3.5 Answer

3.5

1.

(a)

$$\int{\sin^{3}{x}} \ dx = \int{\sin^{2}{x} \sin{x}\ dx} = \int{(1 - \cos^{2}{x})\sin{x} \ dx}$$

$$= \int{(1 - t^2)(-dt)} \ \left(\begin{array}{l} t = \cos{x} \\ dt = -\sin{x}dx \end{array}\right)$$

$$= -(t - \frac{t^3}{3}) + c$$

$$= -\cos{x} + \frac{\cos^{3}{x}}{3} + C$$

(b)

$$\int{\sin^{2}{x}} \ dx = \int{\frac{1 - \cos{6x}}{2}}\ dx = \frac{x}{2} - \frac{\sin{6x}}{12} + c$$

(c)

$$\int{\sin^{3}{x}\cos^{2}{x}} \ dx = \int{\sin^{2}{x}\cos^{2}{x}\sin{x}}\ dx$$

$$= \int{(1 - \cos^{2}{x})\cos^{2}{x}\sin{x}}\ dx$$

$$= \int{(1 - t^2)t^2}(-dt) \ \left(\begin{array}{l} t = \cos{x} \\ dt = -\sin{x}dx \end{array}\right)$$

$$= \frac{t^5}{5} - \frac{t^3}{3} + c = \frac{\cos^{5}{x}}{5} - \frac{\cos^{3}{x}}{3} + c$$

(d) Note that $\sin{\alpha}\cos{\beta} = \frac{1}{2}(\sin{(\alpha + \beta)} - \sin{(\alpha- \beta)})$．

$$\int{\cos{3x}\sin{2x}} \ dx = \frac{1}{2}\int{\sin{5x} - \sin{x}}\ dx$$

$$= \frac{1}{2}[\frac{-\cos{5x}}{5} + \cos{x}] + c$$

(e)

$$\int{\sin^{5}{x}} \ dx = \int{\sin^{4}{x}\sin{x}}\ dx$$

$$= \int{(1 - \cos^{2}{x})^2 \sin{x}}\ dx$$

$$= \int{(1 - t^2)^2} (-dt) \ \left(\begin{array}{l} t = \cos{x} \\ dt = -\sin{t}dt \end{array}\right)$$

$$= -\int{1 -2t^2 + t^4}\ dt = -(t - \frac{2t^3}{3} + \frac{t^5}{5}) + c$$

$$= -\cos{x} + \frac{2\cos^{3}{x}}{3} - \frac{\cos^{5}{x}}{5} + c$$

(f)

$$\int{\sec^{2}{\pi x}} \ dx = \frac{1}{\pi}\int\sec^{2}{t}\ dt \ (t = \pi t, dt = \pi dt)$$

$$= \frac{1}{\pi}\tan{t} + c = \frac{1}{\pi}\tan{\pi x} + c$$

(g)

$$\int{\tan^{3}{ x}} \ dx = \int{\frac{\sin^{3}{x}}{\cos^{3}{x}}}\ dx$$

$$= \int{\frac{\sin^{2}{x}\sin{x}}{\cos^{3}{x}}}\ dx = \int{\frac{(1 - \cos^{2}{x})\sin{x}}{\cos^{3}{x}}}\ dx$$

$$= \int{\frac{(1 - t^2)}{t^3}}(-dt) \ (t = \cos{x}, dt = -\sin{x}\ dx)$$

$$= \int{(\frac{1}{t} - t^{-3})}\ dt = \log\vert t\vert + \frac{t^{-2}}{2} + c$$

$$= \log\vert\cos{x}\vert + \frac{1}{2}\sec^{2}{x} + c$$

(h)

$$I = \int{\tan^{2}{x}\sec^{2}{x}} \ dx \ \left\{\begin{array}{ll} u = ... ...ec^{2}{x}\\ du = \sec^{2}{x}dx & v = \frac{1}{2}\tan^{2}{x} \end{array}\right.$$

