« Section 4.4
Section 4.6 »

Section 4.5

Please post comments if you find any errors, or if you wish to contribute answers for additional problems. You may find information on using \LaTeX here.

1. (a) \displaystyle{\int 3x\sin(x)dx = -3x\cos(x) + 3\sin(x) + c}

(c) \displaystyle{\int 4x\sin(3x)dx = -\frac{4}{3}x\cos(3x) + \frac{4}{9}\sin(3x) + c}

(e) \displaystyle{\int 2x^2\sin(4x)dx = -\frac{1}{2}x^2\cos(4x) + \frac{1}{4}x\sin(4x) + \frac{1}{16}\cos(4x) + c}

(g) \displaystyle{\int 3x^3\sin(2x)dx = -\frac{3}{2}x^3\cos(2x) + \frac{9}{4}x^2\sin(2x) + \frac{9}{4}x\cos(2x)}
\displaystyle{- \frac{9}{8}\sin(2x) + c}

(h) \displaystyle{\int x\sqrt{1+x} \ dx = \frac{2}{3}x(1 + x)^{\frac{3}{2}} - \frac{4}{15}(1 + x)^{\frac{5}{2}} + c}

2. (a) \displaystyle{\int_0^\pi 4x\sin(x)dx = 4\pi}

(c) \displaystyle{\int_0^{\frac{\pi}{3}} 2t\sin(3t)dt = \frac{2\pi}{9}}

(e) \displaystyle{\int_0^{\frac{\pi}{4}} 2x^2\sin(2x)dx = \frac{\pi}{4} -\frac{1}{2}}

3. (a) \displaystyle{\int \sin^2(2x)dx = \frac{x}{2} - \frac{1}{8}\sin(4x) + c}

(c) \displaystyle{\int 5\sin^2(2t)\cos^2(2t)dt = \frac{5}{8}t - \frac{5}{64}\sin(8t) + c}

(e) \displaystyle{\int 6\cos^3(2z)dz = 3\sin(2z) - \sin^3(2z) + c}

(g) \displaystyle{\int \cos^5(2x)dx = \frac{1}{2}\sin(2x) - \frac{1}{3}\sin^3(2x) + \frac{1}{10}\sin^5(2x) + c}

4. (a) \displaystyle{\int_0^\pi \sin^2(x)dx = \frac{\pi}{2}}

(c) \displaystyle{\int_0^{\frac{\pi}{2}} 3\sin^2(z)\cos^2(z)dz = \frac{3\pi}{16}}

(e) \displaystyle{\int_0^\pi \sin^3(3t)dt = \frac{4}{9}}

5. Using wxMaxima:
(a) \displaystyle{\int \cos^6(x)dx = \frac{5}{16}x + \frac{1}{4}\sin(2x) - \frac{1}{48}\sin^3(2x) + \frac{3}{64}\sin(4x) + c}

(c) \displaystyle{\int \sin^4(2t)\cos^4(3t)dt = \frac{9}{64}t - \frac{3}{64}\sin(2t) - \frac{23}{512}\sin(4t)}
\displaystyle{+ \frac{1}{32}\sin(6t) + \frac{1}{512}\sin(8t) - \frac{1}{80}\sin(10t) + \frac{1}{256}\sin(12t)}
\displaystyle{+ \frac{1}{448}\sin(14t) - \frac{1}{512}\sin(16t) + \frac{1}{2560}\sin(20t) + c}

(e) \displaystyle{\int_{-1}^1 \sqrt{1 - x^2} \ dx = \frac{\pi}{2}}

(g) \displaystyle{\int_0^{2\pi} \sin^8(2t)dt = \frac{35\pi}{64}}

6. Using wxMaxima:
(a) \displaystyle{\int_{\pi}^\pi \sin^4(x)dx = \frac{3\pi}{4}}

(c) \displaystyle{\int_0^5 \sin(3x^2)dx = 0.33117} (approximation, rounded to five decimal places)

(e) \displaystyle{\int_1^2 \frac{1}{x} \ dx = \log(2)}

(g) \displaystyle{\int_0^\pi \sqrt{5 - 3\sin^2(t)} \ dt = 5.80675} (approximation, rounded to five decimal places)

7. (a) T = 1.52309 , rounded to fice decimal places.

(c) The approximation is 1.41923 , rounded to five decimal places.

This entry was posted on 23 July 23 2007 at 4:08 pm and is filed under Section 4.5. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

Leave a Reply