Question

For ${{x}^{2}}\ne n\pi +1,n\in N$(the set of natural numbers), the integral
$\int{(x)}\sqrt{\dfrac{2\,\sin \,({{x}^{2}}-1)-\sin 2({{x}^{2}}-1)}{2\,\sin \,({{x}^{2}}-1)+\sin 2({{x}^{2}}-1)}}dx$ is equal to
(where c is a constant of integration).
(a) ${{\log }_{e}}\left| \sec \left( \dfrac{{{x}^{2}}-1}{2} \right) \right|+c$
(b) ${{\log }_{e}}\left| \dfrac{1}{2}{{\sec }^{2}}\left( {{x}^{2}}-1 \right) \right|+c$
(c) ${{\log }_{e}}\left| {{\sec }^{2}}\left( \dfrac{{{x}^{2}}-1}{2} \right) \right|+c$
(d) $\dfrac{1}{2}{{\log }_{e}}\left| {{\sec }^{2}}\left( {{x}^{2}}-1 \right) \right|+c$

Answer 1

Hint: Use general substitution techniques to simplify the given terms. Then use trigonometric formulas to cancel common terms and get a form where you can use basic integration properties. Expand all terms trigonometrically at start and then combine the terms after cancelling the common terms in numerator and denominator. Use the following formulas.

Sin2x=2sinx\text{ }cosx \\\ \begin{aligned} & 1-cos2x=2{{\sin }^{2}}x \\\ & 1+\cos 2x=2{{\cos }^{2}}x \\\ \end{aligned} \\\ \end{array}$$ Complete step-by-step answer: Given integral in the question, can be written in the form of: $\int{(x)}\sqrt{\dfrac{2\,\sin \,({{x}^{2}}-1)-\sin \left( 2({{x}^{2}}-1) \right)}{\sin \,\left( 2({{x}^{2}}-1) \right)+2\sin ({{x}^{2}}-1)}}dx$ Let us assume this integral to be denoted by I, we get: $I=\int{(x)}\sqrt{\dfrac{2\,\sin \,({{x}^{2}}-1)-\sin \left( 2({{x}^{2}}-1) \right)}{\sin \,\left( 2({{x}^{2}}-1) \right)+2\sin ({{x}^{2}}-1)}}dx$ Let us assume a substitution, as follows and its differentiation $u={{x}^{2}},\,du=2xdx$ By multiplying and dividing by 2 on I, we get it as: $$I=\dfrac{1}{2}\int{{}}\sqrt{\dfrac{2\,\sin \,({{x}^{2}}-1)-\sin \left( 2{{x}^{2}}-2 \right)}{\,\sin \,\left( 2{{x}^{2}}-2 \right)+2\sin ({{x}^{2}}-1)}}\left( 2xdx \right)$$ By substituting, we get the equation: $I=\dfrac{1}{2}\int{{}}\sqrt{\dfrac{2\,\sin \,(u-1)-\sin (2u-2)}{\sin \,(2u-2)+2\sin (u-1)}}du$ Now, we move to the next substitution into the integration. Let us assume the second substitution to be v, given by: v=u-1: dv=du By substituting this substitution into our integral, we get: $I=\dfrac{1}{2}\int{{}}\sqrt{\dfrac{\,\sin \,v-\sin 2v}{\sin \,2v+\sin v}}dv$ By basic trigonometric knowledge, we can say the formula: $$Sin2x=2sinx\times cosx$$ By substituting this equation into our I, we get it as: $I=\dfrac{1}{2}\int{{}}\sqrt{\dfrac{\sin \,v-2\sin v\cos v}{\sin \,v+2\sin v\cos v}}dv$ Now take the term sinv common from numerator, denominator: $$I=\dfrac{1}{2}\int{{}}\sqrt{\dfrac{\sin \,v\left( 2-2\cos v \right)}{\sin \,v\left( 2+2\cos v \right)}}dv$$ Now, cancel the term sinv inside the square root, we get: $I=\dfrac{1}{2}\int{\sqrt{\dfrac{1-\cos v}{1+\cos v}}dv}$ So, by trigonometric knowledge we can write formulas: $$1-cosx=2{{\sin }^{2}}\dfrac{x}{2}$$ By substitution this value into our I, we get it as: $I=\dfrac{1}{2}\sqrt{\dfrac{2{{\sin }^{2}}\dfrac{v}{2}}{1+\cos v}}dv$ By basic knowledge of trigonometric, we can write formula as: $1+\cos v=2{{\cos }^{2}}\dfrac{v}{2}$ By substituting this into our equation and simplifying, we get: $$I=\dfrac{1}{2}\int{\dfrac{\sin \dfrac{v}{2}}{\cos \dfrac{v}{2}}}dv$$ Let us take a substitution and its differentiation as $w=\cos \dfrac{v}{2};\,dw=\dfrac{-\sin \dfrac{v}{2}}{2}dv$ Now writing $\dfrac{1}{2}$ inside the integral, we get it as: $I=\int{\dfrac{1}{\cos \dfrac{v}{2}}\left( \dfrac{\sin v\dfrac{v}{2}}{2}dv \right)}$ By substituting the substitution, we get the I into the form of: $I=-\int{\dfrac{1}{w}dw}$ By basic integration, we can say it as follows: $I=-\ln \left( w \right)+c$ By substituting w back into equation, we get: $I=-\ln \left( \cos \dfrac{v}{2} \right)$ By substituting all variables back into equation, we get: $I=-\ln \left( \cos \left( \dfrac{{{x}^{2}}-1}{2} \right) \right)+c$ By sending the “- “inside the logarithm we get it as: $I=\ln \left| \sec \left( \dfrac{{{x}^{2}}-1}{2} \right) \right|+c$ ln is nothing but log e. Therefore option (a) is the correct answer. Note: Be careful with substituting as the number of substitutions increases confusion increases. Instead of u, v you can take one variable as ${{x}^{2}}-1$ and decrease the land of substitution. When there is logarithm then never leave a number single outside just send it as inverse into power of term inside the logarithm