#### 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 ?

$\dfrac{\pi}{12}$

##### 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 $I_1=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \cfrac{dx}{1+\sqrt{cotx}}$
$\Rightarrow I_1 =\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \cfrac{\sqrt{tanx} \ dx}{1+\sqrt{tanx}}$
$I_1+I_1 = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}}dx$
$2I_1=\cfrac{\pi}{3}-\cfrac{\pi}{6}$
$I_1=\cfrac{\pi}{12}$

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Single Correct Medium Published on 17th 09, 2020
Mathematics

# What is $I_1-I_2$ equal to ?

##### 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
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 111

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