Question
Evaluate:$\int\limits_{\pi/6}^{\pi/3}\frac{\sin\text{x}+\cos{\text{x}}}{\sqrt{\sin\text{2x}}}\text{dx}$ .
✓
Answer
Let sin x – cos x = t $\Rightarrow$ (cos x + sin x) dx = dt, Also, sin 2x = 1 – t2
When $\text{x}\frac{\pi}{3},\text{t}=\frac{\sqrt{3}-1}{2},\text{when x}=\frac{\pi}{6},\text{t}=\frac{1-\sqrt{3}}{2}$
$\therefore$ Given integral becomes I = $\int^{\frac{\sqrt{3}-1}{2}}_{\frac{1-\sqrt{3}}{2}}\frac{\text{dt}}{\sqrt{1-\text{t}^{2}}}$
$[\sin^{-1}\text{t}]^{\frac{\sqrt{3}-1}{2}}_{\frac{1-\sqrt{3}}{2}}=\sin^{-1}\bigg(\frac{\sqrt{3}-1}{2}\bigg)-\sin^{-1}\bigg(\frac{1-\sqrt{3}}{2}\bigg)$
OR $2\sin^{-1}\Bigg(\frac{\sqrt{3}-1}{2}\Bigg).$
When $\text{x}\frac{\pi}{3},\text{t}=\frac{\sqrt{3}-1}{2},\text{when x}=\frac{\pi}{6},\text{t}=\frac{1-\sqrt{3}}{2}$
$\therefore$ Given integral becomes I = $\int^{\frac{\sqrt{3}-1}{2}}_{\frac{1-\sqrt{3}}{2}}\frac{\text{dt}}{\sqrt{1-\text{t}^{2}}}$
$[\sin^{-1}\text{t}]^{\frac{\sqrt{3}-1}{2}}_{\frac{1-\sqrt{3}}{2}}=\sin^{-1}\bigg(\frac{\sqrt{3}-1}{2}\bigg)-\sin^{-1}\bigg(\frac{1-\sqrt{3}}{2}\bigg)$
OR $2\sin^{-1}\Bigg(\frac{\sqrt{3}-1}{2}\Bigg).$
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