Calculus Solution Ex#5.3

 

 

EXERCISE $5 \cdot 3$

Evaluate the integrals using appropriate Substitutions

$15-$

$$
\begin{aligned}
\int(4 x-3)^{9} d x \\ \\
\text { Let } u=4 x-3  \quad \quad \quad (1) \\ \\
d u=4 d x \\ \\
\frac{1}{4} d u=d x
\end{aligned}
$$

then the above integral becomes

$$
\begin{aligned}
\frac{1}{4} \int u^{9} d u & =\frac{1}{4}\left[\frac{u^{9+1}}{9+1}\right]+c \\  \\
& =\frac{1}{40} u^{10}+c-(2) \\ \\
\end{aligned}
$$

From (1) substituting $u=4 x-3$ into (2).$$
\int(4 x-3)^{9} d x=\frac{1}{40}(4 x-3)^{10}+c
$$

$16-$
$$
\begin{aligned}
\int x^{3} \sqrt{(5+x^{4})} d x \\ \\
\int x^{3}\left(5+x^{4}\right)^{1 / 2} d x-(1) \\ \\
\text { Let } u=5+x^{4}—-(2) \\ \\
d u=4 x^{3} d x \\ \\
\frac{1}{4} du=x^{3} dx

\end{aligned}
$$

then the integral (1) becomes

$$
\begin{aligned}
\frac{1}{4} \int u^{1 / 2} d u & =\frac{1}{4}\left[\frac{u^{\frac{1}{2}+1}}{\frac{1}{2}+1}\right]+c \\ \\
& =\frac{1}{4}\left[\frac{u^{3 / 2}}{3 / 2}\right]+c \\ \\
& =\frac{1}{6} u^{3 / 2}+c —(3)
\end{aligned}
$$

Now substituting $u=5+x^{4}$ into (3)

$$
\int x^{3} \sqrt{5+x^{4}} d x=\frac{1}{6}\left[5+x^{4}\right]^{3 / 2}+c
$$

$17-$

$$
\begin{aligned}
\int \sin 7 x d x \\ \\
\text {Let} \quad u=7 x \\ \\
d u=7 d x \\ \\
\frac{1}{7} d u=d x
\end{aligned}
$$

Then the above integral becomes

$$
\begin{aligned}
\frac{1}{7} \int \sin u d u & =\frac{1}{7}(-\cos u)+c \\ \\
& =-\frac{1}{7} \cos u+c —–(1)
\end{aligned}
$$

Substituting $u=7 x$ into (1)

$$
\int \sin 7 x d x=-\frac{1}{7} \cos 7 x+c
$$

$18-$
$$
\begin{aligned}
\int \cos \frac{x}{3} d x \\ \\
\text {Let} \quad u=\frac{x}{3} \\ \\
d u=\frac{1}{3} d x \Rightarrow 3 d u=d x
\end{aligned}
$$

Then the above integral becomes

$$
\end{aligned}
3 \int \cos u d u=3 \sin u+c —-(1)
\end{aligned}
$$

Substituting $u=x / 3$ into (1)

$$
\int \cos \frac{x}{3} d x=3 \sin \frac{x}{3}+c
$$

$19-$
$$
\begin{aligned}
\int \sec 4 x \tan 4 x d x \\ \\
\tex {Let} \quad u=4 x \\ \\
d u= 4 dx \Rightarrow \frac{1}{4} d u=dx
\end{aligned}
$$

Then the above integral becomes

$$
\frac{1}{4} \int \sec u \tan u d u=\frac{1}{4} \sec u+c —(1)
$$

Substituting $u=4 x$ into (1)

$$
\int \sec 4 x \tan 4 x d x=\frac{1}{4} \sec 4 x+c
$$

$20-$
$$
\begin{aligned}
\int \sec ^{2} 5 x d x$ \\ \\

\text {Let} \quad u=5 x \Rightarrow d u=5 d x \Rightarrow \frac{1}{5} d u=d x
\end{aligned}
$$
Then the above integral becomes

$$
\frac{1}{5} \int \sec ^{2} u d u=\frac{1}{5} \tan u+c —-(1)
$$

Substituting $u=5 x$ into (1)

$$
\int \sec ^{2} 5 x d x=\frac{1}{5} \tan 5 x+c
$$
$23-$
$$
\begin{aligned}
\int \frac{d x}{\sqrt{1-4 x^{2}}} \\ \\
\text {Let} \quad u=2 x \\ \\
d u=2 d x \Rightarrow \frac{1}{2} d u=d x
\end{aligned}
$$

Then the above integral becomes

$$
\begin{aligned}
\int \frac{\frac{1}{2} d u}{\sqrt{1-u^{2}}} & =\frac{1}{2} \int \frac{d u}{\sqrt{1-u^{2}}} \\ \\
& =\frac{1}{2} \sin ^{-1}(u)+c —-(1)
\end{aligned}
$$

Substituting $u=2 x$ into (1)

$$
\int \frac{d x}{\sqrt{1-4 x^{2}}}=\frac{1}{2} \sin ^{-1}(2 x)+c
$$

$28-$
$$
\begin{aligned}
\int \frac{x^{2}+1}{\sqrt{x^{3}+3 x}} d x \\ \\

\text {Let} \quad u=x^{3}+3 x \\ \\

d u & =(3 x^{2}+3) d x \Rightarrow \frac{1}{3} d u=(x^{2}+1) d x \\ \\
\text {Then the above integral becomes} \\ \\
\frac{1}{3} \int \frac{d u}{\sqrt{u}}=\frac{1}{3} \int \frac{d u}{u^{1 / 2}} \\ \\
=\frac{1}{3} \int u^{-1 / 2} d u \\ \\
=\frac{1}{3} \left[\frac{u^{\frac{-1}{2}+1}}{\frac{-1}{2}+1} \right]+c \\ \\
=\frac{1}{3}\left[\frac{u^{\frac{1}{2}}}{\frac{1}{2}}\right]+c \\ \\
=\frac{2}{3} u^{\frac{1}{2}}+c ——(1)

\end{aligned}
$$

Substituting $u=x^{3}+3 x$ into (1)

$$
\int \frac{x^{3}+3 x}{\sqrt{x^{3}+3 x}} d x=\frac{2}{3}\left(x^{3}+3 x\right)^{\frac{1}{2}}+c
$$