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Step-by-Step Solution
Step 1: Rewrite the integrand
We start with the integral:
\displaystyle I = \int \frac{\sin x}{\sin^3 x + \cos^3 x} \, dx.
Observe that \sin^3 x + \cos^3 x can be factored if we consider the factorization approach, or we can try to separate in terms of \tan x . A useful trick is to express everything in terms of \tan x . Notice that:
\sin x = \frac{\tan x}{\sqrt{1 + \tan^2 x}}, \quad
\cos x = \frac{1}{\sqrt{1 + \tan^2 x}}.
One direct approach is to factor out \cos^3 x in the denominator and then use the substitution t = \tan x . Let us apply that strategy.
Step 2: Factor out \cos^3 x and simplify
Rewrite
\sin^3 x + \cos^3 x = \cos^3 x \bigl(\underbrace{ \frac{\sin^3 x}{\cos^3 x} + 1 }_{\tan^3 x + 1} \bigr)
= \cos^3 x \, (1 + \tan^3 x).
Thus, the integral becomes:
I = \int \frac{\sin x}{\cos^3 x (1 + \tan^3 x)} \, dx.
Note that
\frac{\sin x}{\cos^3 x} \;=\; \frac{\sin x}{\cos x} \cdot \frac{1}{\cos^2 x}
= \tan x \cdot \sec^2 x.
Hence,
I = \int \frac{\tan x \, \sec^2 x}{(1 + \tan^3 x)} \, dx.
We also notice 1 + \tan^3 x can be factorized or rearranged suitably, but the key substitution is clear.
Step 3: Apply the substitution t = \tan x
Let
t = \tan x \quad \Longrightarrow \quad dt = \sec^2 x \, dx.
Therefore,
I = \int \frac{t}{1 + t^3} \, dt.
However, a common factorization for 1 + t^3 is:
1 + t^3 = (t + 1)\bigl(t^2 - t + 1\bigr).
Thus,
I = \int \frac{t}{(t + 1)(t^2 - t + 1)} \, dt.
Step 4: Decompose the integrand into partial fractions
We want to find constants A , B , C such that
\frac{t}{(t+1)(t^2 - t + 1)}
= \frac{A}{t+1}
+ \frac{B(2t - 1) + C}{t^2 - t + 1}.
A more systematic approach is:
\frac{t}{(t+1)(t^2 - t + 1)}
= \frac{A}{t+1}
+ \frac{B(2t - 1)}{t^2 - t + 1}
+ \frac{C}{t^2 - t + 1}.
After combining the fractions, we match coefficients of like powers of t . We get the system:
A + 2B = 0, \quad
-A + B + C = 1, \quad
A - B + C = 0.
Solving, we find:
A = -\frac{1}{3}, \quad
B = \frac{1}{6}, \quad
C = \frac{1}{2}.
Step 5: Integrate each partial fraction term
So the integral splits as:
I
= \int \left(
-\frac{1}{3} \cdot \frac{1}{t+1}
+ \frac{1}{6} \cdot \frac{2t - 1}{t^2 - t + 1}
+ \frac{1}{2} \cdot \frac{1}{t^2 - t + 1}
\right) \, dt.
We can write that as:
I
= -\frac{1}{3} \int \frac{dt}{t+1}
+ \frac{1}{6} \int \frac{2t - 1}{t^2 - t + 1} \, dt
+ \frac{1}{2} \int \frac{1}{t^2 - t + 1} \, dt.
Step 6: Integrate each term separately
\displaystyle -\frac{1}{3} \int \frac{dt}{t+1}
= -\frac{1}{3} \ln|t + 1|.
To handle \displaystyle \int \frac{2t - 1}{t^2 - t + 1} \, dt , notice that
\frac{d}{dt}[\,t^2 - t + 1\,] = 2t - 1.
Hence,
\int \frac{2t - 1}{t^2 - t + 1} \, dt
= \ln|\,t^2 - t + 1\,|.
