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This is the second of our many quizzes on Integration by parts: tabular integration. Click on this to access more such competitive quizzes on Calculus.

Integration by parts: tabular integration Quiz 2

mathematics

 What will be the value $g(x)=\dfrac{10}{3}\sqrt[3]{x}$ in the integral $\int{\dfrac{10}{3}\sqrt[3]{x}\sin 3xdx}$ after integrating it thrice using tabular integration.

 Form a table using tabular integration such that the first column gets differentiated and the second column gets integrated by going down. The third element in the second column is the required answer.

Take a table of $f\left( x \right)=\sin 3x$ and $g\left( x \right)=\dfrac{10}{3}\sqrt[3]{x}$ such that $\\ f\left( x \right)$ is differentiated and $g\left( x \right)$ is integrated. $ \\ \begin{array}{|c|c|} \hline f(x) & g(x) \\ \hline \sin 3 x & \dfrac{10}{3} \sqrt[3]{x} \\ \hline 3 \cos 3 x & \dfrac{5}{2} x^{\frac{4}{3}} \\ \hline-9 \sin 3 x & \dfrac{15}{14} x^{\frac{7}{3}} \\ \hline-27 \cos 3 x & \dfrac{9}{28} x^{\frac{10}{3}} \\ \hline \end{array} \\ $ After integrating thrice there is $\dfrac{9}{28}{{x}^{\frac{10}{3}}}$ in place of $\dfrac{10}{3}\sqrt[3]{x}$

 The given below is tabular integration of some integral. Fill in the missing values. $ \\ \begin{array}{|c|c|} \hline f(x) & g(x) \\ \hline x^{4} & -- \\ \hline 4 x^{3} & -\cos 2 x \\ \hline 12 x^{2} & -\dfrac{1}{2} \sin 2 x \\ \hline 24 x & -- \\ \hline 24 & \dfrac{1}{8} \sin 2 x \\ \hline 0 & -\dfrac{1}{16} \cos 2 x \\ \hline \end{array} $

 The $g\left( x \right)$ in the tabular method gets integrated whereas $f\left( x \right)$ gets differentiated. Find integration or differentiation of required parts to get missing parts.

The second column will get integrated by going down $ \\ $ Since $-\cos 2x$ is the integration of the first part. $ \\ $ So, differentiating $-\cos 2x$ the first missing term is $2\sin 2x \\$. $ \\ $ Now, by integrating $-\dfrac{1}{2}\sin 2x$ the next missing part is, $\\ \int{-\dfrac{1}{2}\sin 2x=\dfrac{1}{4}\cos 2x} \\ $.

 Find the product of the first positive terms in first column while integrating  $\int{\dfrac{1}{{{x}^{2}}}\sin \left( \dfrac{4}{3}x \right)dx}$ using tabular integration method at least once.

 The $g\left( x \right)$ in the tabular method gets integrated whereas $f\left( x \right)$ gets differentiated. Then find the first positive term in the first column and do multiplication diagonally to get the required answer.

 Use tabular integration for given integral $ \\ \begin{array}{|c|c|} \hline f(x) & g(x) \\ \hline \dfrac{1}{x^{2}} & \sin \dfrac{4}{3} x \\ \hline-\dfrac{2}{x^{3}} & -\dfrac{3}{4} \cos \dfrac{4}{3} x \\ \hline \dfrac{6}{x^{4}} & -\dfrac{9}{16} \sin \dfrac{4}{3} x \\ \hline-\dfrac{24}{x^{5}} & \dfrac{27}{64} \cos \dfrac{4}{3} x \\ \hline \end{array} \\ $ The product in tabular integration is done only diagonally. The first positive term in the first column is $\dfrac{6}{{{x}^{4}}}$ $ \\ $ So, the required product is $\\ \Rightarrow P=\dfrac{6}{{{x}^{4}}}\times \dfrac{27}{64}\cos \left( \dfrac{4}{3}x \right)\\ \Rightarrow P=\dfrac{81}{32{{x}^{4}}}\cos \left( \dfrac{4}{3}x \right)$

 What is the value opposite to the constant term in the tabular integration of $\int{\left( {{x}^{4}}+4{{x}^{2}}+5 \right){{e}^{3x+4}}dx}$

 Create a table of tabular integration with respective functions. Then check the value in the table that is opposite to the constant in the first column.

 The tabular integration of $f\left( x \right)={{x}^{4}}+4{{x}^{2}}+5$ and $g\left( x \right)={{e}^{3x+4}}$ is $ \\ \begin{array}{|c|c|} \hline f(x) & g(x) \\ \hline x^{4}+4 x^{2}+5 & e^{3 x+4} \\ \hline 4 x^{3}+8 x & \dfrac{1}{3} e^{3 x+4} \\ \hline 12 x^{2}+8 & \dfrac{1}{9} e^{3 x+4} \\ \hline 24 x & \dfrac{1}{27} e^{3 x+4} \\ \hline 24 & \dfrac{1}{81} e^{3 x+4} \\ \hline \end{array} \\ $ The term opposite to constant $'24'$ is $\dfrac{1}{81}{{e}^{3x+4}}$

 What is the product of constant after tabular integration by parts of  ${{x}^{2}}\left( 1-{{x}^{4}} \right)$ and $\sin x$.

 Develop the tabular integration until the first column becomes $0$ and then find the product of the constant in the first column as a diagonal product of the constant.

 Create a table of tabular integration for given functions as $ \\ \begin{array}{|c|c|} \hline f(x) & g(x) \\ \hline x^{2}-x^{6} & \sin x \\ \hline 2 x-6 x^{5} & -\cos x \\ \hline 4 -30 x^{4} & -\sin x \\ \hline -120 x^{3} & \cos x \\ \hline-360 x^{2} & \sin x \\ \hline-720 x & -\cos x \\ \hline-720 & -\sin x \\ \hline 0 & \cos x \\\hline \end{array} \\ $ The product is always done in diagonally, So the required product of constant is $ \\ \Rightarrow P=\left( -720 \right)\times \cos x \\ \Rightarrow P=-720\cos x $ 