EXERCISE 7.3

 

Exercise 7.3

$1-52$ Evaluate the integral.

(1) $$ \int \cos ^{3} x \sin x d x —(1)$$

$$
\begin{aligned}
\text { Let } \quad u & =\cos x –(2)\\ \\
d u & =-\sin x d x \\ \\
-d u & =\sin x d x
\end{aligned}
$$

then (1) becomes

$$
-\int u^{3} d u=-\frac{u^{4}}{4}+c—(3)
$$

From (2) substituting $u=\cos x$ into (3).

$$
\begin{aligned}
\therefore & \int \cos ^{3} x \sin x d x=-\frac{\cos ^{4} x}{4}+c \end{aligned} $$

(2)

$$

\begin{aligned}
\int \sin ^{5} 3 x \cos 3 x d x & -(1) \\ \\
\text { Let } u & =\sin 3 x-(2) \\ \\
d u & =3 \cos 3 x d x \\ \\
\frac{1}{3} d u & =\cos 3 x d x
\end{aligned}
$$

then (1) becomes

$$
\frac{1}{3} \int u^{5} d u=\frac{1}{3} \frac{u^{6}}{6}+c=\frac{1}{18} u^{6}+c—(3)
$$

From (2) substituting $u=\cos 3x$ into (3)

$$
\int \sin ^{5} 3 x \cos 3 x d x=\frac{1}{18} \sin ^{6} 3 x+c
$$

 

(3)

$$\int \sin ^{2} 5 \theta d \theta—(1)$$

We know that

$$
\sin ^{2} 5 \theta=\frac{1-\cos 10 \theta}{2}
$$

then (1) becomes

$$
\begin{aligned}
& \int \frac{1-\cos 10 \theta}{2} d \theta \\ \\ &=\frac{1}{2} \int d \theta-\frac{1}{2} \int \cos 10 \theta d \theta \\ \\
& =\frac{1}{2} \theta-\frac{1}{2} \frac{\sin 10 \theta}{10}+c \\ \\
& =\frac{1}{2} \theta-\frac{1}{20} \sin 10 \theta+c
\end{aligned}
$$

(4)

$$\int \cos ^{2} 3 x d x —(1)$$

We know that

$$
\cos ^{2} 3 x=\frac{1+\cos 6 x}{2}
$$

then (1) becomes

$$
\begin{aligned}
& \int \frac{1+\cos 6 x}{2} d x=\frac{1}{2} \int d x+\frac{1}{2} \int \cos 6 x d x \\ \\
& =\frac{1}{2} x+\frac{1}{2} \cdot \frac{\sin 6 x}{6}+c \\ \\
& =\frac{1}{2} x+\frac{1}{12} \sin 6 x+c
\end{aligned}
$$

5.

$$
\int \sin ^{3} a \theta d \theta
$$

We can write it as

$$
\begin{aligned}
& \int \sin a \theta \sin ^{2} a \theta d \theta \\ \\
& =\int \sin a \theta\left(1-\cos ^{2} a \theta\right) d \theta-(1) \\ \\
& \operatorname{Let} u=\cos a \theta-(2) \\ \\
& d u=-d \sin a \theta d \theta \Rightarrow-\frac{1}{a} d u=\sin a \theta d \theta
\end{aligned}
$$

Then (1) becomes

$$
\begin{aligned}
& =\int\left(1-u^{2}\right)\left(-\frac{1}{a} d u\right) \\ \\
& =-\frac{1}{a} \int d u+\frac{1}{a} \int u^{2} d u \\ \\
& =-\frac{1}{a} u+\frac{1}{a} \cdot \frac{u^{3}}{3}+c—(3)
\end{aligned}
$$

From (2) Substituting $u=\cos a \theta$ into (3)

$$
\int \sin ^{3} a \theta d \theta=-\frac{\cos a \theta}{a}+\frac{1}{3 a} \cos ^{3} a \theta+c
$$

(6)

$$\quad \int \cos ^{3} a t dt $$

This can be written as

$$
\begin{aligned}
& \int \cos ^{2} a t \cos a t d t \\
& =\int\left(1-\sin ^{2} a t\right) \cos a t d t—(1)
\end{aligned}
$$

$$
\text { Let } u=\sin a t-(2)
$$

$$
\begin{aligned}
d u & =a \cos a t d t \\
\frac{1}{a} d u & =a \cos a t d t
\end{aligned}
$$

then (1) becomes

$$
\begin{aligned}
& \int\left(1-u^{2}\right) \frac{1}{d} d u \\
= & \frac{1}{a} \int d u-\frac{1}{a} \int u^{2} d u \\
= & \frac{1}{a} u-\frac{1}{a} \frac{u^{3}}{3}+c—(3)
\end{aligned}
$$

From (2) Substituting $u= \sin at$ into (3)

$$
\int \cos ^{3} a t d t=\frac{1}{a} \sin a t-\frac{1}{3 a} \sin ^{3} a t+c
$$

7.

$$
\begin{aligned}
\int \sin a x \cos a x d x — (1) \\ \\
\text { Let } u & =\sin a x \\ \\
d u & =a \cos a x d x \\ \\
\frac{1}{a} d u & =\cos a x d x
\end{aligned}
$$

Then (1) becomes

$$
\frac{1}{a} \int u d u=\frac{1}{a} \frac{u^{2}}{2}+c—-(3)
$$

From (2) Substituting $u=\sin a x$ into (3)

$$
\begin{aligned}
& \therefore \quad \sin a x \cos a x d x=\frac{1}{2 a} \sin ^{2} a x+c
\end{aligned}
$$

(8)

$$
\begin{aligned}
& \int \sin ^{3} x \cos ^{3} x d x \\ \\
& =\int \sin ^{3} x \cos ^{2} x \cos x d x \\ \\
& =\int \sin ^{3} x\left(1-\sin ^{2} x\right) \cos x d x \\ \\
& =\int\left(\sin ^{3} x-\sin ^{5} x\right) \cos x d x-(1) \\ \\
& \text { Let } u=\sin x-(2) \\ \\
& d u=\cos x d x
\end{aligned}
$$

Then (1) becomes

$$
\begin{aligned}
& \int\left(u^{3}-u^{5}\right) d u \\
& =\frac{u^{4}}{4}-\frac{u^{6}}{6}+c—-(3)
\end{aligned}
$$

From (2) substituting $u=\sin x$ into (3)

$$
\int \sin ^{3} x \cos ^{3} x d x=\frac{1}{4} \sin ^{4} x-\frac{1}{6} \sin ^{6} x+c
$$