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