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Definite Integration question

2021 · 20 Jul · Shift 1 · Q24
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  5. /2021 · 20 Jul · Shift 1 · Q24

Definite Integration question

2021 · 20 Jul · Shift 1 · Q24

JEE MainMathematicsDefinite IntegrationMCQ+4 / −1
The value of the integral ∫−11log⁡e(1−x+1+x)dx\int\limits_{ - 1}^1 {{{\log }_e}(\sqrt {1 - x} + \sqrt {1 + x} )dx}−1∫1​loge​(1−x​+1+x​)dx is equal to:
  1. A
    12log⁡e2+π4−32{1 \over 2}{\log _e}2 + {\pi \over 4} - {3 \over 2}21​loge​2+4π​−23​
  2. B
    2log⁡e2+π4−12{\log _e}2 + {\pi \over 4} - 12loge​2+4π​−1
  3. C
    log⁡e2+π2−1{\log _e}2 + {\pi \over 2} - 1loge​2+2π​−1
  4. D
    2log⁡e2+π2−122{\log _e}2 + {\pi \over 2} - {1 \over 2}2loge​2+2π​−21​
View written solutionFree

Correct answer: C

  1. Let I=∫−11ln⁡(1−x+1+x) dx.I=\int_{-1}^{1}\ln\big(\sqrt{1-x}+\sqrt{1+x}\big)\,dx.I=∫−11​ln(1−x​+1+x​)dx.

We simplify the expression inside the logarithm.

  1. Put x=cos⁡2θ,x=\cos 2\theta,x=cos2θ, so that as xxx goes from −1-1−1 to 111, we can take 2θ:π→0⇒θ:π2→0.2\theta: \pi \to 0 \quad \Rightarrow \quad \theta: \frac{\pi}{2}\to 0.2θ:π→0⇒θ:2π​→0. Also, dx=−2sin⁡2θ dθ=−4sin⁡θcos⁡θ dθ.dx=-2\sin 2\theta\,d\theta=-4\sin\theta\cos\theta\,d\theta.dx=−2sin2θdθ=−4sinθcosθdθ.

Now, 1−x=1−cos⁡2θ=2sin⁡2θ,1-x=1-\cos 2\theta=2\sin^2\theta,1−x=1−cos2θ=2sin2θ, 1+x=1+cos⁡2θ=2cos⁡2θ.1+x=1+\cos 2\theta=2\cos^2\theta.1+x=1+cos2θ=2cos2θ. Hence 1−x=2sin⁡θ,1+x=2cos⁡θ\sqrt{1-x}=\sqrt{2}\sin\theta, \qquad \sqrt{1+x}=\sqrt{2}\cos\theta1−x​=2​sinθ,1+x​=2​cosθ for θ∈[0,π/2]\theta\in[0,\pi/2]θ∈[0,π/2]. So 1−x+1+x=2(sin⁡θ+cos⁡θ).\sqrt{1-x}+\sqrt{1+x}=\sqrt{2}(\sin\theta+\cos\theta).1−x​+1+x​=2​(sinθ+cosθ). Using sin⁡θ+cos⁡θ=2sin⁡(θ+π4),\sin\theta+\cos\theta=\sqrt{2}\sin\left(\theta+\frac{\pi}{4}\right),sinθ+cosθ=2​sin(θ+4π​), we get 1−x+1+x=2sin⁡(θ+π4).\sqrt{1-x}+\sqrt{1+x}=2\sin\left(\theta+\frac{\pi}{4}\right).1−x​+1+x​=2sin(θ+4π​). Therefore, ln⁡(1−x+1+x)=ln⁡2+ln⁡sin⁡(θ+π4).\ln(\sqrt{1-x}+\sqrt{1+x})=\ln 2+\ln\sin\left(\theta+\frac{\pi}{4}\right).ln(1−x​+1+x​)=ln2+lnsin(θ+4π​).

  1. Substitute into the integral: I=∫π/20[ln⁡2+ln⁡sin⁡(θ+π4)](−4sin⁡θcos⁡θ) dθ.I=\int_{\pi/2}^{0}\left[\ln 2+\ln\sin\left(\theta+\frac{\pi}{4}\right)\right](-4\sin\theta\cos\theta)\,d\theta.I=∫π/20​[ln2+lnsin(θ+4π​)](−4sinθcosθ)dθ. Since 4sin⁡θcos⁡θ=2sin⁡2θ4\sin\theta\cos\theta=2\sin 2\theta4sinθcosθ=2sin2θ, I=∫0π/24sin⁡θcos⁡θ[ln⁡2+ln⁡sin⁡(θ+π4)]dθ.I=\int_0^{\pi/2}4\sin\theta\cos\theta\left[\ln 2+\ln\sin\left(\theta+\frac{\pi}{4}\right)\right]d\theta.I=∫0π/2​4sinθcosθ[ln2+lnsin(θ+4π​)]dθ. This is manageable, but there is an easier algebraic simplification.

