Definite IntegrationhardFree

Definite Integration: Let Real Number Integral

JEE Maths question with a full step-by-step solution.

Question
Let α\alpha be a real number, then the integral, (n∈I)\left(n \in I\right)
I(α)=∫−11sin⁡α dx1−2xcos⁡α+x2=I\left(\alpha\right) = \int_{-1}^{1}\frac{\sin\alpha\,dx}{1-2x\cos\alpha+x^2} =
Aπ2\dfrac\pi2 if α∈[2nπ,(2n+1)π]\alpha \in \left[2n\pi,\left(2n+1\right)\pi\right]
Bπ2\dfrac\pi2 if α∈((2n+1)π,(2n+2)π)\alpha \in \left(\left(2n+1\right)\pi,\left(2n+2\right)\pi\right)
C−π2\dfrac{-\pi}{2} if α∈((2n+1)π,(2n+2)π)\alpha \in \left(\left(2n+1\right)\pi,\left(2n+2\right)\pi\right)correct
Dπ2\dfrac\pi2 if α∈(2nπ,(2n+1)π)\alpha \in \left(2n\pi,\left(2n+1\right)\pi\right)correct
Solution
Step 1: If α\alpha is an integer multiple of π\pi then sin⁡α=0\sin\alpha = 0, the numerator vanishes identically, and
I(α)=0I\left(\alpha\right) = 0
 The denominator is then (x∓1)2\left(x\mp1\right)^2, which vanishes at an endpoint, but the integrand is identically zero on the open interval. Step 2: For sin⁡α≠0\sin\alpha \ne 0, completing the square,
1−2xcos⁡α+x2=(x−cos⁡α)2+(1−cos⁡2α)=(x−cos⁡α)2+sin⁡2α1-2x\cos\alpha+x^2 = \left(x-\cos\alpha\right)^2+\left(1-\cos^2\alpha\right) = \left(x-\cos\alpha\right)^2+\sin^2\alpha
∫sin⁡α dx(x−cos⁡α)2+sin⁡2α=arctan⁡(x−cos⁡αsin⁡α)+C\int\frac{\sin\alpha\,dx}{\left(x-\cos\alpha\right)^2+\sin^2\alpha} = \arctan\left(\frac{x-\cos\alpha}{\sin\alpha}\right)+C
I(α)=arctan⁡(1−cos⁡αsin⁡α)−arctan⁡(−1−cos⁡αsin⁡α)I\left(\alpha\right) = \arctan\left(\frac{1-\cos\alpha}{\sin\alpha}\right) -\arctan\left(\frac{-1-\cos\alpha}{\sin\alpha}\right)
 Step 3:
1−cos⁡αsin⁡α=tan⁡α2,−1−cos⁡αsin⁡α=−1+cos⁡αsin⁡α=−cot⁡α2\frac{1-\cos\alpha}{\sin\alpha} = \tan\frac\alpha2 ,\qquad \frac{-1-\cos\alpha}{\sin\alpha} = -\frac{1+\cos\alpha}{\sin\alpha} = -\cot\frac\alpha2
 so, writing u=tan⁡α2u = \tan\dfrac\alpha2,
I(α)=arctan⁡u+arctan⁡1uI\left(\alpha\right) = \arctan u+\arctan\frac1u
 Step 4: The two arguments have product 11, so the two arctangents differ by exactly a right angle:
arctan⁡u+arctan⁡1u={  π2,u>0,−π2,u<0.\arctan u+\arctan\frac1u = \begin{cases}\ \ \dfrac\pi2 , & u>0 ,\\[2mm] -\dfrac\pi2 , & u<0 .\end{cases}
 The sign of II is therefore the sign of tan⁡α2\tan\dfrac\alpha2. Step 5:
tan⁡α2>0  ⟺  α2∈(nπ, nπ+π2)  ⟺  α∈(2nπ, (2n+1)π)\tan\frac\alpha2>0 \iff \frac\alpha2 \in \left(n\pi,\ n\pi+\frac\pi2\right) \iff \alpha \in \left(2n\pi,\ \left(2n+1\right)\pi\right)
tan⁡α2<0  ⟺  α∈((2n+1)π, (2n+2)π)\tan\frac\alpha2<0 \iff \alpha \in \left(\left(2n+1\right)\pi,\ \left(2n+2\right)\pi\right)
I(α)={  π2,α∈(2nπ,(2n+1)π),−π2,α∈((2n+1)π,(2n+2)π),  0,α=kπ.I\left(\alpha\right) = \begin{cases} \ \ \dfrac\pi2 , & \alpha \in \left(2n\pi,\left(2n+1\right)\pi\right),\\[2mm] -\dfrac\pi2 , & \alpha \in \left(\left(2n+1\right)\pi,\left(2n+2\right)\pi\right),\\[2mm] \ \ 0 , & \alpha = k\pi . \end{cases}
 Answer: (3) and (4)
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