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Properties of integrals Quiz 2

mathematics

 Find the value of $\int_{1}^{a}{\sqrt{1+\dfrac{1}{{{x}^{2}}}}}dx+\ln \left( \sqrt{{{a}^{2}}+1}+1 \right)+\sqrt{2}$

 First substitute $\tan \theta $ in place of $x$ then use the property $\int\limits_{p}^{q}{f\left( a \right)da}=\int\limits_{p}^{q}{f\left( t \right)dt}$.

  $\text{Let, }x=\tan \theta \Rightarrow dx={{\sec }^{2}}\theta d\theta \\  \text{Using the property, } \int\limits_{p}^{q}{f\left( a \right)da}=\int\limits_{p}^{q}{f\left( t \right)dt}. \\  \text{Now, } \\   \int{\sqrt{1+\dfrac{1}{{{x}^{2}}}}}dx=\int{\dfrac{\sec \theta }{\tan \theta }{{\sec }^{2}}\theta }d\theta  \\   =\int{\dfrac{\sec \theta (1+{{\tan }^{2}}\theta )}{\tan \theta }}d\theta =\int{\left( cosec \theta +\sec \theta \tan \theta  \right)}d\theta  \\   =-\ln (cosec \theta +\cot \theta )+\sec \theta +c=-\ln \left( \sqrt{{{x}^{2}}+\dfrac{1}{x}}+\dfrac{1}{x} \right)+\sqrt{{{x}^{2}}+1}+c \\   =-\ln \left( \sqrt{{{a}^{2}}+1}+1 \right)+\ln a+\sqrt{{{a}^{2}}+1}+\ln \left( \sqrt{2}+1 \right)-\sqrt{2}$ 

 Which of the following can be a possible value of $f\left( x \right)$ if $\int{f(x)dx=1+\dfrac{{{x}^{2}}}{2!}}-\dfrac{3{{x}^{4}}}{4!}+\dfrac{5{{x}^{6}}}{6!}-...$  

 First find the value of integration by using By-parts formula, $\int{udv=uv+\int{vdu}}$, then put the values for the trigonometric functions.  

  $ \text{Integrating option C by parts, } \\  \int{x\cos xdx=x\int{\cos xdx-\int{1\int{\cos xdx=x\sin x+\cos x}}}} \\   \text{Now, }\sin x=x-\dfrac{{{x}^{3}}}{3!}+\dfrac{{{x}^{5}}}{5!}-...\text{ and }\cos x=1-\dfrac{{{x}^{2}}}{2!}+\dfrac{{{x}^{4}}}{4!}-... \\   \text{Putting these values above,} \\  x\sin x+\cos x=x\left( x-\dfrac{{{x}^{3}}}{3!}+\dfrac{{{x}^{5}}}{5!}-... \right)+1\left( 1-\dfrac{{{x}^{2}}}{2!}+\dfrac{{{x}^{4}}}{4!}-... \right) \\   \Rightarrow 1+\dfrac{{{x}^{2}}}{2!}-\dfrac{3{{x}^{4}}}{4!}+\dfrac{5{{x}^{6}}}{6!}-... $  

 Find the value of I, such that $I=\int_{0}^{\pi }{\dfrac{{{\sin }^{8}}x-{{\cos }^{8}}x}{{{\sin }^{4}}x-{{\cos }^{4}}x}}dx$ 

 First divide the integration into intervals using the property $\int\limits_{p}^{q}{f(a)d(a) = \int\limits_{p}^{r}{f(a)d(a)+\int\limits_{r}^{q}{f(a)d(a)}}}$ to get separate values then combines them at last. 

 $\text{The denominator is zero when } x=\dfrac{\pi }{4}\text{ and }x=\dfrac{3\pi }{4},\text{ so} \\   I=\int_{0}^{\frac{\pi }{4}}{\dfrac{{{\sin }^{8}}x-{{\cos }^{8}}x}{{{\sin }^{4}}x-{{\cos }^{4}}x}}dx+\int_{\frac{\pi }{4}}^{\frac{3\pi }{4}}{\dfrac{{{\sin }^{8}}x-{{\cos }^{8}}x}{{{\sin }^{4}}x-{{\cos }^{4}}x}}dx+\int_{\frac{3\pi }{4}}^{\pi }{\dfrac{{{\sin }^{8}}x-{{\cos }^{8}}x}{{{\sin }^{4}}x-{{\cos }^{4}}x}}dx \\   \text{Now, }\int_{0}^{\frac{\pi }{4}}{\dfrac{{{\sin }^{8}}x-{{\cos }^{8}}x}{{{\sin }^{4}}x-{{\cos }^{4}}x}}dx=-\dfrac{1}{16}+\dfrac{6\pi +2}{32} \\   \int_{\frac{\pi }{4}}^{\frac{3\pi }{4}}{\dfrac{{{\sin }^{8}}x-{{\cos }^{8}}x}{{{\sin }^{4}}x-{{\cos }^{4}}x}}dx=\dfrac{1}{8}+\dfrac{12\pi -4}{32} \\   \int_{\frac{3\pi }{4}}^{\pi }{\dfrac{{{\sin }^{8}}x-{{\cos }^{8}}x}{{{\sin }^{4}}x-{{\cos }^{4}}x}}dx=-\dfrac{1}{16}+\dfrac{6\pi +2}{32} \\   \therefore A=-\dfrac{1}{16}+\dfrac{6\pi +2}{32}-\dfrac{1}{16}+\dfrac{6\pi +2}{32}+\dfrac{1}{8}+\dfrac{12\pi -4}{32}=\dfrac{3\pi }{4} $   

