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Indefinite integrals Quiz 2

mathematics

 What is the coefficient  of ${{x}^{5}}$ after twice integration of  $\dfrac{\sqrt[3]{5}}{2}{{x}^{3}}+{{x}^{2}}+1$.

 To find the coefficient we need to integrate the function twice and after that look for the coefficient of ${{x}^{5}}$.

  $ f(x)=\dfrac{\sqrt[3]{5}}{2}{{x}^{3}}+{{x}^{2}}+1 \\ \text{Integrating with respect to x:} \\  I=\int{\left(\dfrac{\sqrt[3]{5}}{2}{{x}^{3}}+{{x}^{2}}+1\right)dx} \\ I=\dfrac{\sqrt[3]{5}}{2\times 4}{{x}^{4}}+\dfrac{{{x}^{3}}}{3}+x \\ \text{Integrating with respect to x:} \\  I'=\dfrac{\sqrt[3]{5}}{2\times 4\times 5}{{x}^{5}}+\dfrac{{{x}^{4}}}{3\times 4}+\dfrac{{{x}^{2}}}{2} \\ \text{Therefore coefficient of }{x}^{5}=\dfrac{\sqrt[3]{5}}{40} $ 

 Add the integrals  $\int{{{x}^{4}}+{{x}^{2}}+\sqrt{x}+8}$ and $\int{2{{x}^{4}}+2x+4\sqrt{x}}$

 First integrate the function 1 and then integrate the function 2  and then add both the values obtained after Integration.

  $ {{I}_{1}}=\int{{{x}^{4}}+{{x}^{2}}+\sqrt{x}+8} \\ \Rightarrow \dfrac{{{x}^{5}}}{5}+\dfrac{{{x}^{3}}}{3}+\dfrac{2{{x}^{\frac{3}{2}}}}{3}+8x \\ {{I}_{2}}=\int{2{{x}^{4}}+2x+4\sqrt{x}} \\ \Rightarrow \dfrac{2}{5}{{x}^{5}}+{{x}^{2}}+\dfrac{8}{3}{{x}^{\frac{3}{2}}} \\ {{I}_{1}}+{{I}_{2}}=\dfrac{3{{x}^{5}}}{5}+\dfrac{{{x}^{3}}}{3}+\dfrac{10{{x}^{\frac{3}{2}}}}{3}+{{x}^{2}}+8x+C $ 

 Find  $\int{\dfrac{2\sin 2x}{3\cos x\tan x}+\cos 2x}$

 Integrate the terms by taking 2 terms and integrating them separately and then simplify the terms and using the basic principles of integration find its integral.

  $ I=\int{\dfrac{2\sin 2x}{3\cos x\tan x}+\cos 2x} \\ \Rightarrow \int{\left( \dfrac{4\sin x\cos x\cos x}{3\cos x\sin x}+\cos 2x \right)}dx \\ \Rightarrow \int{\left(\dfrac{4}{3} \cos x+\cos 2x \right)}dx \\ \Rightarrow \dfrac{4}{3}\sin x+\dfrac{\sin 2x}{2}+c $ 

 If  ${{I}_{1}}=\int{4{{x}^{3}}+2x}$  and  ${{I}_{2}}=\int{-3{{x}^{2}}+2x}$  .Find  $\dfrac{{{I}_{1}}+{{I}_{2}}}{{{I}_{1}}-{{I}_{2}}}$.

 First find the integral of equation 1 and then for equation 2 and after that place the values of it in the equation given and compute the value.

  $ {{I}_{1}}=\int{4{{x}^{3}}+2x} \\ {{I}_{1}}=\int{\left( 4{{x}^{3}}+2x \right)dx} \\ \Rightarrow {{x}^{4}}+{{x}^{2}} \\ {{I}_{2}}=\int{-3{{x}^{2}}+2x} \\ {{I}_{2}}=\int{\left( -3{{x}^{2}}+2x \right)dx} \\ \Rightarrow -{{x}^{3}}+{{x}^{2}} \\ \dfrac{{{I}_{1}}+{{I}_{2}}}{{{I}_{1}}-{{I}_{2}}}=\dfrac{{{x}^{4}}+{{x}^{2}}-{{x}^{3}}+{{x}^{2}}}{{{x}^{4}}+{{x}^{2}}+{{x}^{3}}-{{x}^{2}}} \\ \Rightarrow \dfrac{{{x}^{4}}+2{{x}^{2}}-{{x}^{3}}}{{{x}^{4}}+{{x}^{3}}}+c $ 

 Apply integration by parts and then solve the equation to get the value of the integral as mentioned.

  $ I=\int{{{e}^{-x}}\sin 2xdx} \\ \text{Use integration by parts:} \\ u=e^{-x}, v’=sin(2x) \\ I=-\dfrac{1}{2}e^{-x}\cos \left(2x\right)-\int \dfrac{1}{2}e^{-x}\cos \left(2x\right)dx \\ \text{Again apply integration by parts:} \\ u=e^{-x}, v’=cos(2x) \\ I=-\dfrac{1}{2}e^{-x}\cos \left(2x\right)-\dfrac{1}{2}\left(\dfrac{1}{2}e^{-x}\sin \left(2x\right)-\left(-\dfrac{1}{2}\cdot \int \:e^{-x}\sin \left(2x\right)dx\right)\right)\\I=\dfrac{-{{e}^{-x}}\left( \sin 2x+2\cos 2x \right)}{5}+C $ 