3.6 Answer

3.6

1.

(a) Let $\sqrt{\frac{ax + b}{cx + d}}$ be equal to $t$. Then

$$\int{x\sqrt{1+x}} \ dx = \int{(t^2 -1)t(2t)}\ dt \ \left(\begin{array}{ll} t = \sqrt{1 + x} \mbox{implies} t^2 = 1 + x\\ x = t^2 - 1, dx = 2t dt \end{array}\right)$$

$$= 2\int(t^4 - t^2)\ dt = 2(\frac{t^5}{5} - \frac{t^3}{3}) + c$$

$$= 2[\frac{(1 + x)^{5/2}}{5} - \frac{(1+x)^{3/2}}{3}] + c$$

(b) Let $t = \sqrt{x}$. Then $t^2 = x$ and $2t dt = dx$. Thus,

$$\int{\frac{\sqrt{x}}{\sqrt{x} - 1}}\ dx = \int{\frac{t}{t-1} 2t}\ dt$$

$$= 2\int{\frac{t^2}{t-1}}\ dt$$

Now the degree of the numerator is greater than the degree of the denominator. So,

$$\frac{t^2}{t-1} = t + 1 + \frac{1}{t-1}.$$

Thus,

$$\int{\frac{\sqrt{x}}{\sqrt{x} - 1}}\ dx = 2\int{(t + 1 + \frac{1}{t-1})}\ dt$$

$$= 2[\frac{t^2}{2} + t + \log{\vert t-1\vert}] + c$$

$$= t^2 + 2t + 2\log{\vert t-1\vert} + c$$

$$= x + 2\sqrt{x} + 2\log{\vert\sqrt{x} - 1\vert} + c$$

(c) Let $t = \sqrt{1 + e^{x}}$. Then $t^2 = 1 + e^{x}$ and $2t dt = e^{x}dx$．Now express the integrand interms ot $t$ and $dt$. Then

$$dx = \frac{2t dt}{e^{x}} = \frac{2t dt}{t^2 - 1}.$$

Therefore,

$$\int{\frac{dx}{\sqrt{1 + e^{x}}}} = \int{\frac{2t}{t(t^2 - 1)}}\ dt$$

$$= 2\int{\frac{1}{t^2 -1}}\ dt \ \left(\begin{array}{l} \int{\frac{1... ...2}}\ dx \\ = \frac{1}{2a}\log{\vert\frac{x-a}{x+a}\vert} +c \end{array}\right)$$

$$= 2\cdot \frac{1}{2} \log{\vert\frac{t-1}{t+1}\vert} + c$$

$$= \log{\vert\frac{\sqrt{1+e^x} -1}{\sqrt{1+e^{x} + 1}}\vert} +c$$

(d) Let $t = \sqrt{\frac{x+1}{x-1}}$. Then $t^2 = \frac{x+1}{x-1}$． To find $dx$, rewrite the above equation in $x$ and solve for $x$.

$$t^2 = \frac{x+1}{x-1}$$

$$xt^2 - t^2 = x + 1$$

$$x(t^2 - 1) = t^2 + 1$$

$$x = \frac{t^2 + 1}{t^2 - 1}$$

Thus,

$$dx = \frac{2t(t^2 - 1) - (t^2 + 1)(2t)}{(t^2 - 1)^2} dt$$

$$= \frac{-4t}{(t^2 - 1)^2} dt$$

Put this back into the original equation. Then

$$\int{\sqrt{\frac{x+1}{x-1}}}\ dx = \int{\frac{-4t^2}{(t^2 - 1)^2}}\ dt$$

To integrate this, we use the partial fraction expansion.．

$$\frac{-4t^2}{(t^2 - 1)^{2}} = \frac{-4t^2}{(t+1)^{2}(t-1)^{2}}$$

$$= \frac{A}{t+1} + \frac{B}{(t+1)^{2}} + \frac{C}{t-1} + \frac{D}{(t-1)^{2}}$$

Clear the denominator.

