Integration with Trigonometric Identities

## **Problem**

**Part 1:** Evaluate the indefinite integral

∫

a
2

−

x
2

x
2

d
x
.

{\displaystyle \int \frac{\sqrt{a^2-x^2}}{x^2} \,dx. }

**Part 2:** Evaluate the definite integral

∫

0

π

/

8

cos

11

⁡
(
4
x
)

sin
3

⁡
(
4
x
)

d
x
.

{\displaystyle \int\_{0}^{\pi/8} \cos^{11} (4x) \sin^3(4x) \,dx. }

## **Solution**

**Part 1**

Begin with the trigonometric substitution

# x
a
sin
⁡
θ

{\displaystyle x = a\sin \theta}

 and

d
# x
a
cos
⁡
θ

d
θ

{\displaystyle dx = a\cos \theta \, d\theta }

.

Hence,

∫

a
2

−

x
2

x
2

d
x

=
∫

a
2

−

a
2

sin
2

⁡
θ

a
2

sin
2

⁡
θ

a
cos
⁡
θ

d
θ

=
∫

a
2

cos
2

⁡
θ

a
2

sin
2

⁡
θ

d
θ

=
∫

cot
2

⁡
θ

d
θ
.

{\displaystyle \begin{align}
\int \frac{\sqrt{a^2-x^2}}{x^2} \,dx &= \int \frac{\sqrt{a^2-{a^2\sin^2 \theta}}}{a^2 \sin^2 \theta} \,a\cos \theta \,d\theta \\[5 pt]
&= \int \frac{a^2 \cos^2 \theta}{a^2 \sin^2 \theta }\, d\theta \\[5 pt]
&= \int \cot^2 \theta \, d\theta.
\end{align} }

Since

1

cot
2

⁡
# θ

csc
2

⁡
θ

{\displaystyle 1 + \cot^2 \theta = \csc^2 \theta}

, we have

∫

csc
2

⁡
θ
−
1

d
# θ
−
cot
⁡
θ
−
θ
C
.

{\displaystyle \int \csc^2 \theta - 1\, d\theta = -\cot \theta - \theta + C. }

From the trigonometric substitution,

# θ
arcsin
⁡

(

x
a

)

{\displaystyle \theta = \arcsin\left(\frac{x}{a} \right)}

; therefore,

∫

a
2

−

x
2

x
2

d
# x
−
cot
⁡

(

arcsin
⁡

(

x
a

)

)

−
arcsin
⁡

(

x
a

)

C

{\displaystyle \int \frac{\sqrt{a^2-x^2}}{x^2}\, dx = -\cot \left(\arcsin\left(\frac{x}{a} \right)\right) - \arcsin\left(\frac{x}{a} \right) + C }

or alternatively

∫

a
2

−

x
2

x
2

d
# x
−

a
2

−

x
2

x

−
arcsin
⁡

(

x
a

)

C
.

{\displaystyle \int \frac{\sqrt{a^2-x^2}}{x^2}\, dx = -\frac{\sqrt{a^2 - x^2}}{x} - \arcsin\left(\frac{x}{a} \right) + C. }

**Part 2**

Begin by making the substitution

# u
4
x

{\displaystyle u = 4x}

 and

d
# u
4

d
x

{\displaystyle du = 4\, dx}

. The integration bounds then become

[
0
,
π

/

2
]

{\displaystyle [0, \pi/2] }

.

∫

0

π

/

8

cos

11

⁡
(
4
x
)

sin
3

⁡
(
4
x
)

d
# x

1
4

∫

0

π

/

2

cos

11

⁡
u

sin
3

⁡
u

d
u

{\displaystyle \int\_{0}^{\pi/8} \cos^{11} (4x) \sin^3(4x) \,dx = \frac{1}{4}\int\_{0}^{\pi/2} \cos^{11} u \sin^3 u \,du }

Since

sin
2

⁡
# u
1
−

cos
2

⁡
u

{\displaystyle \sin^2 u = 1 - \cos^2 u }

, we can express the integral as

∫

0

π

/

8

cos

11

⁡
(
4
x
)

sin
3

⁡
(
4
x
)

d
x

=

1
4

∫

0

π

/

2

cos

11

⁡
u
(
1
−

cos
2

⁡
u
)
sin
⁡
u

d
u

=

1
4

∫

0

π

/

2

(

cos

11

⁡
u
−

cos

13

⁡
u
)
sin
⁡
u

d
u
.

{\displaystyle \begin{align}
\int\_{0}^{\pi/8} \cos^{11} (4x) \sin^3(4x) \,dx &= \frac{1}{4}\int\_{0}^{\pi/2} \cos^{11} u (1-\cos^2 u)\sin u \,du \\[5 pt]
&= \frac{1}{4}\int\_{0}^{\pi/2} (\cos^{11} u - \cos^{13} u) \sin u \,du.
\end{align} }

Make the substitution

# v
cos
⁡
u

{\displaystyle v = \cos u }

 and

d
# v
−
sin
⁡
u

d
u

{\displaystyle dv = -\sin u \, du }

. The integration bounds then become

[
1
,
0
]

{\displaystyle [1, 0] }

.

∫

0

π

/

8

cos

11

⁡
(
4
x
)

sin
3

⁡
(
4
x
)

d
x

=
−

1
4

∫

1

0

v

11

−

v

13

d
v

=

1
4

∫

0

1

v

11

−

v

13

d
v

=

1
4

[

v

12

12

−

v

14

14

0

1

]

=

1
4

[

(

1
12

−

1
14

)

−

(

0
12

−

0
14

)

]

=

1
336

{\displaystyle \begin{align}
\int\_{0}^{\pi/8} \cos^{11} (4x) \sin^3(4x) \,dx &= -\frac{1}{4}\int\_{1}^{0} v^{11} - v^{13} \, dv \\[5 pt]
&= \frac{1}{4}\int\_{0}^{1} v^{11} - v^{13} \, dv \\[5 pt]
&= \frac{1}{4} \left[\frac{v^{12}}{12} - \frac{v^{14}}{14} \Biggr|\_{0}^{1} \right] \\[5 pt]
&= \frac{1}{4} \left[\left(\frac{1}{12} - \frac{1}{14}\right) - \left( \frac{0}{12} - \frac{0}{14} \right) \right] \\[5 pt]
&=\frac{1}{336}
\end{align} }