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Common integrals Quiz 2

mathematics

 Let $\int_{0}^{1}{z\,d\ell \,\,\,=\,a}$ where  $z\,=\,5\ell +123$. Write the value of $2a$.

 Integrate and substitute value in the given equation to find $'a'$.

  $\text{Consider, }\int_{0}^{1}{zd\ell =a} \\ \text{ Substitute }z=5\ell +123\\ \Rightarrow \int_{0}^{1}{\left( 5\ell +123 \right)}d\ell =a\\ \text{Integrate with respect to }'\ell '\text{ in L.H.S.} \\ \left[ \dfrac{5{{\ell }^{2}}}{2}+123\ell  \right]_{0}^{1}=a\\\Rightarrow \dfrac{5}{2}+123=a\Rightarrow a=\dfrac{251}{2}\\\Rightarrow 2a=2\times \dfrac{251}{2}=251$

 Write the value of the integral $\int{\dfrac{{{\left( 295 \right)}^{{{\tan }^{-1}}2m}}}{1+4{{m}^{2}}}dm}$.

 Use substitution method and put ${{\tan }^{-1}}2m=t$ to solve the given integral.

  $\text{Consider the given integral as “I”.} \\ I=\int{\dfrac{{{\left( 295 \right)}^{{{\tan }^{-1}}2m}}}{1+4{{m}^{2}}}dm} ----(1) \\ \text{Put }{{\tan }^{-1}}2m=t\\\Rightarrow \dfrac{1}{1+4{{m}^{2}}}.2dm=dt\\\Rightarrow \dfrac{dm}{1+4{{m}^{2}}}=\dfrac{1}{2}dt -----(2)\\ \text{Use equation }(1)\text{ in }(2)\\I=\dfrac{1}{2}\int{{{\left( 295 \right)}^{t}}dt}\\\Rightarrow I=\dfrac{1}{2}.\dfrac{{{\left( 295 \right)}^{t}}}{\log 295}+c \\ \text{Substitute }t={{\tan }^{-1}}2m\\\Rightarrow I=\dfrac{{{\left( 295 \right)}^{{{\tan }^{-1}}2m}}}{2\log 295}+c $

 Integrate  $\int{\dfrac{\sin \,100p+\sin \,200p}{\cos \,100p\,-\,\cos \,200p}dp}$.

 Use suitable trigonometric formulae to reduce the function in simplest form. Then integrate.

  $\text{Let }I=\int{\dfrac{\sin 100p+\sin 200p}{\cos 100p-\cos 200p}}dp\\ \text{Apply “C-D” formulae.} \\ I=\int{\dfrac{2\sin \left( \dfrac{100p+200p}{2} \right)\cos \left( \dfrac{100p-200p}{2} \right)}{-2\sin \left( \dfrac{100p+200p}{2} \right)\sin \left( \dfrac{100p+200p}{2} \right)}}dp\\ \Rightarrow I=\int{\dfrac{2\sin 150p\cos \left( -50p \right)}{2\sin 150p\sin \left( -50p \right)}}dp\\ \Rightarrow I=\int{\cot 50pdp} \,\,\,\,\,\,\,\,[ \because \cos \left( -\theta  \right)=\cos \theta  \\ \sin \left( -\theta  \right)=-\sin \theta] \\ \text{Solve further, }I=-\dfrac{1}{50}\operatorname{logsin}\left| 50p \right|+c $

 Evaluate the integral $\int{{{e}^{{{\sec }^{-1}}x+{{\operatorname{cosec}}^{-1}}x}}dx}$.

 Put ${{\sin }^{-1}}x+{{\cos }^{-1}}x=\dfrac{\pi }{2}$ in the function and convert it into lowest terms. Then integrate.

  $\text{Consider the given integral as “I”}. \\ I=\int{{{e}^{{{\sec }^{-1}}x+\cos e{{c}^{-1}}x}}dx.} -----(1) \\ \text{Substitute }{{\sec }^{-1}}x+\cos e{{c}^{-1}}x=\dfrac{\pi }{2}\text{ in equation (1)} \\ I=\int{{{e}^{\pi /2}}dx}\\\Rightarrow I=x.{{e}^{\pi /2}}+c$

 Write the value of the integral $\int{\log \,\left( \dfrac{2s}{1+{{s}^{2}}} \right).\dfrac{2-2{{s}^{2}}}{{{\left( 1+{{s}^{2}} \right)}^{2}}}ds}$.

 Use substitution method and reduce the givens function in simple form. The integrate using by parts.

  $\text{Let }I=\int{\log \left( \dfrac{2s}{1+{{s}^{2}}} \right).\dfrac{2-2{{s}^{2}}}{{{\left( 1+{{s}^{2}} \right)}^{2}}}ds} -----(1)\\ \text{Substitute }\dfrac{2s}{1+{{s}^{2}}}=t\\\Rightarrow \dfrac{2-2{{s}^{2}}}{{{\left( 1+{{s}^{2}} \right)}^{2}}}ds=dt -----(2) \\ \text{Use equation }(2)\text{ in equation (1)}\\I=\int{\log tdt}\text{ Integrate using by parts }\log t\int{1dt-}\int{\left( \dfrac{d}{dt}\left( \log t \right).\int{1dt} \right)dt}\\\Rightarrow I=t\log t-\int{\left( \dfrac{1}{t}\times t \right)dt} \\ \text{Substitute }t=\dfrac{2s}{1+{{s}^{2}}}\\I=\left( \dfrac{2s}{1+{{s}^{2}}} \right)\log \left( \dfrac{2s}{1+{{s}^{2}}} \right)-\left( \dfrac{2s}{1+{{s}^{2}}} \right)+c$