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Definite Integration question

2022 · 25 Jun · Shift 1 · Q24
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  5. /2022 · 25 Jun · Shift 1 · Q24

Definite Integration question

2022 · 25 Jun · Shift 1 · Q24

JEE MainMathematicsDefinite IntegrationMCQ+4 / −1
The value of ∫0πecos⁡xsin⁡x(1+cos⁡2x)(ecos⁡x+e−cos⁡x)dx\int\limits_0^\pi {{{{e^{\cos x}}\sin x} \over {(1 + {{\cos }^2}x)({e^{\cos x}} + {e^{ - \cos x}})}}dx}0∫π​(1+cos2x)(ecosx+e−cosx)ecosxsinx​dx is equal to:
  1. A
    π24{{{\pi ^2}} \over 4}4π2​
  2. B
    π22{{{\pi ^2}} \over 2}2π2​
  3. C
    π4{\pi \over 4}4π​
  4. D
    π2{\pi \over 2}2π​
View written solutionFree

Correct answer: C

  1. We need to evaluate I=∫0πecos⁡xsin⁡x(1+cos⁡2x)(ecos⁡x+e−cos⁡x) dx.I=\int_0^\pi \frac{e^{\cos x}\sin x}{(1+\cos^2 x)(e^{\cos x}+e^{-\cos x})}\,dx.I=∫0π​(1+cos2x)(ecosx+e−cosx)ecosxsinx​dx.

  2. Use the substitution t=cos⁡x  ⟹  dt=−sin⁡x dx.t=\cos x \implies dt=-\sin x\,dx.t=cosx⟹dt=−sinxdx. When x=0x=0x=0, t=1t=1t=1; when x=πx=\pix=π, t=−1t=-1t=−1. So,

=\int_{-1}^{1} \frac{e^t}{(1+t^2)(e^t+e^{-t})}\,dt.$$ 3. Define $$f(t)=\frac{e^t}{(1+t^2)(e^t+e^{-t})}.$$ Then $$f(-t)=\frac{e^{-t}}{(1+t^2)(e^{-t}+e^t)}.$$ Hence, $$f(t)+f(-t)=\frac{e^t+e^{-t}}{(1+t^2)(e^t+e^{-t})}=\frac{1}{1+t^2}.$$ 4. Since the interval is symmetric, $$\int_{-1}^1 f(t)\,dt = \frac12 \int_{-1}^1 \big(f(t)+f(-t)\big)\,dt =\frac12 \int_{-1}^1 \frac{1}{1+t^2}\,dt.$$ Therefore, $$I=\frac12\int_{-1}^1 \frac{dt}{1+t^2}.$$ 5. Now evaluate: $$\int \frac{dt}{1+t^2}=\tan^{-1} t.$$ So, $$I=\frac12\left[\tan^{-1} t\right]_{-1}^{1} =\frac12\left(\tan^{-1}1-\tan^{-1}(-1)\right).$$ Using $$\tan^{-1}1=\frac\pi4,\qquad \tan^{-1}(-1)=-\frac\pi4,$$ we get $$I=\frac12\left(\frac\pi4+\frac\pi4\right)=\frac12\cdot\frac\pi2=\frac\pi4.$$ 6. Therefore the correct option is $$\boxed{\frac\pi4}.$$
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