  2. Observe directly: (1−x+1+x)2=(1−x)+(1+x)+21−x2=2+21−x2.(\sqrt{1-x}+\sqrt{1+x})^2=(1-x)+(1+x)+2\sqrt{1-x^2}=2+2\sqrt{1-x^2}.(1−x​+1+x​)2=(1−x)+(1+x)+21−x2​=2+21−x2​. Hence 1−x+1+x=2+21−x2.\sqrt{1-x}+\sqrt{1+x}=\sqrt{2+2\sqrt{1-x^2}}.1−x​+1+x​=2+21−x2​​. So ln⁡(1−x+1+x)=12ln⁡(2+21−x2).\ln(\sqrt{1-x}+\sqrt{1+x})=\frac12\ln\big(2+2\sqrt{1-x^2}\big).ln(1−x​+1+x​)=21​ln(2+21−x2​). Factor 222: =12ln⁡2+12ln⁡(1+1−x2).=\frac12\ln 2+\frac12\ln\big(1+\sqrt{1-x^2}\big).=21​ln2+21​ln(1+1−x2​). Thus

=\ln 2+\frac12\int_{-1}^1\ln(1+\sqrt{1-x^2})\,dx.$$ 5. Now use $x=\sin t$, with $t\in[-\pi/2,\pi/2]$. Then $$\sqrt{1-x^2}=\cos t, \qquad dx=\cos t\,dt.$$ So $$\int_{-1}^1\ln(1+\sqrt{1-x^2})dx =\int_{-\pi/2}^{\pi/2}\ln(1+\cos t)\cos t\,dt.$$ Since the integrand is even, $$=2\int_0^{\pi/2}\ln(1+\cos t)\cos t\,dt.$$ Therefore $$I=\ln 2+\int_0^{\pi/2}\ln(1+\cos t)\cos t\,dt.$$ 6. Use $$1+\cos t=2\cos^2\frac t2.$$ Hence $$\ln(1+\cos t)=\ln 2+2\ln\cos\frac t2.$$ Therefore $$I=\ln 2+\ln 2\int_0^{\pi/2}\cos t\,dt+2\int_0^{\pi/2}\cos t\ln\cos\frac t2\,dt.$$ Now $$\int_0^{\pi/2}\cos t\,dt=1,$$ so $$I=2\ln 2+2J,$$ where $$J=\int_0^{\pi/2}\cos t\ln\cos\frac t2\,dt.$$ 7. Put $u=t/2$. Then $t=2u$, $dt=2du$, and when $t:0\to\pi/2$, $u:0\to\pi/4$. Also, $$\cos t=\cos 2u=2\cos^2 u-1.$$ Thus $$J=2\int_0^{\pi/4}(2\cos^2 u-1)\ln(\cos u)\,du.$$ This route is possible but lengthy. Instead, use integration by parts on the earlier form: $$K=\int_0^{\pi/2}\ln(1+\cos t)\cos t\,dt.$$ Take $$u=\ln(1+\cos t), \qquad dv=\cos t\,dt.$$ Then $$du=-\frac{\sin t}{1+\cos t}dt=-\tan\frac t2\,dt, \qquad v=\sin t.$$ So $$K=\left[\sin t\ln(1+\cos t)\right]_0^{\pi/2}-\int_0^{\pi/2}\sin t\left(-\frac{\sin t}{1+\cos t}\right)dt.$$ Boundary term: $$\left[\sin t\ln(1+\cos t)\right]_0^{\pi/2}=0,$$ because at $t=\pi/2$, $\sin t=1$ and $\ln(1+\cos t)=\ln 1=0$; at $t=0$, $\sin 0=0$. Thus $$K=\int_0^{\pi/2}\frac{\sin^2 t}{1+\cos t}dt.$$ Now, $$\frac{\sin^2 t}{1+\cos t}=\frac{1-\cos^2 t}{1+\cos t}=1-\cos t.$$ Hence $$K=\int_0^{\pi/2}(1-\cos t)dt=\left[t-\sin t\right]_0^{\pi/2}=\frac\pi2-1.$$ 8. Therefore $$I=\ln 2+K=\ln 2+\frac\pi2-1.$$ 9. Compare with options: - A: $\frac12\ln 2+\frac\pi4-\frac32$ ❌ - B: $2\ln 2+\frac\pi4-1$ ❌ - C: $\ln 2+\frac\pi2-1$ ✅ - D: $2\ln 2+\frac\pi2-\frac12$ ❌ So the correct option is **C**.
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