$$-4t^2 = A(t+1)(t-1)^2 + B(t-1)^2 + C(t+1)^{2}(t-1) + D(t+1)^2$$

Set $t = 1$. Then

$$-4 = 4D \Rightarrow D = -1$$

Set $t = -1$. Then

$$-4 = 4B \Rightarrow B = -1$$

Match the coefficients of $t^3$. Then

$$0 = A + C$$

Match the coefficients of $t^0$. Then

$$-4 = A + B - C + D \Rightarrow A - C = -2$$

Thus, we have

$$A = -1, \ C = 1$$

Therefore,

$$\int\frac{-4t^2}{(t^2 - 1)^2}\ dt = \int{\frac{-1}{t+1}}\ dt + \int{\frac{-1}{(t+1)^2}}\ dt$$

$$+ \int{\frac{1}{t-1}}\ dt + \int{\frac{-1}{(t-1)^2}}\ dt$$

$$= -\log\vert t+1\vert + \frac{1}{t+1} + \log\vert t-1\vert + \frac{1}{t-1} + c$$

$$= -\log\vert t+1\vert + \log\vert t-1\vert + \frac{2t}{t^2 - 1} + c$$

$$= \log\vert\sqrt{\frac{x+1}{x-1}} + 1\vert - \log\vert\sqrt{\frac{x+1}{x-1}}\vert + \frac{(x-1)}{2}\sqrt{\frac{x+1}{x-1}} + c$$

2.

(a) Let $t = x^2 - 4$. Then $dt = 2x dx$.

$$\int{\frac{x}{\sqrt{x^2 - 4}}}\ dx = \frac{1}{2}\int{\frac{dt}{\sqrt{t}}}$$

$$= \frac{1}{2}\int{t^{-1/2}}\ dt = \frac{1}{2}\cdot 2t^{1/2} + c$$

$$= \sqrt{x^2 - 4} + c$$

(b) Let $t = \sqrt{4 - x^2}$. Then $t^2 = 4 - x^2$ implies $2tdt = -2xdx$. Now consider the traiangle with the angle $t$ and whose opposite side is $x$ and whose hypotenuse is 2. Then the bottom is $\sqrt{4 - x^2}$．Now let $x = 2\sin{t}$. Then $dx = 2\cos{t}dt$ and

$$\sqrt{4 - x^2} = \sqrt{4 - 4\sin^{2}{t}} = \sqrt{4\cos^{2}{t}} = 2\cos{t}$$

Thus,

$$\int{\frac{x^2}{\sqrt{4 - x^2}}}\ dx = \int{\frac{4\sin^{2}{t}}{2\cos{t}}2\cos{t}}\ dt = 4\int{\sin^{2}{t}}\ dt$$

$$= 4\int{\frac{1 - \cos{2t}}{2}}\ dt = 2[t - \frac{\sin{2t}{2}}] + c$$

$$= 2[t - \sin{t}\cos{t}] + c = 2[\sqrt{4 - x^2} - \frac{x}{2}\frac{\sqrt{4 - x^2}}{2} + c$$

(c) Let $t = e^{x}$. Then $dt = e^{x} dx$. Then

$$\int{\frac{e^{x}}{9 - e^{2x}}}\ dx = \int{\frac{1}{9 - t^2}}\ dt$$

$$= \frac{1}{6}\int{\frac{1}{3 - t} + \frac{1}{3+t}}\ dt$$

$$= \frac{1}{6}[-\log\vert 3-t\vert + \log\vert 3 + t\vert] +c = \frac{1}{6}\log\vert\frac{3+t}{3-t}\vert + c$$

$$= \frac{1}{6}\log\vert\frac{3 + e^{x}}{3 - e^{x}}\vert + c$$

(d) Let $t = \sqrt{1 - x^2}$. Then $t^2 = 1 - x^2$ and $2tdt = -2xdx$. Now consider the triangle with the angle $t$ and whose opposite side is $x$ and thee hypotenuse is 1. Then the bottom is $\sqrt{1 - x^2}$. Now $x = \sin{t}$ implies $dx = \cos{t}dt$ and ，