$$= \frac{1}{2}\tan^{3}{x} - \frac{1}{2}\int{\tan^{2}{x}\sec^{2}{x}}\ dx$$

$$= \frac{1}{2}\tan^{3}{x} - \frac{1}{2}I$$

Thus $I$ is given by

$$I = \int{\tan^{2}{x}\sec^{2}{x}} \ dx = \frac{1}{3}\tan^{3}{x} + c$$

(i)

$$\int{\tan^{3}{x}\sec^{3}{x}} \ dx = \int{\tan^{2}{x}\sec^{2}{x}\sec{x}\tan{x}} \ dx$$

$$= \int{(sec^{2}{x} -1)\sec^{2}{x}\sec{x}\tan{x}} \ dx$$

$$= \int{(t^2 - 1)t^2} \ dt \ (t = \sec{x}, \ dt = \sec{x}\tan{x}dx)$$

$$= \frac{t^5}{5} - \frac{t^3}{3} + c = \frac{\sec^{5}{x}}{5} - \frac{\sec^{3}{x}}{3} + c$$

Alternate solution

$$\int{\tan^{3}{x}\sec^{3}{x}} \ dx = \int{\frac{\sin^{3}{x}}{\cos^{6}{x}}} \ dx$$

$$= \int{\frac{\sin^{2}{x}}{\cos^{6}{x}}\sin{x}} \ dx = \int{\frac{(1 - \cos^{2}{x})\sin{x}}{\cos^{6}{x}}}\ dx$$

$$= \int{\frac{1 - t^2}{t^6}} \ (-dt) \ (t = \cos{x}, \ dt = -\sin{x}dx)$$

$$= -\int{(\frac{1}{t^{-6}} - \frac{1}{t^{-4}}}\ dt = \frac{t^{-5}}{5} - \frac{t^{-3}}{3} + c$$

$$= \frac{\sec^{5}{x}}{5} - \frac{\sec^{3}{x}}{3} + c$$

(j)

$$I = \int{\sec^{5}{x}} \ dx = \int{\sec^{3}{x}\sec^{2}{x}}\ dx \ \left... ...dv = \sec^{2}{x}\\ du = 3\sec^{3}{x}\tan{x}dx & v = \tan{x} \end{array}\right.$$

$$= \sec^{3}{x}\tan{x} - 3\int{\tan^{2}{x}\sec^{3}{x}}\ dx$$

$$= \sec^{3}{x}\tan{x} - 3\int{(\sec^{2}{x} - 1)\sec^{3}{x}}\ dx$$

$$= \sec^{3}{x}\tan{x} - 3\int{\sec^{5}{x}}dx + 3\int{\sec^{3}{x}}\ dx$$

$$= \sec^{3}{x}\tan{x} - 3I + 3\int{\sec^{3}{x}}\ dx$$

Thus $I$ is given by

$$I = \int{\sec^{5}{x}} \ dx = \frac{1}{4}[\sec^{3}{x}\tan{x} + 3\int{\sec^{3}{x}}\ dx + c$$

Thus, we need to find $\int{\sec^{3}{x}}\ dx$．

$$J = \int{\sec^{3}{x}} \ dx = \int{\sec{x}\sec^{2}{x}}\ dx \ \left\{\b... ...x} & dv = \sec^{2}{x}\\ du = \sec{x}\tan{x}dx & v = \tan{x} \end{array}\right.$$

$$= \sec{x}\tan{x} - \int{\tan^{2}{x}\sec{x}}\ dx$$

$$= \sec{x}\tan{x} - \int{(\sec^{2}{x} - 1)\sec{x}}\ dx$$

$$= \sec{x}\tan{x} - \int{\sec^{3}{x}}dx + \int{\sec{x}}\ dx$$

$$= \sec{x}\tan{x} - J + \log\vert\sec{x}+\tan{x}\vert$$

Thus,

$$J = \frac{1}{2}[\sec{x}\tan{x} + \log\vert\sec{x}+\tan{x}\vert + c$$

Therefore，

$$I = \frac{1}{4}[\sec^{3}{x}\tan{x} + \frac{3}{2}(\sec{x}\tan{x} + \log\vert sec{x} + \tan{x}] + c$$

