Find the value of integral \[\int_{0}^{{\pi }/{2}\;}{\dfrac{\sin 2x}{1+2{{\cos }^{2}}x}}\]
a) \[\dfrac{1}{2}\log 2\]
b) \[\log 2\]
c) \[\dfrac{1}{2}\log 3\]
d) \[\log 3\]
e) \[\dfrac{1}{3}\log 3\]

Hint: To solve the question, we have to apply the trigonometric formulae to convert the above expression in terms of variables which will ease the procedure of solving. Then apply integration formulae to arrive at the solution.

Complete step-by-step answer:

We know that the formula for \[\sin 2x\] is given by \[2\sin x\cos x\]. By substituting this value in the given expression, we get

\[\int_{0}^{{\pi }/{2}\;}{\dfrac{2\sin x\cos x}{1+2{{\cos }^{2}}x}}dx\]

By rearranging the terms of the expression, we get

\[=\int_{0}^{{\pi }/{2}\;}{\dfrac{2\cos x}{1+2{{\cos }^{2}}x}}\sin xdx\]

Let a be equal to \[\cos x\].

\[a=\cos x\]

By differentiating the above equation, we get

\[da=\dfrac{d\left( \cos x \right)}{dx}=-\sin xdx\]

Since we know the derivative of \[\cos x\] is equal to \[-\sin xdx\]

\[\Rightarrow \sin xdx=-da\]

Since the variable of differentiation is changed the limits of integration also changes. By

substituting the limits in \[\cos x\] we get the limits for the expression in a.

\[\cos \left( \dfrac{\pi }{2} \right)=0,\cos (0)=1\]

Thus, 0 and 1 are the upper limit and lower limit of the new expression formed respectively.

By substituting this value in the given expression, we get

\[=\int_{1}^{0}{\dfrac{2a}{1+2{{a}^{2}}}}\left( -da \right)\]

\[=\int_{1}^{0}{\dfrac{-2a}{1+2{{a}^{2}}}}da\]

\[=\left( -\int_{0}^{1}{\dfrac{-2a}{1+2{{a}^{2}}}}da \right)\]

Since I know that the sign of expression changes when limits are interchanged.
\[=\int_{0}^{1}{\dfrac{2a}{1+2{{a}^{2}}}}da\]

By rearranging the terms of expression, we get

\[=\int_{0}^{1}{\dfrac{2a}{2\left( \dfrac{1}{2}+{{a}^{2}} \right)}}da\]

\[=\dfrac{1}{2}\int_{0}^{1}{\dfrac{2a}{\left( \dfrac{1}{2}+{{a}^{2}} \right)}}da\]

\[=\dfrac{1}{2}\int_{0}^{1}{\dfrac{1}{\left( \dfrac{1}{2}+{{a}^{2}} \right)}2a}da\] ……. (1)

Let the expression \[\dfrac{1}{2}+{{a}^{2}}=t\]

By differentiating the above equation, we get

\[dt=\dfrac{d\left( \dfrac{1}{2}+{{a}^{2}} \right)}{da}=\left( 0+2{{a}^{2-1}} \right)da=2ada\]

Since we know that derivative of a constant is 0 and derivative of \[{{x}^{n}}=n{{x}^{n-1}}\]

Since the variable of differentiation is changed the limits of integration also changes. By

substituting the limits in \[\left( \dfrac{1}{2}+{{a}^{2}} \right)\]we get the limits for the

expression in t.

At a = 1

\[t=\dfrac{1}{2}+{{1}^{2}}=\dfrac{1}{2}+1=\dfrac{3}{2}\]

At a = 0

\[t=\dfrac{1}{2}+{{0}^{2}}=\dfrac{1}{2}\]

Thus, \[\dfrac{3}{2}\] and \[\dfrac{1}{2}\] are the upper limit and lower limit of the new expression formed respectively.

By substituting this value in the equation (1), we get

\[=\dfrac{1}{2}\int_{{1}/{2}\;}^{{3}/{2}\;}{\dfrac{1}{t}}dt\]

\[=\int_{{1}/{2}\;}^{{3}/{2}\;}{\dfrac{1}{2t}}dt\]

We know that integral of \[\dfrac{1}{t}\]is equal to \[\log t\]

\[=\left. \log (2t) \right|_{{1}/{2}\;}^{{3}/{2}\;}\]

We know the formula \[\int_{b}^{a}{{{f}^{1}}(x)}dx=f(a)-f(b)\]

By applying the above formula for the expression, we get

\[=\log \left( 2\times \dfrac{3}{2} \right)-\log \left( 2\times \dfrac{1}{2} \right)\]

\[=\log 3-\log 1\]

We know that \[\log 1\]is equal to 0.

Thus, \[\int_{0}^{{\pi }/{2}\;}{\dfrac{\sin 2x}{1+2{{\cos }^{2}}x}}=\log 3\]

Hence, option (c) is the right answer.

Note: The possibility of mistake can be not applying the integration formulae, trigonometric formulae to solve the given expression. The other possibility of mistake is to confuse among the assigned variables at different points of solving the expression.