$\int\limits_{0}^{\pi/4}\sqrt{\tan x}$ + $\sqrt{\cot x}$ dx = $\int\limits_{0}^{\pi/4}$ $\left(\frac{\sqrt{\sin x}}{\sqrt{\cos x}} + \frac{\sqrt{\cos x}}{\sqrt{\sin x}}\right)$ dx = $\int\limits_{0}^{\pi/4}$ $\left(\frac{\sin x + \cos x}{\sqrt{\sin x \cos x}}\right)$ dx = $\sqrt{2} \int\limits_{0}^{\pi/4}$ $\left(\frac{\sin x + \cos x}{\sqrt{2 \sin x \cos x}}\right)$ dx $\sqrt{2} \int\limits_{0}^{\pi/4}$ $\left(\frac{\sin x + \cos x}{\sqrt{1 - \sin x - \cos x^{2}}}\right)$ dx Put sin x - cos x = t ⇒ (cos x + sin x) dx = dt If x = 0 , t = 0 - 1 = - 1 and if x = $\frac{\pi}{4}$ , t = $\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}$ = 0 ∴ $\int\limits_{0}^{\pi/4}$ $\sqrt{\tan x} + \sqrt{\cot x}$ dx = $\sqrt{2} \int\limits_{-1}^{0} \frac{dt}{\sqrt{1 - t^{2}}}$ = $\sqrt{2} \left| \sin^{-1} t \right|_{-1}^{0}$ = $\sqrt{2} \left| \sin^{-1} 0 - \sin^{-1}(-1) \right|$ = $\sqrt{2} \left| 0 + \frac{\pi}{2} \right|$ = $\sqrt{2} \times \frac{\pi}{2}$ OR $\int\limits_{1}^{3} 2x^{2}$ + 5x dx Here, a = 1 , b = 3 , f (x) = $2x^{2}$ + 5x ∴ nh = b - a = 3 - 1 = 2 Now $\int\limits_{a}^{b}$ f (x) dx = $\lim\limits_{h\to 0}$ f (a) + f (a + h) + f (a + 2h) + ... + f (a + (n - 1)h) ∴ $\int\limits_{1}^{3} 2x^{2}$ + 5x dx = $\lim\limits_{h\to 0}$ h |2 $(1)^{2}$ +5 (1) + 2 $(1+h)^{2}$ + 5 (1 + h) + [2$(1+2h)^{2}$ + 5 (1 + 2h)] ... + [2 $(1+(n-1)h)^{2}$ + 5 (1 + (n - 1) h)]| = $\lim\limits_{h\to 0}$ |7 + ($2h^{2}$ + 9h + 7) + ($8h^{2}$ + 18h + 7) + ... + (2 $(n-1)^{2} h^{2}$ + 9 (n - 1) h + 7)| = $\lim\limits_{h\to 0}$ |7n + $2h^{2}$ ($1^{2} + 2^{+} \dots + (n-1)^{2}$) + 9h (1 + 2 + ... + (n - 1))| = $\lim\limits_{h\to 0}$ = $\lim\limits_{h\to 0}$ = $\lim\limits_{h\to 0}$ = 14 + $\frac{16}{3}$ + 18 = $\frac{112}{3}$