$$\sqrt{1 - x^2} = \sqrt{1 - \sin^{2}{t}} = \sqrt{\cos^{2}{t}} = \cos{t}$$

Then

$$\int{\frac{\sqrt{1 - x^2}}{x^4}}\ dx = \int{\frac{\cos{t}}{\sin^{4}{t}}\cos{t}}\ dt = \int{\frac{\cos^{2}{t}}{\sin^{4}{t}}}\ dt$$

Let $u = \tan{t}$. Then $t = \tan^{-1}{u}$ and

$$dt = \frac{1}{1 + u^2}\ du$$

$$\sin{t} = \frac{u}{\sqrt{1 + u^2}}$$

$$\cos{t} = \frac{1}{\sqrt{1 + u^2}}$$

Thus,

$$\int{\frac{\cos^{2}{t}}{\sin^{4}[t}}\ dt = \int{\frac{\frac{1}{1+u^2}}{\frac{u^4}{(1 + u^2)^{2}}}\frac{1}{1 + u^2}}\ du$$

$$= \int{\frac{1}{u^{4}}}\ du = \int{u^{-4}}\ du$$

$$= -\frac{1}{3}u^{-3} + c = -\frac{1}{3}\frac{1}{\tan^{3}{t}} + c$$

$$= -\frac{1}{3}(\frac{\cos{t}}{\sin{t}})^{3} + c = -\frac{(1 - x^2)^{3/2}}{3x^3} + c$$

(e) Let $t = \sqrt{x^2 - a^2}$と. Then $t^2 = x^2 - a^2$ and $2tdt = 2xdx$. Now consider the triangle with the angle $t$ and whose hypotenuse is $x$ and the bottom is $a$. Then we have the opposite side is $\sqrt{x^2 - a^2}$．Then $x = a\sec{t}$ implies $dx = a\sec{t}\tan{t}dt$ and

$$\sqrt{x^2 - a^2} = \sqrt{a^{2}\sec^{2}{t} - a^2} = \sqrt{a^{2}\tan^{2}{t}} = a\tan{t}$$

Thus,

$$\int{\frac{\sqrt{dx}}{x^2 \sqrt{x^2 - 4}}} = \int{\frac{a\sec{t}\tan{t}}{a^{2}\sec^{2}{t}a\tan{t}}}\ dt = \frac{1}{a^{2}}\int{\frac{1}{\sec{t}}}\ dt$$

$$= \frac{1}{a^2}\int{\cos{t}}\ dt = \frac{1}{a^2}\sin{t} + c$$

$$= \frac{\sqrt{x^2 - a^2}}{a^{2}x} + c$$

(f) Let $t = \sqrt{4 + e^{2x}}$. Then $t^2 = 4 + e^{2x}$ and $2tdt = 2e^{2x}dx$. Consider the triangle with the angle $t$ and whose opposite side is $e^{x}$ and the bottom is 2. Then the hypotenuse is $\sqrt{4 + e^{2x}}$．Now $e^{x} = 2\tan{t}$ and $e^{x}dx = 2\sec^{2}{t}dt$．Also, ，

$$\sqrt{4 + e^{2x}} = \sqrt{4 + (2\tan{t})^{2}} = \sqrt{4\sec^{2}{t}} = 2\sec{t}$$

Then

$$\int{\frac{\sqrt{dx}}{e^{x}\sqrt{4 + e^{2x}}}} = \int{\frac{1}{2\tan{t}2\sec{t}}\frac{2\sec^{2}{t}}{2\sec{t}}}\ dt$$

$$= \frac{1}{4}\int{\frac{\sec{t}}{\tan^{2}{t}}}\ dt = \frac{1}{4}\int{\frac{\frac{1}{\cos{t}}}{\frac{\sin^{2}{t}}{\cos^{2}{t}}}}\ dt$$

$$= \frac{1}{4}\int{\frac{\cos{t}}{\sin^{2}{t}}}\ dt = \frac{1}{4}\int{\frac{du}{u^{2}}}$$

$$= -\frac{1}{4u} + c = -\frac{1}{4\sin{t}} + c = -\frac{\sqrt{4 + e^{2x}}}{4e^{x}} + c$$