(k) Let $x = 2\tan^{-1}{t}$. Then $dx = \frac{2}{1 + t^2}dt$, $\cos{x} = \frac{1- t^2}{1 + t^2}$, $\sin{x} = \frac{2t}{1+t^2}$. Thus any trigonometric function can be written using rational functions．

$$\int{\frac{dx}{3 - 2\cos{x}}} = \int{\frac{1}{3 - 2\frac{1-t^2}{1+t^2}}\frac{2}{1+t^2}}\ dt$$

$$= \int{\frac{2}{3 + 3t^2 - 2 + 2t^2}}\ dt = \int{\frac{2}{5t^2 + 1}}\ dt$$

$$= \frac{2}{5}\int{\frac{1}{t^2 + \frac{1}{5}}}\ dt = \frac{2}{5}\sqrt{5}\tan^{-1}{(\sqrt{5}t)} + c$$

$$= \frac{2\sqrt{5}}{5}\tan^{-1}(\sqrt{5}\tan{\frac{x}{2}}) + c$$

(l) Let $x = 2\tan^{-1}{t}$. Then $dx = \frac{2}{1 + t^2}dt$, $\cos{x} = \frac{1- t^2}{1 + t^2}$, $\sin{x} = \frac{2t}{1+t^2}$

$$\int{\frac{\sin{x}}{2 - \sin{x}}}\ dx = \int{\frac{\frac{2t}{1+t^2}}{2 - \frac{2t}{1+t^2}}\cdot\frac{2}{1+t^2}}\ dt$$

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

Now using partial fraction expansion on $\frac{2t}{(1 +t^2)(t^2 - t + 1)}$.

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

Clear the denominator. Then

$$2t = (At+B)(t^2 - t+1) + (t^2+1)(Ct +D)$$

$$= (A + C)t^3 +(-A + B+D)t^2 +(A -B+C)t + B + D$$

Now matching the degee. Then we have

$$\left\{\begin{array}{l} A+C = \\ -A+B+D = 0\\ A -B+C = 2\\ B+D = 0 \end{array}\right.$$

Then we have $A = 0$ which implies that $c =0, B = -2, D = 2$．Therefore，

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

First we find $\int{\frac{-2}{t^2 + 1}} dt$．

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

$$= -2 \tan^{-1}{(t)} + c$$

Next we find $\int{\frac{2}{t^2 -t +1}}\ dt$．

$$\int{\frac{2}{t^2 -t +1}}\ dt = \int{\frac{2}{(t - \frac{1}{2})^2 + \frac{3}{4}}}\ dt$$

$$= \int{\frac{2}{u^2 + \frac{3}{4}}} \ du \ \left(\begin{array}{l} u = t - \frac{1}{2} \\ du = dt \end{array}\right)$$

$$= 2\frac{2}{\sqrt{3}}\tan^{-1}{u} = \frac{4}{\sqrt{3}}\tan^{-1}{(t - \frac{1}{2})} + c$$

Now substitute $t = \tan{(\frac{x}{2})}$. Then

$$\int{\frac{\sin{x}}{2 - \sin{x}}}\ dx = - 2\tan^{-1}{(t)} + +\frac{4}{\sqrt{3}} \tan^{-1}{(t - \frac{1}{2})} + c$$

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

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

(m) Let $x = 2\tan^{-1}{t}$. Then $dx = \frac{2}{1 + t^2}dt$, $\cos{x} = \frac{1- t^2}{1 + t^2}$, $\sin{x} = \frac{2t}{1+t^2}$.