Therefore,
\frac{1}{6} \int \frac{2t - 1}{t^2 - t + 1} \, dt
= \frac{1}{6} \ln|\,t^2 - t + 1\,|.
For \displaystyle \int \frac{1}{t^2 - t + 1} \, dt , complete the square:
t^2 - t + 1
= t^2 - t + \frac{1}{4} + \frac{3}{4}
= \left(t - \frac{1}{2}\right)^2 + \frac{3}{4}.
Hence,
\int \frac{dt}{t^2 - t + 1}
= \int \frac{dt}{\left(t - \frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2}.
This is of the form
\int \frac{d\bigl(u\bigr)}{u^2 + a^2} = \frac{1}{a} \tan^{-1}\left(\frac{u}{a}\right).
Here, u = t - \frac{1}{2} and a = \frac{\sqrt{3}}{2} . Therefore,
\int \frac{dt}{t^2 - t + 1}
= \int \frac{du}{u^2 + \left(\frac{\sqrt{3}}{2}\right)^2}
= \frac{2}{\sqrt{3}} \tan^{-1}\left(\frac{2u}{\sqrt{3}}\right)
= \frac{2}{\sqrt{3}} \tan^{-1}\left(\frac{2t - 1}{\sqrt{3}}\right).
Multiplying by \frac{1}{2} in front,
\frac{1}{2} \int \frac{1}{t^2 - t + 1} \, dt
= \frac{1}{2} \cdot \frac{2}{\sqrt{3}} \tan^{-1}\left(\frac{2t - 1}{\sqrt{3}}\right)
= \frac{1}{\sqrt{3}} \tan^{-1}\left(\frac{2t - 1}{\sqrt{3}}\right).
Combining all parts, we get:
\displaystyle
I = -\frac{1}{3}\,\ln|t + 1|
+ \frac{1}{6}\,\ln|t^2 - t + 1|
+ \frac{1}{\sqrt{3}} \tan^{-1}\!\Bigl(\frac{2t - 1}{\sqrt{3}}\Bigr)
+ C.
Step 7: Substitute back t = \tan x
Recall t = \tan x . Substituting back yields the final form of the antiderivative:
\displaystyle
I = -\frac{1}{3} \ln\bigl|1 + \tan x\bigr|
+ \frac{1}{6} \ln\bigl|\tan^2 x - \tan x + 1\bigr|
+ \frac{1}{\sqrt{3}} \tan^{-1}\!\Bigl(\frac{2\tan x - 1}{\sqrt{3}}\Bigr)
+ C.
Step 8: Identify \alpha , \beta , \gamma and compute 18(\alpha + \beta + \gamma^2)
From the general form
\alpha \ln|1 + \tan x| + \beta \ln|1 - \tan x + \tan^2 x| + \gamma \tan^{-1}\!\Bigl(\tfrac{2\tan x - 1}{\sqrt{3}}\Bigr) ,
we see:
\alpha = -\tfrac{1}{3}.
\beta = \tfrac{1}{6}. (Note that 1 - \tan x + \tan^2 x is the same as \tan^2 x - \tan x + 1 up to rearranging terms.)
\gamma = \frac{1}{\sqrt{3}}.
Hence,
\displaystyle
\alpha + \beta + \gamma^2
= -\frac{1}{3} + \frac{1}{6} + \left(\frac{1}{\sqrt{3}}\right)^2
= -\frac{1}{3} + \frac{1}{6} + \frac{1}{3}
= 0.
Wait, let's compute carefully:
-\tfrac{1}{3} + \tfrac{1}{6} = -\tfrac{1}{6}.
-\tfrac{1}{6} + \tfrac{1}{3} = \tfrac{1}{6}.
So actually,
\alpha + \beta + \gamma^2 = \frac{1}{6}.
Now multiply by 18:
\displaystyle
18 \bigl(\alpha + \beta + \gamma^2\bigr)
= 18 \times \frac{1}{6} = 3.
Therefore, the value of 18(\alpha + \beta + \gamma^2) is 3.