(g) Since

$$x^2 - 2x - 3 = (x-1)^2 - 4 = (x-1)^2 - 2^2$$

consider the triangle with the angle $t$ and whose hypotenuse is $x-1$ and the bottom is $2$. Then the opposite side is $\sqrt{(x-1)^2 - 4}$. Now $x-1 = 2\sec{t}$ implies $dx = 2\sec{t}\tan{t}dt$ and

$$\sqrt{(x-1)^2 - 4} = \sqrt{4\sec^{2}{t} - 4} = \sqrt{4\tan^{2}{t}} = 2\tan{t}$$

Thus,

$$\int{\frac{dx}{\sqrt{x^2 - 2x - 3}}} = \int{\frac{2\sec{t}\tan{t}}{2\tan{t}}}\ dt$$

$$= \int{\sec{t}}\ dt = \log\vert sec{t} + \tan{t}\vert + c$$

$$= \log\vert\frac{x-1}{2} + \frac{sqrt{(x-1)^2 - 4}}{2}\vert + c$$

We have used the formula such as $\int{\sec{t}\ dt} = \log\vert sec{t} + \tan{t}\vert + c$. Without the formula, we can integrate the following way.

$$\int{\sec{t}}\ dt = \int{\frac{1}{\cos{t}}}\ dt = \int{\frac{\cos{t}}{\cos^{2}{t}}}\ dt$$

$$= \int{\frac{\cos{t}}{1 - \sin^{2}{t}}}\ dt$$

$$= \int{\frac{du}{1 - u^{2}}}\ du = \frac{1}{2}\log\vert\frac{1+u}{1-u}\vert + c$$

$$= \frac{1}{2}\log\vert\frac{1+\cos{t}}{1 - \cos{t}}\vert + c$$

(h) Completing the square,

$$6x - x^2 = 9 - 9 + 6x - x^2 = 9 - (9 - 6x + x^2) = 3^2 - (3-x)^2$$

From this, we have the right triangle with the angle $t$ and whose hypotenuse is $3$，the bottom is $\sqrt{3^2 - (3-x)^2}$，the height is $3 - x$. Then $3 - x = 3\sin{t}$ implies $-dx = 3\cos{t}dt$ and

$$\sqrt{3^2 - (3-x)^2} = \sqrt{9 - 9\sin^{2}{t}} = \sqrt{9\cos^{2}{t}} = 3\cos{t}$$

Thus,

$$\int{\frac{x}{\sqrt{3^2 - (3-x)^2}}}\ dx = \int{\frac{3 - 3\sin{t}}{3\cos{t}} (-3\cos{t})}\ dt$$

$$= -\int{3 - 3\sin{t}}\ dt = -[3t + 3\cos{t}] + c$$

$$= -3\sin^{(\frac{3-x}{3})} - 3\frac{\sqrt{9 - (3-x)^2}}{3} + + c$$

$$= -3\sin^{(\frac{3-x}{3})} - \sqrt{6x - x^2} + c$$

(i) Completing the square, we have

$$x^2 - 2x - 3 = (x-1)^2 -2^2$$

Then consider the right triangle with the angle $t$ and whose hypotenuse is $x-1$，the bottom is $2$，the height is $\sqrt{(x-1)^2 - 4}$. Then $x-1 = 2\sec{t}$implies $dx = 2\sec{t}\tan{t}dt$ and

