Question

Please help solve this tricky integral. ​

Answer 1

Question :

$\displaystyle \sf\int\limits_{0}^{\pi/2} \dfrac{1}{1 + \sqrt{tan(x)}}dx$

Solution :

Let,

$:\implies \sf y = \displaystyle \sf\int\limits_{0}^{\pi/2} \dfrac{1}{1 + \sqrt{tan(x)}}dx$

We know that tan(x) = sin(x)/cos(x), so by substituting it in the equation, we get :

$:\implies \sf y = \displaystyle \sf\int\limits_{0}^{\pi/2} \dfrac{1}{1 + \sqrt{\dfrac{sin(x)}{cos(x)}}}dx$

$:\implies \sf y = \displaystyle \sf\int\limits_{0}^{\pi/2} \dfrac{1}{1 + \dfrac{\sqrt{sin(x)}}{\sqrt{cos(x)}}}dx$

$:\implies \sf y = \displaystyle \sf\int\limits_{0}^{\pi/2} \dfrac{1}{\dfrac{\sqrt{cos(x)} + \sqrt{sin(x)}}{\sqrt{cos(x)}}}dx$

$:\implies \sf y = \displaystyle \sf\int\limits_{0}^{\pi/2} \dfrac{\sqrt{cos(x)}}{\sqrt{cos(x)} + \sqrt{sin(x)}}dx$

$\therefore \sf y = \displaystyle \sf\int\limits_{0}^{\pi/2} \dfrac{\sqrt{cos(x)}}{\sqrt{cos(x)} + \sqrt{sin(x)}}dx\quad\quad ...(i)$

We know the sum formula for definite integral,i.e,

$\boxed{\displaystyle \sf\int\limits_{a}^{b} f(x) = \displaystyle \sf\int\limits_{a}^{b} f(a + b - x)}$

So by using the above formula in the equation, we get :

$:\implies \sf y = \displaystyle \sf\int\limits_{0}^{\pi/2} \dfrac{1}{1 + \sqrt{tan(x)}}dx$

$:\implies \sf y = \displaystyle \sf\int\limits_{0}^{\pi/2} \dfrac{1}{1 + \sqrt{tan\left(\dfrac{\pi}{2} + 0 - x\right)}}dx$

$:\implies \sf y = \displaystyle \sf\int\limits_{0}^{\pi/2} \dfrac{1}{1 + \sqrt{tan\left(\dfrac{\pi}{2} + x\right)}}dx$

Now by substituting the values of tan(π/2 - x) [i.e, cot(x)] in the equation, we get :

$:\implies \sf y = \displaystyle \sf\int\limits_{0}^{\pi/2} \dfrac{1}{1 + \sqrt{cot(x)}}dx$

We know that cot(x) = cos(x)/sin(x), so by substituting it in the equation, we get :

$:\implies \sf y = \displaystyle \sf\int\limits_{0}^{\pi/2} \dfrac{1}{1 + \sqrt{\dfrac{cos(x)}{sin(x)}}}dx$

$:\implies \sf y = \displaystyle \sf\int\limits_{0}^{\pi/2} \dfrac{1}{1 + \dfrac{\sqrt{cos(x)}}{\sqrt{sin(x)}}}dx$

$:\implies \sf y = \displaystyle \sf\int\limits_{0}^{\pi/2} \dfrac{1}{\dfrac{\sqrt{sin(x)} + \sqrt{cos(x)}}{\sqrt{sin(x)}}}dx$

$:\implies \sf y = \displaystyle \sf\int\limits_{0}^{\pi/2} \dfrac{\sqrt{sin(x)}}{\sqrt{sin(x)} + \sqrt{cos(x)}}dx$

$\therefore \sf y = \displaystyle \sf\int\limits_{0}^{\pi/2} \dfrac{\sqrt{sin(x)}}{\sqrt{sin(x)} + \sqrt{cos(x)}}dx\quad\quad ...(ii)$

Now by adding Equation.(i) and Equation.(ii), we get :-

$:\implies \sf y + y = \displaystyle \sf\int\limits_{0}^{\pi/2} \dfrac{\sqrt{cos(x)}}{\sqrt{cos(x)} + \sqrt{sin(x)}}dx + \displaystyle \sf\int\limits_{0}^{\pi/2} \dfrac{\sqrt{sin(x)}}{\sqrt{cos(x)} + \sqrt{sin(x)}}dx$

By using the sum rule of definite integral, we get :

$:\implies \sf \displaystyle \sf\int\limits_{a}^{b} \sf f(x)dx + \displaystyle \sf\int\limits_{a}^{b} g(x)dx = \displaystyle \sf\int\limits_{a}^{b} \sf [f(x) + g(x)]dx$

$:\implies \sf y + y = \displaystyle \sf\int\limits_{0}^{\pi/2} \left(\dfrac{\sqrt{cos(x)}}{\sqrt{cos(x)} + \sqrt{sin(x)}} + \dfrac{\sqrt{sin(x)}}{\sqrt{cos(x)} + \sqrt{sin(x)}}\right)dx$

$:\implies \sf 2y = \displaystyle \sf\int\limits_{0}^{\pi/2} \dfrac{\sqrt{cos(x)} + \sqrt{sin(x)}}{\sqrt{cos(x)} + \sqrt{sin(x)}}dx$

$:\implies \sf 2y = \displaystyle \sf\int\limits_{0}^{\pi/2} 1dx$

$:\implies \sf 2y = \sf \big[x\big]_{0}^{\pi/2}$

$:\implies \sf 2y = \sf \left(\dfrac{\pi}{2} - 0\right)$

$:\implies \sf y = \sf\dfrac{\pi}{4}$

$\boxed{\therefore \sf y = \displaystyle \sf\int\limits_{0}^{\pi/2} \dfrac{1}{1 + \sqrt{tan(x)}}dx = \dfrac{\pi}{4}}$