Let \(f(x)= \) \( \int_0^x (\frac{sin^2u}{sin x^3})du\)
And, we have to find \(\lim_{x \to 0} f(x)\)
First, let's solve f(x)
\(\int_0^x (\frac{sin^2u}{sin x^3}) du = \frac{1}{sin x^3} \int_0^x (sin^2u) du\)
[because since we are differentiating wrt 'u' , 1/sin x^3 can be taken out of the integral.]
Now,
\(cos2x=1-2sin^2x \)
So, \(sin^2u = \frac{1-cos2u}{2}\)
Therefore,
\(f(x)= \frac{1}{sinx^3}\int_0^x (\frac{1-cos2u}{2}) du\)
\(f(x)=\frac{1}{sinx^3}( \int_0^x (\frac{1}{2}) du - \int_0^x (\frac{cos2u}{2})du)\)
\(f(x)=\frac{1}{sinx^3}\left[\frac{u}{2}\right]_0^x - \frac{1}{sinx^3}\left[\frac{\frac{sin2u}{2}}{2}\right]_0^x\)
\(f(x)=\frac{1}{sinx^3} (\frac{x}{2}-0)-\frac{1}{sinx^3}(\frac{sin2x}{4}-0)\)
\(f(x)=\frac{2x-sin2x}{4sinx^3}\)
Now, onto taking the limit
\(\lim_{x \to 0} f(x)=\lim_{x \to 0} \frac{2x-sin2x}{4sinx^3}\)
Putting x=0 gives us the 0/0 form, so using L'Hospital Rule
\(\lim_{x \to 0} \frac{2-2cos2x}{4cosx^3.3x^2} =\lim_{x \to 0}\frac{1-cos2x}{6x^2cosx^3}\)
Putting x=0 gives us the 0/0 form, so using L'Hospital Rule
\(\lim_{x \to 0}\frac{2sin2x}{(-6x^2sinx^3.3x^2)+(cosx^3.12x)} = \lim_{x \to 0}\frac{sin2x}{(-9x^4sinx^3)+(6xcosx^3)}\)
Putting x=0 again gives us the 0/0 form, so using L'Hospital Rule
\(\lim_{x \to 0}\frac{2cos2x}{[(6cosx^3)-(6xsinx^3.3x^2)]-[(9x^4cosx^3.3x^2)+(sinx^3.36x^3)]}\)
\( = \lim_{x \to 0}\frac{2cos2x}{6cosx^3-18x^2sinx^3-27x^6cosx^3-36x^3sinx^3}\)
Now, on putting x=0, we get \(\frac{2cos0}{6cos0-0-0-0} = \frac{2}{6} = \frac{1}{3}\)
Therefore, The answer is 1/3.
Hope this helps:)