$$\sqrt{(x-1)^2 - 4} = \sqrt{4\sec^{2}{t} - 4} = \sqrt{4\tan^{t}} = 2\tan{t}$$

より

$$\int{\frac{x}{\sqrt{x^2 - 2x - 3}}}\ dx = \int{\frac{2\sec{t} + 1}{2\tan{t}} (2\sec{t}\tan{t})}\ dt$$

$$= \int{(2\sec^{2}{t} + \sec{t})}\ dt$$

$$= 2\tan{t} + \log\vert\sec{t} + \tan{t}\vert + c$$

$$= 2(\frac{\sqrt{(x-1)^{2} - 4}}{2}) + \log\vert\frac{x-1}{2} + \frac{\sqrt{(x-1)^2 - 4}}{2} + c$$

$$= \sqrt{x^2 - 2x - 3} + \log\vert\frac{x-1 + \sqrt{(x-1)^2 - 4}}{2}\vert + c$$

(j) Completing the square.

$$6x - x^2 - 8 = 9 - 9 + 6x - x^2 - 8 = 1 - (9 - 6x + x^2) = 1 - (3-x)^2$$

Then consider the triangle with the angle $t$ and whose hypotenuse is $1$，the bottom is $\sqrt{1 - (3-x)^2}$，and the height is $3 - x$. Then $3-x = \sin{t}$ implies $-dx = \cos{t}dt$ and，

$$\sqrt{1 - (3-x)^2} = \sqrt{1 - \sin^{2}{t}} = \sqrt{\cos^{2}{t}} = \cos{t}$$

より

$$\int{\sqrt{6x - x^2 - 8}}\ dx = \int{\cos{t}(-\cos{t})}\ dt$$

$$= -\int{\cos^{2}{t}}\ dt$$

$$= -\int{\frac{1 + \cos{t}}{2}}\ dt = -[\frac{t}{2} + \frac{\sin{2t}}{4}] + c$$

$$= -[\frac{t}{2} + \frac{\sin{t}\cos{t}}{2}] + c$$

$$= -\frac{\sin^{-1}{(3-x)}}{2} - \frac{(3-x)\sqrt{1 - (3-x)^2}}{2} + c$$

(k) Completing the square.

$$x^2 + 6x = (x+3)^2 - 9$$

Then consider the right triangle with the angle $t$ and whose hypotenuse is $x+3$，the bottom is $3$，and the height is $\sqrt{(x+3)^2 - 9}$. Then $x+3 = 3\sec{t}$ implies $dx = 3\sec{t}\tan{t}dt$ and

$$\sqrt{(x+3)^2 - 9} = \sqrt{9\sec^{2}{t} - 9} = \sqrt{9\tan^{2}{t}} = 3\tan{t}$$

Thus,

$$\int{x\sqrt{x^2 + 6x}}\ dx = \int{(3\sec{t} - 3)3\tan{t}3\sec{t}\tan{t}}\ dt$$

$$= 27\int{(\sec^{2}{t}\tan^{2}{t} - \sec{t}\tan^{2}{t}}\ dt$$

$$= 27\int{(\frac{\sin^{2}{t}}{\cos^{4}{t}} - \frac{\sin^{2}{t}}{\cos^{3}{t}})}\ dt$$

Now integrate $int{\frac{\sin^{2}{t}}{\cos^{4}{t}}}\ dt$． In this case, the degree of the numerator and the denominator is even. So, we let $u = \tan{t}$. Then $t = \tan^{-1}{u}$ and

$$dt = \frac{1}{1 + u^2}\ du$$

$$\sin{t} = \frac{u}{\sqrt{1 + u^2}}$$

$$\cos{t} = \frac{1}{\sqrt{1 + u^2}}$$

Thus,

$$\int{\frac{\sin^{2}{t}}{\cos^{4}[t}}\ dt = \int{\frac{\frac{u^2}{1+u^2}}{\frac{1}{(1 + u^2)^{2}}}\frac{1}{1 + u^2}}\ du$$

