$\int\limits_{0}^{\pi/4}$ $\frac{\sin x + \cos x}{16 + 9 \sin 2x}$ dx Now, sin2x = 2sinx cosx ∴ 1- sin2x = 1 - 2sinx cosx ∴ 1 - sin2x = $(\sin x - \cos x)^2$ Put sin x - cos x = t ⇒ (sin x + cos x) dx = dt x = $\frac{\pi}{4}$ , t = 0 x = 0 , t = - 1 So, $\int\limits_{-1}^{0}$ $\frac{1}{16-9(1-t^2)}$ dt = $\int\limits_{-1}^{0}$ $\frac{1}{16-9-9t^2}$ dt = $\int\limits_{-1}^{0}$ $\frac{1}{25-9t^2}$ dt = $\frac{1}{9} \int\limits_{-1}^{0}$ $\frac{1}{\frac{25}{9} - t^2}$ dt = = $\frac{1}{30}$ $\left(\log |1| - \log \frac{1}{4}\right)$ = $\frac{1}{30}$ log 4 OR Given $\int\limits_{1}^{3}(x^2+3x+e^x)$ dx ⇒ a = 1 , b = 3 ⇒ h = $\frac{3-1}{n}$ and f (x) = $x^2$ + 3x + $e^x$ I = $\int\limits_{1}^{3}(x^2+3x+e^x)$ dx I = $\lim\limits_{h\to 0}$ h [f (1) + f (1 + h) + ... + f (1 + (n - 1)h)] I = $\lim\limits_{h\to 0}$ h [4 + e + [$(1+h)^2$ + 3 (1 + h) + $e^{1+h}$]] + [$(1+2h)^2$ + 3 (1 + 2h) + $e^{1+2h}$] + ... + [$(1 + (n-1)h)^2$ + 3 (1 + (n - 1) h + $e^{1+(n-1)h}$)] I = $\lim\limits_{h\to 0}$ h I = $\lim\limits_{n\to \infty}$ 4n × $\frac{2}{n}$ + e × $\frac{2}{n} + \frac{8}{n^3}$ $\frac{(n-1)(n-2)(n-3)}{6}$ + $\frac{5 \times 4}{n^2} \left( \frac{n^2 - n}{2} \right)$ + $e^{\frac{2}{n+1}}$ $\left(e^{\frac{2}{n}(n-1)} - 1\right)$ I = 8 + $\frac{8}{3}$ + 10 + e ($e^2$ - 1) I = $\frac{62}{3}$ + $e^3$ - e