 Find the value of $-\int\limits_{\frac{5\pi }{12}}^{0}{\left[ \dfrac{1}{\cos x}-\dfrac{(\sin x+1)}{\cos x} \right]}dx$.

 First simplify the expression given, then find the value of the trigonometric ratio at the limits given, at last put the limit values.

 $\text{Let }I=-\int _{\frac{5\pi }{12}}^0\dfrac{1}{\cos \left(x\right)}-\dfrac{\sin \left(x\right)+1}{\cos \left(x\right)}dx\\ \Rightarrow -\int _{\frac{5\pi }{12}}^0\dfrac{\sin \left(x\right)}{\cos \left(x\right)} \\ \Rightarrow -\int _{\frac{5\pi }{12}}^0 tanx \\ \Rightarrow -\left[-\ln \left|\cos \left(x\right)\right|\right]_{\frac{5\pi }{12}}^0 \\ \Rightarrow -\left(\dfrac{1}{2}\ln \left(2-\sqrt{3}\right)-\ln \left(2\right)\right) \\ \Rightarrow -\dfrac{1}{2}\ln \left(2-\sqrt{3}\right)+\ln \left(2\right)$

 Evaluate $\int_{\pi }^{0}{\left( \dfrac{1}{7+5cosx} \right)dx}+\int_{2}^{-2}{l}og\left( \dfrac{5-x}{5+x} \right)dx$.

 First solve the first integration by substituting the formula, $\cos x=\left( \dfrac{1-{{\tan }^{2}}\dfrac{x}{2}}{1+{{\tan }^{2}}\dfrac{x}{2}} \right)$, then solve the integration further by putting the limits, and the second integration will be 0 as it is an odd function. 

 $\text{Let us suppose }I={{I}_{1}}+{{I}_{2}} \\  \text{Again }{{I}_{1}}=\int_{\pi }^{0}{\dfrac{1}{7+5cosx}dx}=\int_{\pi }^{0}{\dfrac{1}{7+5\left( \dfrac{1-{{\tan }^{2}}\dfrac{x}{2}}{1+{{\tan }^{2}}\dfrac{x}{2}} \right)}dx}=\int_{\pi }^{0}{\dfrac{{{\sec }^{2}}\dfrac{x}{2}}{7+7{{\tan }^{2}}\dfrac{x}{2}+5\left( 1-{{\tan }^{2}}\dfrac{x}{2} \right)}dx}=\int_{\pi }^{0}{\dfrac{{{\sec }^{2}}\dfrac{x}{2}}{12+2{{\tan }^{2}}\dfrac{x}{2}}dx} \\  \text{Let }\tan \dfrac{x}{2}=t,dt=\dfrac{1}{2}{{\sec }^{2}}\dfrac{x}{2} \\  {{I}_{1}}=\int_{0}^{\infty }{\dfrac{dt}{6+{{t}^{2}}}}=\dfrac{1}{6}\left[ {{\tan }^{-1}}\dfrac{t}{6} \right]_{0}^{\infty }=\dfrac{1}{6}\times \dfrac{\pi }{2}=\dfrac{\pi }{12} \\  \text{Let, }{{I}_{2}}=\int_{2}^{-2}{l}og\left( \dfrac{5-x}{5+x} \right)dx \\  g\left( x \right)=log\left( \dfrac{5x}{5+x} \right) \\  \text{Now, } g\left( -x \right)=log\left( \dfrac{5(-x)}{5+\text{ (-}x)} \right)=log\left( \dfrac{5x}{5+x} \right)=g\left( x \right) \\  \text{So from here we see that }g\left( x \right)\text{ is an odd function} \\  \text{Therefore, }\int_{2}^{-2}{g\left( x \right)dx}=0\Rightarrow {{I}_{2}}=0 \\  I={{I}_{1}}+{{I}_{2}}=\dfrac{\pi }{12}+0=\dfrac{\pi }{12} $  