$$= \int{u^2}\ du = \frac{u^3}{3} + c$$

$$= \frac{\tan^{3}{t}}{3} + c$$

Next we integrate $int{\frac{\sin^{2}{t}}{\cos^{3}{t}}}\ dt$．In this case the degree of the denominator is odd. Then we multiply $\cos{t}$ to both the numerator and the denominator. Then we have

$$\int{\frac{\sin^{2}{t}}{\cos^{3}{t}}}\ dt = \int{\frac{\sin^{2}{t}\cos{t}}{\cos^{4}{t}}}\ dt$$

$$= \int{\frac{\sin^{2}{t}\cos{t}}{(1 - \sin^{2}{t})^{2}}}\ dt$$

Let $u = \sin{t}$, Then $du = \cos{t}\ dt$ and

$$\int{\frac{\sin^{2}{t}\cos{t}}{(1 - \sin^{2}{t})^{2}}}\ dt = \int{\frac{u^2}{(1 - u^2)^{2}}}\ du = \int{\frac{u^2}{(1+u)^2 (1 -u)^2}}\ du$$

To integrate this, we use the partial fraction expansion.．

$$\frac{u^2}{(1+u)^2 (1 -u)^2} = \frac{A}{1+u} + \frac{B}{(1+u)^{2}} + \frac{C}{1-u} + \frac{D}{(1-u)^{2}}$$

Clear the denominator,

$$u^2 = A(1+u)(1-u)^2 + B(1-u)^2 + C(1+u)^{2}(1-u) + D(1+u)^2$$

Set $u=1$. Then

$$1 = 4D \Rightarrow D = \frac{1}{4}$$

Set $u=-1$. Then

$$1 = 4B \Rightarrow B = \frac{1}{4}$$

Match the coefficients of $t^3$. Then

$$0 = A - C$$

Match the coefficients of $t^0$. Then

$$0 = A + B + C + D \Rightarrow A + C = \frac{-1}{2}$$

Thus,

$$A = \frac{-1}{4}, \ C = \frac{-1}{4}．$$

Hence,

$$\int\frac{u^2}{(1 - u^2)^2}\ du = \int{\frac{-1/4}{1+u}}\ du + \int{\frac{1/4}{(1+u)^2}}\ du$$

$$- \int{\frac{1/4}{1-u}}\ du + \int{\frac{1/4}{(1-u)^2}}\ du$$

$$= -\frac{1}{4}\log\vert 1+u\vert - \frac{1}{4}(\frac{1}{1+u}) + \frac{1}{4}(\log\vert 1-u\vert) + \frac{1}{4}(\frac{1}{1-u}) + c$$

$$= \frac{1}{4}\log\vert\frac{1-u}{1+u}\vert - \frac{1}{4}(\frac{1}{1+u} - \frac{1}{1-u}) + c$$

$$= \frac{1}{4}\log\vert\frac{1-u}{1+u}\vert + \frac{1}{4}(\frac{2u}{1-u^2}) + c$$

$$= \frac{1}{4}\log\vert\frac{1-\sin{t}}{1+\sin{t}}\vert + \frac{1}{4}(\frac{2\sin{t}}{1-\sin^{2}{t}}) + c$$

Therefore，

$$\int{x\sqrt{x^2 + 6x}}\ dx = 27\int{\frac{\sin^{2}{t}}{\cos^{4}{t}} - \frac{\sin^{2}{t}}{\cos^{3}{t}}}\ dt$$

$$= 27[\frac{\tan^{3}{t}}{3} + \frac{1}{4}\log\vert\frac{1-\sin{t}}{1+\sin{t}}\vert + \frac{1}{4}(\frac{2\sin{t}}{1-\sin^{2}{t}}) ]+ c$$

$$= 27[\frac{1}{3}(\frac{\sqrt{x^2 + 6x}}{3})^{3} + \frac{1}{4}\log\v... ...(\frac{2\frac{\sqrt{x^2 + 6x}}{x+3}}{1-(\frac{\sqrt{x^2 + 6x}}{x+3})^{2}})] + c$$