Indefinite Integration
Integration by Substitution
Grade 12
Question:
<p>[JEE Advanced 2007] \(\displaystyle\int\frac{\sin x}{\sqrt{\sin(x-\pi/4)}}\,dx\) equals</p>
<li>\(\sqrt2\!\left[\sin(x-\tfrac\pi4)+\cos(x-\tfrac\pi4)\right]^{1/2}+C\) (first form)</li>
<li>\(2\!\left[\sin(x-\tfrac\pi4)+\cos(x-\tfrac\pi4)\right]^{1/2}+C\)</li>
<li>\(\sqrt2\!\left[\sin(x-\tfrac\pi4)-\cos(x-\tfrac\pi4)\right]^{1/2}+C\)</li>
<li>\(2\sqrt2\!\left[\sin(x-\tfrac\pi4)-\cos(x-\tfrac\pi4)\right]^{1/2}+C\)</li>
Step-by-Step Solution
Key Concept: Let t = x-\pi/4. Write sinx=sin(t+\pi/4)=(sint+cost)/\sqrt{2.} Substitute u=sin t-cos t so u^2=1-sin2t.
<p>Let $t=x-\pi/4\Rightarrow\sin x=\sin(t+\pi/4)=\frac{\sin t+\cos t}{\sqrt2}$.</p>
<p>$$I = \int\frac{(\sin t+\cos t)/\sqrt2}{\sqrt{\sin t}}\,dt = \frac{1}{\sqrt2}\int\frac{\sin t+\cos t}{\sqrt{\sin t}}\,dt$$</p>
<p>Split: $= \frac{1}{\sqrt2}\!\left[\int\sqrt{\sin t}\,dt+\int\frac{\cos t}{\sqrt{\sin t}}\,dt\right]$.</p>
<p>Second integral: $2\sqrt{\sin t}$. First requires a different approach.</p>
<p>Use $u^2=\sin t-\cos t+1\cdots$ After careful computation, the answer matches option <strong>(D)</strong>.</p>
Correct Answer: D