

(1)

]

=

8

15

Now try the following exercise.

Exercise 172 Further problems on reduc-

tion formulae for integrals of the form∫

sinnxdxand

∫

cosnxdx

1. Use a reduction formula to determine∫
   sin^7 xdx.
   ⎡

⎢

⎣

−

1

7

sin^6 xcosx−

6

35

sin^4 xcosx

−

8

35

sin^2 xcosx−

16

35

cosx+c

⎤

⎥

⎦

2. Evaluate

∫π

0 3 sin

(^3) xdx using a reduction
formula. [4]

3. Evaluate

∫π

2

0

sin^5 xdx using a reduction

formula.

[

8

15

]

4. Determine, using a reduction formula,∫

cos^6 xdx.

⎡

⎢

⎣

1

6

cos^5 xsinx+

5

24

cos^3 xsinx

+

5

16

cosxsinx+

5

16

x+c

⎤

⎥

⎦

5. Evaluate

∫ π

2

0

cos^7 xdx.

[

16

35

]

44.5 Further reduction formulae

The following worked problems demonstrate further

examples where integrals can be determined using

reduction formulae.

Problem 13. Determine a reduction formula

for

∫

tannxdxand hence find

∫

tan^7 xdx.

LetIn=

∫

tannxdx≡

∫

tann−^2 xtan^2 xdx

by the laws of indices

=

∫

tann−^2 x(sec^2 x−1) dx

since 1+tan^2 x=sec^2 x

=

∫

tann−^2 xsec^2 xdx−

∫

tann−^2 xdx