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

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

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

$$= t + \log{\vert 1 + t^2\vert} + c = \tan(\frac{x}{2}) + \log{\vert 1 + \tan^{2}{(\frac{x}{2})}\vert} + c$$

(n) You might set $x = 2\tan^{-1}{t}$. But the all terms are square. So we let $x = \tan^{-1}{t}$. Then $dx = \frac{1}{1 + t^2}dt$, $\cos^{2}{x} = \frac{1}{1 + t^2}$, $\sin{x} = \frac{t^2}{1+t^2}$．

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

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

Now we use the partial fraction expansion.．

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

Clear the denominator.

$$t^2 = A(t-1)(1+t^2) + B(t+1)(1+t^2) + (Ct+D)(t+1)(t-1)$$

Here, we set $t = 1$. Then

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

Set $t = -1$. Then

$$1 = -4A より A = - \frac{1}{4}$$

Now compare the coefficients of $t^3$, we have $C = 0$. Finally, we compare the coefficients of $t^2$. Then

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

Thus,

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

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

Note that $t = \tan{x}$. Then

$$\int{\frac{1 + \sin{x}}{1 + \cos{x}}}\ dx = -\frac{1}{4}\log{\vert\tan{x} + 1\vert} + \frac{1}{4}\log{\vert\tan{x} -1\vert} + \frac{1}{2}\tan^{-1}{(\tan{x})} + c$$

$$= \frac{1}{4}\log{\vert\frac{\tan{x} - 1}{\tan{x} + 1}\vert} + \frac{x}{2} + c$$

(o) Let $x = 2\tan^{-1}{t}$. Then $dx = \frac{2}{1 + t^2}dt$, $\cos{x} = \frac{1- t^2}{1 + t^2}$, $\sin{x} = \frac{2t}{1+t^2}$.

$$\int{\frac{1}{1 + \tan{x}}}\ dx = \int{\frac{1}{1 + \frac{\sin{x}}{\cos{x}}}}\ dx = \int{\frac{\cos{x}}{\cos{x} + \sin{x}}}\ dx$$

$$= \int{\frac{\frac{1 - t^2}{1+ t^2}}{\frac{1 - t^2}{1+t^2} + \frac{... ...rac{2}{1+t^2}}\ dt = \int{\frac{1 - t^2}{1 - t^2 + 2t}\cdot\frac{2}{1+t^2}}\ dt$$

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

Using the partial fraction expansion, we have．

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

Clear the denominator.

$$-2 + 2t^2 = (At +B)(1+t^2) + (Ct +D)(t^2 - 2t -1)$$

$$= At^3 + Bt^2 + At + B + Ct^3 -(C - D)t -(C + 2D)$$

Now we use the coefficient matching.

Coefficieints of $t^3$

$$A + C = 0$$

Coefficients of $t^2$

$$B- 2C +D = 2$$

Coefficients of $t$

$$A-C-2D = 0$$

Constant term

$$B - D = 0$$

Then we have

$$A = 1, B = 1, C= -1, D = 1$$

Thus,，

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

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

Now let $u = t-1$. Then $du = dt$. Thus,

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

$$= \int{\frac{A}{u+ \sqrt{2}} + \frac{B}{u - \sqrt{2}}}\ du$$

$$= \frac{1 - \sqrt{2}}{2}\int{\frac{1}{u + \sqrt{2}}}\ du + \frac{1 + \sqrt{2}}{2}\int{\frac{1}{u - \sqrt{2}}}\ du$$

$$= \frac{1 - \sqrt{2}}{2}\log{\vert u + \sqrt{2}\vert} + \frac{1 + \sqrt{2}}{2}\log{\vert u - \sqrt{2}\vert}$$

$$= \frac{1}{2}\log{\vert u^2 - 2\vert} - \frac{\sqrt{2}}{2}\log{\vert\frac{u + \sqrt{2}}{u - \sqrt{2}}\vert}$$

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

Then

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

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

Finally, substitute $t = \tan{(\frac{x}{2})}$ and we are done.

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

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

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