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Definite Integration question

2021 · 20 Jul · Shift 2 · Q28
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  5. /2021 · 20 Jul · Shift 2 · Q28

Definite Integration question

2021 · 20 Jul · Shift 2 · Q28

JEE MainMathematicsDefinite IntegrationMCQ+4 / −1
If the real part of the complex number (1−cos⁡θ+2isin⁡θ)−1{(1 - \cos \theta + 2i\sin \theta )^{ - 1}}(1−cosθ+2isinθ)−1 is 15{1 \over 5}51​ for θ∈(0,π)\theta \in (0,\pi )θ∈(0,π), then the value of the integral ∫0θsin⁡xdx\int_0^\theta {\sin x} dx∫0θ​sinxdx is equal to:
  1. A
    1
  2. B
    2
  3. C
    −-− 1
  4. D
    0
View written solutionFree

Correct answer: A

  1. We are given z=(1−cos⁡θ+2isin⁡θ)−1z=\left(1-\cos\theta+2i\sin\theta\right)^{-1}z=(1−cosθ+2isinθ)−1 and the real part of zzz is 15\frac1551​.

We need to find ∫0θsin⁡x dx.\int_0^\theta \sin x\,dx.∫0θ​sinxdx.

  1. First, write the reciprocal in standard form.

Let a=1−cos⁡θ,b=2sin⁡θ.a=1-\cos\theta,\qquad b=2\sin\theta.a=1−cosθ,b=2sinθ. Then z=1a+ib=a−iba2+b2.z=\frac{1}{a+ib}=\frac{a-ib}{a^2+b^2}.z=a+ib1​=a2+b2a−ib​. So, ℜ(z)=aa2+b2.\Re(z)=\frac{a}{a^2+b^2}.ℜ(z)=a2+b2a​.

Given ℜ(z)=15\Re(z)=\frac15ℜ(z)=51​, we get 1−cos⁡θ(1−cos⁡θ)2+4sin⁡2θ=15.\frac{1-\cos\theta}{(1-\cos\theta)^2+4\sin^2\theta}=\frac15.(1−cosθ)2+4sin2θ1−cosθ​=51​.

  1. Simplify the denominator.

Using sin⁡2θ=1−cos⁡2θ\sin^2\theta=1-\cos^2\thetasin2θ=1−cos2θ,

=1-2\cos\theta+\cos^2\theta+4(1-\cos^2\theta). $$ So, $$=1-2\cos\theta+\cos^2\theta+4-4\cos^2\theta =5-2\cos\theta-3\cos^2\theta.$$ Hence, $$\frac{1-\cos\theta}{5-2\cos\theta-3\cos^2\theta}=\frac15.$$ 4. Cross-multiply: $$5(1-\cos\theta)=5-2\cos\theta-3\cos^2\theta.$$ This gives $$5-5\cos\theta=5-2\cos\theta-3\cos^2\theta.$$ So, $$-5\cos\theta=-2\cos\theta-3\cos^2\theta.$$ $$0=3\cos\theta-3\cos^2\theta.$$ $$3\cos\theta(1-\cos\theta)=0.$$ Thus, $$\cos\theta=0 \quad \text{or} \quad \cos\theta=1.$$ Since $\theta\in(0,\pi)$, $\cos\theta=1$ is not possible. Therefore, $$\cos\theta=0 \implies \theta=\frac\pi2.$$ 5. Now evaluate the integral: $$\int_0^\theta \sin x\,dx=\int_0^{\pi/2}\sin x\,dx.$$ Since $$\int \sin x\,dx=-\cos x,$$ we get $$\int_0^{\pi/2}\sin x\,dx=[-\cos x]_0^{\pi/2}$$ $$=-\cos\frac\pi2-(-\cos0)=0-(-1)=1.$$ 6. Therefore, the correct option is $$\boxed{A: 1}.$$
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