The question is: If x satisfies the equation $$\left(\int_0^1 \frac{dt}{t^2+2t\cos \alpha +1}\right)x^2 - \left(\int_{-3}^3 \frac{t^2 \sin {2t}dt}{t^2+1}\right)x - 2=0$$ $$(0<\alpha<\pi)$$, then the value of x is:
I know the coefficient of x will equate to zero. And the answer that I got after solving the question does not match with any of the options. So, I'll be grateful if someone can help me out on this one. Thank You.
Well, the second integral is zero as the integrand function is an odd one and the integration's interval is symmetric wrt zero, so your equation actually is;
$$\left(\int_0^1\frac{dt}{\left(t+\cos\alpha\right)^2+\sin^2\alpha}\right)x^2=2$$
Can you take it from here?
Added as further hint:
$$\int\frac{dt}{\sin^2\alpha+(t+\cos\alpha)^2}=\frac1{\sin\alpha}\int\frac{\frac1{\sin\alpha}dt}{1+\left(\frac t{\sin\alpha}+\cot\alpha\right)^2}=\ldots$$