EXERCISE 7.1

 

1-30 Evaluate the integrals by making appropriate u-substitution.

  1. $$\quad \int(4-2 x)^{3} d x ——–(1)$$

Let

$$
u=4-2 x——-(2)
$$

$$ du=-2dx \Rightarrow \frac{-du}{2}=dx$$

then (1) be comes

$$
\int u^{3}\left(-\frac{1}{2} d u\right)=-\frac{1}{2} \int u^{3} d u
$$

Integrating w.r.t $u$, we get

$$
=\frac{-1}{2}\left[\frac{u^{4}}{4}\right]+c=-\frac{1}{8} u^{4}+c—–(3)
$$

$$
\text {From (2) substituting } u=4-2 x \quad \text {into}  \quad (3)
$$

$$
\int(4-2 x)^{3} d x=-\frac{1}{8}(4-2 x)^{4}+c
$$

2. $$\int 3 \sqrt{4+2 x} d x—(1)$$

$$
\begin{aligned}
& \text {Let} \quad  u=4+2 x —-(2)  \\ \\
& d u=2 d x \Rightarrow d x=\frac{1}{2} d u
\end{aligned}
$$

then (1) becomes

$$
\int 3 u^{1 / 2} \frac{d u}{2}=\frac{3}{2}\left[\frac{u^{3 / 2}}{3 / 2}\right]+c
$$

$$
=u^{3 / 2}+c——(3)
$$

From (2) Substituting $u=4+2 x$ into (3), we get

$$
\int 3 \sqrt{4+2 x} d x=(4+2 x)^{3 / 2}+c
$$

3.  $$\int x \sec ^{2}\left(x^{2}\right) d x \quad-(1)$$

Let $$u=x^{2}——(2)$$

$$

d u=2 x d x \Rightarrow x d x=\frac{d u}{2}
$$

then (1) becomes

$$
\begin{aligned}
\int \sec ^{2} u \frac{d u}{2} & =\frac{1}{2} \int \sec ^{2} u d u=\frac{1}{2} \tan u+c—–(3) \\ \\
\text { Substituting } u & =x^{2} \text { in (3) } \\ \\
& \int x \sec ^{2}\left(x^{2}\right) d x=\frac{1}{2} \tan x^{2}+c
\end{aligned}
$$

4.  $$\int 4 x \tan \left(x^{2}\right) d x—(1)$$

Let $$u=x^{2}—(2)$$

$$ du=2dx \Rightarrow dx=\frac{du}{2}$$

then (1) becomes

$$
\int 4 \tan u \frac{d u}{2}=2 \int \tan u d u=-2 \ln |\cos u|+c——(3)
$$

Substituting $u=x^{2}$ into (3)

 

$$
\int 4 x \tan \left(x^{2}\right) d x=-2 \ln \left|\cos x^{2}\right|+c
$$

5 .

$$
\begin{aligned}
\int \frac{\sin 3 x}{2+\cos 3 x} d x & —(1) \\ \\
\operatorname{Let} \quad & u =2+\cos 3 x —-(2) \\ \\
d u & =-3 \sin 3 x d x \\
-\frac{d u}{3} & =\sin 3 x d x
\end{aligned}
$$

then (1) becomes

$$
\int \frac{1}{u}\left(-\frac{d u}{3}\right)=-\frac{1}{3} \int \frac{1}{u} d u=-\frac{1}{3} \ln |u|+C
$$

Substituting $u=2+\cos 3 x$ into (3) we get

$$
\int \frac{\sin 3 x}{2+\cos 3 x} d x=\frac{-1}{3} \ln |2+\cos 3 x|+c
$$

6.

$$
\begin{aligned}
& \int \frac{1}{9+4 x^{2}} d x —–(1)\\ \\
& =\frac{1}{9} \int \frac{1}{1+\frac{4}{9} x^{2}} d x=\frac{1}{9} \int \frac{1}{1+\left(\frac{2}{3} x\right)^{2}} d x \\ \\
& \text { Let } u=\frac{2}{3} x—-(2) \\ \\
& d u=\frac{2}{3} d x \Rightarrow \frac{3}{2} d u=d x
\end{aligned}
$$

then (1) becomes

$$
\begin{aligned}
& =\frac{1}{9} \int \frac{1}{1+u^{2}} \cdot \frac{3}{2} d u=\frac{3 / 2}{9} \int \frac{1}{1+u^{2}} d u=\frac{1}{6} \int \frac{1}{1+u^{2}} d u \\ \\
& =\frac{1}{6} \tan ^{-1} u+c \quad-(3) \\ \\

& \text { Substititing } \quad u=\frac{2}{3} x \quad \text { into (3) } \\ \\

& \int \frac{1}{9+4 x^{2}} d x=\frac{1}{6} \tan ^{-1} \left(\frac{2}{3}x\right )+c
\end{aligned}
$$

7.

$$
\int e^{x} \sinh \left(e^{x}\right) d x
$$

Let

$$
\begin{aligned}
& u=e^{x}-(2) \\ \\
& d u=e^{x} d x
\end{aligned}
$$

then (1) becomes

$$
\int \sinh u d u=\cosh u+c
$$

Substitinting $u=e^{x}$ into (3), we get

$$
\int e^{x} \sinh \left(e^{x}\right) d x=\cosh \left(e^{x}\right)+c
$$

8. $$\int \frac{\sec (\ln x) \tan (\ln x)}{x} d x—(1)$$

Let $$u=\ln x \quad— (2)$$

$$
d u=\frac{1}{x} d x
$$

then (1) becomes

$$
\int \sec u \tan u d u
$$

integrating w.r.t $u$

$$
=\sec u+c——(3)
$$

Substituting $u=\ln x$ into (3)

$$ \int \frac{\sec (\ln x) \tan (\ln x)}{x} d x=\sec (\ln x)+c$$

9.

$$
\begin{aligned}

& \int e^{\tan x} \sec ^{2} x d x—-(1)
\end{aligned}
$$

Let $$u=\tan x—-(2)$$

$$
d u=\sec ^{2} x d x
$$

then (1) becomes

$$
\begin{aligned}
\int e^{u} d u & =e^{u}+c \\ \\
& =e^{\tan x}+c \quad \therefore u=\tan x .
\end{aligned}
$$

10. $$\int \frac{x}{\sqrt{1-x^{4}}} d x—-(1)$$

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

Let $$u=x^{2}—–(2)$$

then (1) becomes

$$
\begin{aligned}
\int \frac{d u / 2}{\sqrt{1-u^{2}}} & =\frac{1}{2} \sin ^{-1} u+c . \\ \\
& =\frac{1}{2} \sin ^{-1}\left(x^{2}\right)+c \quad \therefore u=x^{2}
\end{aligned}
$$