Passage

For the next two (02) items that follow :
Consider the integrals $$I_1=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\frac{dx}{1+\sqrt{tan x}}$$ and $$I_2=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{sin x}dx}{\sqrt{sin }x+\sqrt{cos}x}$$
Mathematics

What is $$I_1$$ equal to ?


ANSWER

$$\dfrac{\pi}{12}$$


SOLUTION
$$I_1= \int_{\frac{\pi}{6}}^{\frac{\pi}{3}}$$ $$\cfrac{dx}{1+\sqrt{tanx}}$$
Using the property of definite integrals and substituting x by $$x-\cfrac{\pi}{6} - \cfrac{\pi}{3}$$

We get $$I_1=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \cfrac{dx}{1+\sqrt{cotx}}$$
$$\Rightarrow I_1 =\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \cfrac{\sqrt{tanx} \ dx}{1+\sqrt{tanx}} $$
on adding both, we get
$$I_1+I_1 = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}}dx$$
$$2I_1=\cfrac{\pi}{3}-\cfrac{\pi}{6}$$
$$I_1=\cfrac{\pi}{12}$$
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Single Correct Medium Published on 17th 09, 2020
Mathematics

What is $$I_1-I_2$$ equal to ?


ANSWER


SOLUTION
$$I_1 = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\cfrac{dx}{1+\sqrt{tan x}}$$
Using property of definite integrals and replacing $$x $$ with $$\cfrac{\pi}{6}+ \cfrac{\pi}{3}-x$$
$$I_1=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \cfrac{dx}{1+\sqrt{cot x}}$$
Breaking $$cot x$$ to $$\cfrac{cos x}{sinx}$$, we get
$$I_1=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \cfrac{\sqrt{sin x} dx}{\sqrt{sinx}+\sqrt{cos x}}$$         which is equal to $$I_2$$
$$\therefore \ I_1= I_2$$
$$I_1-I_2=0$$ 
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Single Correct Hard Published on 17th 09, 2020
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