Definite Integration
Grade 12
Question:
<p>The value of the integral <span class="math-tex">\(\int_\limits{-1}^{1} \log _{e}(\sqrt{1-x}+\sqrt{1+x}) d x\)</span> is equal to:</p>
<p style="display:inline"><span class="math-tex">\(\frac{1}{2} \log _{e} 2+\frac{\pi}{4}-\frac{3}{2}\)</span></p>
<p style="display:inline"><span class="math-tex">\(2 \log _{e} 2+\frac{\pi}{2}-\frac{1}{2}\)</span></p>
<p style="display:inline"><span class="math-tex">\(2 \log _{e} 2+\frac{\pi}{4}-1\)</span></p>
<p style="display:inline"><span class="math-tex">\(\log _{e} 2+\frac{\pi}{2}-1\)</span></p>
Step-by-Step Solution
Key Concept: Recognize that f(x) = ln(√(1-x) + √(1+x)) is an even function, so ∫_{-1}^{1} f(x)dx = 2∫_0^1 f(x)dx. Then use substitution x = sin(θ) to convert √(1-x) + √(1+x) into trigonometric form, which simplifies using the identity √(1-sin²θ) = cos(θ).
<p>Given <span class="math-tex">$f(x)=\ln (\sqrt{1-x}+\sqrt{1+x})$</span><br />
<span class="math-tex">$f(-x)=\ln (\sqrt{1-(-x)}+\sqrt{1-x})$</span><br />
<span class="math-tex">$\Rightarrow f(-x)=\ln (\sqrt{1+x}+\sqrt{1-x})=f(x)$</span><br />
<span class="math-tex">$\Rightarrow f(-x)=f(x)$</span><br />
<span class="math-tex">$\therefore f$</span> is even function<br />
Now, <span class="math-tex">${I}=\int_\limits{-1}^{1} \ln (\sqrt{1-x}+\sqrt{1+x}) d x$</span><br />
<span class="math-tex">$\Rightarrow {I}=2 \int_\limits{0}^{1} \ln (\sqrt{1-x}+\sqrt{1+x}) d x$</span><br />
Put, <span class="math-tex">$x=\cos 2 \theta \Rightarrow d x=-2 \sin 2 \theta d \theta$</span><br />
also <span class="math-tex">$\cos 2 \theta=2 \cos ^{2} \theta-1=1-2 \sin ^{2} \theta$</span><br />
Limits,<br />
<span class="math-tex">$x=0, \theta=\frac{\pi}{4}$</span><br />
<span class="math-tex">$x=1, \theta=0$</span><br />
<span class="math-tex">$\Rightarrow {I}= -4 \int_\limits{\pi / 4}^{0}[\ln \{(\sin \theta+\cos \theta) \sqrt{2}\}] \sin 2 \theta d \theta$</span><br />
<span class="math-tex">$= 4 \int_\limits{0}^{\pi / 4}[\ln \{(\sin \theta+\cos \theta) \sqrt{2}\}] \sin 2 \theta d \theta$</span><br />
<span class="math-tex">$= 4 \int_\limits{0}^{\pi / 4} \ln (\sin \theta+\cos \theta) \sin 2 \theta d \theta$</span> <span class="math-tex">$+4 \ln \sqrt{2} \int_{0}^{\pi / 4} \sin 2 \theta d \theta$</span><br />
<span class="math-tex">$=4\left[0+\frac{1}{2} \int_{0}^{\pi / 4}(\cos \theta-\sin \theta)^{2} d \theta\right]$</span><strong><span class="math-tex">$+4 \ln \sqrt{2}\left(0+\frac{1}{2}\right)$</span></strong><br />
<span class="math-tex">$=4\left[0+\frac{1}{2} \int_{0}^{\pi / 4}(1-\sin 2 \theta) d \theta\right]+2 \ln \sqrt{2}$</span><br />
<span class="math-tex">$=2\left[\theta+\frac{\cos 2 \theta}{2}\right]_{0}^{\pi / 4}+\ln 2$</span><br />
<span class="math-tex">$=2\left[\frac{\pi}{4}-\frac{1}{2}\right]+\ln 2$</span><br />
<span class="math-tex">$\therefore I =\frac{\pi}{2}-1+\ln 2$</span></p>
Correct Answer: D