Question

Find the integration \[\int {\dfrac{{\sin 2x}}{{({{\sin }^2}x + 1)({{\sin }^2}x + 3)}}} dx\]

Answer 1

Hint: We will start by converting \[\sin 2x\] in the form of \[\sin x\] and \[\cos x\] by using the formula \[\sin 2x = 2\sin x\cos x\]. Then, we will use the substitution method of integration, here, we will be substituting \[\sin x = t\]. This simplifies the equation and makes it easier to integrate the equations in terms of t.

Complete step by step answer:
Given,
\[\int {\dfrac{{\sin 2x}}{{({{\sin }^2}x + 1)({{\sin }^2}x + 3)}}} dx\]
So, by using the formula mentioned above, i.e., by changing \[\sin 2x = 2\sin x\cos x\], we get
\[ \Rightarrow \int {\dfrac{{2\sin x\cos x}}{{({{\sin }^2}x + 1)({{\sin }^2}x + 3)}}} dx\]
By substituting \[\sin x = t\] and differentiating on both the sides, we get
\[\dfrac{d}{{dx}}\sin x = \dfrac{d}{{dx}}t\]
\[\cos x = \dfrac{{dt}}{{dx}}\] \[\left( {\because \dfrac{d}{{dx}}\sin x = \cos x} \right)\]
By transferring \[dx\] to other side, we get
\[\cos x{\text{ }}dx = dt\]
So, finally by replacing \[\cos x{\text{ }}dx\] with \[dt\], and \[\sin x\] with \[t\], we will be able to convert the equation in terms of t, we get
\[ \Rightarrow \int {\dfrac{{2t}}{{({t^2} + 1)({t^2} + 3)}}dt} \] \[ \to \left( 1 \right)\]
Now, we can rewrite this equation by taking the parts separately by using partial fraction, so that it becomes easier for us to integrate
\[ \Rightarrow \dfrac{{2t}}{{({t^2} + 1)({t^2} + 3)}} = \dfrac{{At + B}}{{({t^2} + 1)}} + \dfrac{{Ct + D}}{{({t^2} + 3)}}\]
By using cross multiplication in the RHS so that the denominator becomes the same as LHS
\[ \Rightarrow \dfrac{{2t}}{{({t^2} + 1)({t^2} + 3)}} = \dfrac{{(At + B)({t^2} + 3) + (Ct + D)({t^2} + 1)}}{{({t^2} + 1)({t^2} + 3)}}\]
Now, since the denominators are same, we can simply compare the numerators to get the result,
\[ \Rightarrow 2t = (At + B)({t^2} + 3) + (Ct + D)({t^2} + 1)\]
Now, by opening the brackets, we get
\[ \Rightarrow 2t = (A{t^3} + B{t^2} + 3At + 3B) + (C{t^3} + D{t^2} + Ct + D)\]
Taking the common terms on one side,
\[ \Rightarrow 2t = {t^3}(A + C) + {t^2}(B + D) + t(3A + C) + (3B + D)\]
Here, we can see that on the LHS the equation is only in terms of t. That means, the on RHS the variables having coefficient other than t is 0. Hence, we can say that,
\[{t^3}(A + C) = 0\] and \[t(3A + C) = 2t\]
Therefore, by solving these two equations simultaneously, we get
\[{t^3}(A + C) = 0 \Rightarrow A = - C\],
\[3A + C = 2\]
By putting the value C in terms of A, we get
\[3A - A = 2\]
\[A = 1\] and \[C = - 1\]
Similarly, we will repeat these steps for B and D and get their values as
\[{t^2}(B + D) = 0\] and \[3B + D = 0\]
Solving them simultaneously the values we will get are
\[B = 0\] and \[D = 0\]
Therefore, we can write the whole equation as
\[ \Rightarrow \dfrac{{2t}}{{({t^2} + 1)({t^2} + 3)}} = \dfrac{t}{{({t^2} + 1)}} - \dfrac{t}{{({t^2} + 3)}}\]
Now, multiplying and dividing 2 on the RHS and integrating them, we get
\[ \Rightarrow \dfrac{1}{2}\int {\dfrac{{2t}}{{({t^2} + 1)}}dt - \dfrac{1}{2}} \int {\dfrac{{2t}}{{({t^2} + 3)}}} dt\] \[ \to \left( 2 \right)\]
Here, after solving the equation (2) we will get the same equation as (1)
Now, by taking \[{t^2} = u\] and differentiating it, we get
\[\dfrac{d}{{dt}}{t^2} = \dfrac{d}{{dt}}u\]
\[2t = \dfrac{{du}}{{dt}}\]
\[2t{\text{ }}dt = du\]
By replacing \[2t{\text{ }}dt\] with \[du\] and \[{t^2}\] with \[u\] in equation (2), we get
\[ \Rightarrow \dfrac{1}{2}\int {\dfrac{{du}}{{(u + 1)}}} - \dfrac{1}{2}\int {\dfrac{{du}}{{(u + 3)}}} \]
Now, we already know that \[\int {\dfrac{1}{x}} dx = \log x\], so using the same formula we get,
\[ = \dfrac{1}{2}\log (u + 1) - \dfrac{1}{2}\log (u + 3)\]
By putting the values of u again as \[{t^2}\],
\[ = \dfrac{1}{2}[\log ({t^2} + 1) - \log ({t^2} + 3)]\]
Again, by putting the values of t as \[\sin x\] we get,
\[ = \dfrac{1}{2}[\log ({\sin ^2}x + 1) - \log ({\sin ^2}x + 3)]\]
Hence, the final solution is derived.

Note: Always remember that while applying the substitution method, we must first differentiate it to replace dx with dt. In addition, we will always use partial fractions to convert one form of the equation to another; all we need to do is ensure that after solving equation (2), it becomes the same as equation (1). Make sure when in the denominator the variable has squares in it then the numerator will come in terms of \[At + B\] and \[Ct + D\].