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

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