8.7: Additional exercises
- Page ID
- 149521
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Exercises \(\PageIndex{}\)
\(\displaystyle\int(t+4)^3\,dt\)
- Answer
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\(\dfrac{(t+4)^4}{4}+C\)
\(\displaystyle\int t(t^2-9)^{3/2}\,dt\)
- Answer
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\(\dfrac{(t^2-9)^{5/2}}{5}+C\)
\(\displaystyle\int(e^{t^2}+16)te^{t^2}\,dt\)
- Answer
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\(\dfrac{(e^{t^2}+16)^2}{4}+C\)
\(\displaystyle\int\sin t\cos 2t\,dt\)
- Answer
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\(\cos t-\dfrac{2\cos^3 t}{3}+C\)
\(\displaystyle\int \tan t\sec^2t\,dt\)
- Answer
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\(\dfrac{\tan^2 t}{2}+C\)
\(\displaystyle\int {2t+1\over t^2+t+3}\,dt\)
- Answer
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\(\ln\big|t^2+t+3\big|+C\)
\(\displaystyle\int {1\over t(t^2-4)}\,dt\)
- Answer
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\(\dfrac{1}{8}\ln\left(\big|1-\dfrac{4}{t^2}\big|\right)+C\)
\(\displaystyle\int {1\over (25-t^2)^{3/2}}\,dt\)
- Answer
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\(\dfrac{1}{25}\tan\big(\arcsin(t/5)\big)+C=\dfrac{t}{25\sqrt{25-t^2}}+C\)
\(\displaystyle\int {\cos 3t\over\sqrt{\sin3t}}\,dt\)
- Answer
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\(\dfrac{2\sqrt{\sin 3t}}{3}+C\)
\(\displaystyle\int t\sec^2 t\,dt\)
- Answer
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\(t\tan t-\ln\big|\sec t\big|+C\)
\(\displaystyle\int {e^t\over \sqrt{e^t+1}}\,dt\)
- Answer
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\(2\sqrt{e^t+1}+C\)
\(\displaystyle\int \cos^4 t\,dt\)
- Answer
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\(\dfrac{3t}{8}+\dfrac{\sin 2t}{4}+\dfrac{\sin 4t}{32}+C\)
\(\displaystyle\int {1\over t^2+3t}\,dt\)
- Answer
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\(\dfrac{1}{3}\ln\left|\dfrac{t}{t+3}\right|+C\)
\(\displaystyle\int {1\over t^2\sqrt{1+t^2}}\,dt\)
- Answer
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\(\dfrac{-1}{\sin(\arctan t)}+C=-\dfrac{\sqrt{1+t^2}}{t}+C\)
\(\displaystyle\int {\sec^2t\over (1+\tan t)^3}\,dt\)
- Answer
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\(\dfrac{-1}{2(1+\tan t)^2}+C\)
\(\displaystyle\int t^3\sqrt{t^2+1}\,dt\)
- Answer
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\(\dfrac{(t^2+1)^{5/2}}{5}-\dfrac{(t^2+1)^{3/2}}{3}+C=\dfrac{t^2(t^2+1)^{3/2}}{3}-\dfrac{2(t^2+1)^{5/2}}{15}+C\)
\(\displaystyle\int e^t\sin t\,dt\)
- Answer
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\(\dfrac{e^t}{2}(\sin t-\cos t)+C\)
\(\displaystyle\int (t^{3/2}+47)^3\sqrt{t}\,dt\)
- Answer
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\(\dfrac{(t^{3/2}+47)^4}{6}+C\)
\(\displaystyle\int {t^3\over (2-t^2)^{5/2}}\,dt\)
- Answer
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\(\dfrac{2}{3(2-t^2)^{3/2}}-\dfrac{1}{(2-t^2)^{1/2}}+C=\dfrac{t^2-(4/3)}{(2-t^2)^{3/2}}+C\)
\(\displaystyle\int{1\over t(9+4t^2)}\,dt\)
- Answer
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\(\dfrac{\ln\big|\sin\big(\arctan(2t/3)\big)\big|}{9}+C=\dfrac{1}{18}\left(\ln(4t^2)-\ln(9+4t^2)\right)+C\)
\(\displaystyle\int{\arctan 2t\over 1+4t^2}\,dt\)
- Answer
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\(\dfrac{1}{4}\big(\arctan(2t)\big)^2+C\)
\(\displaystyle\int {t\over t^2+2t-3}\,dt\)
- Answer
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\(\dfrac{3\ln\big|t+3\big|}{4}+\dfrac{\ln\big|t-1\big|}{4}+C\)
\(\displaystyle\int \sin^3 t\cos^4 t\,dt\)
- Answer
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\(\dfrac{\cos^7 t}{7}-\dfrac{\cos^5 t}{5}+C\)
\(\displaystyle\int{1\over t^2-6t+9}\,dt\)
- Answer
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\(\dfrac{1}{3-t}+C\)
\(\displaystyle\int{1\over t(\ln t)^2}\,dt\)
- Answer
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\(\dfrac{-1}{\ln t}+C\)
\(\displaystyle\int t(\ln t)^2\,dt\)
- Answer
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\(\dfrac{t^2(\ln t)^2}{2}-\dfrac{t^2\ln t}{2}+\dfrac{t^2}{4}+C\)
\(\displaystyle\int t^3e^{t}\,dt\)
- Answer
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\((t^3-3t^2+6t-6)e^t+C\)
\(\displaystyle\int{t+1\over t^2+t-1}\,dt\)
- Answer
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\(\dfrac{5-\sqrt{5}}{10}\ln\left(t+\dfrac{1+\sqrt{5}}{2}\right)+\dfrac{5+\sqrt{5}}{10}\ln\left(t+\dfrac{1-\sqrt{5}}{2}\right)+C\)