Let $$I=\int \frac{\sqrt{x}}{\sqrt{a^3-x^3}} d x=\int \frac{\sqrt{x}}{\sqrt{\left(a^{3 / 2}\right)^2-\left(x^{3 / 2}\right)^2}}$$

Put $\quad x^{3 / 2}=t \Rightarrow \frac{3}{2} x^{1 / 2} d x=d t$

$\therefore \quad I=\frac{2}{3} \int \frac{d t}{\sqrt{\left(a^{3 / 2}\right)^2-t^2}}=\frac{2}{3} \sin ^{-1} \frac{t}{a^{3 / 2}}+C$

$=\frac{2}{3} \sin ^{-1} \frac{x^{3 / 2}}{a^{3 / 2}}+C=\frac{2}{3} \sin ^{-1} \sqrt{\frac{x^3}{a^3}}+C$

Let $$\begin{aligned} I & =\int \frac{\cos x-\cos 2 x}{1-\cos x} d x=\int \frac{2 \sin \frac{3 x}{2} \cdot \sin \frac{x}{2}}{1-1+2 \sin ^2 \frac{x}{2}} d x \\ & =2 \int \frac{\sin \frac{3 x}{2} \cdot \sin \frac{x}{2}}{2 \sin ^2 \frac{x}{2}} d x=\int \frac{\sin \frac{3 x}{2}}{\sin \frac{x}{2}} d x \end{aligned}$$

$=\int \frac{3 \sin \frac{x}{2}-4 \sin ^3 \frac{x}{2}}{\sin \frac{x}{2}} d x \quad\left[\because \sin 3 x=3 \sin x-4 \sin ^3 x\right]$

$$\begin{aligned} & =3 \int d x-4 \int \sin ^2 \frac{x}{2} d x=3 \int d x-4 \int \frac{1-\cos x}{2} d x \\ & =3 \int d x-2 \int d x+2 \int \cos x d x \\ & =\int d x+2 \int \cos x d x=x+2 \sin x+C=2 \sin x+x+C \end{aligned}$$

Let $$I=\int \frac{d x}{x \sqrt{x^4-1}}$$

Put $$x^2=\sec \theta \Rightarrow \theta=\sec ^{-1} x^2$$

$$\begin{aligned} \Rightarrow \quad 2 x d x & =\sec \theta \cdot \tan \theta d \theta \\ \therefore \quad I & =\frac{1}{2} \int \frac{\sec \theta \cdot \tan \theta}{\sec \theta \tan \theta} d \theta=\frac{1}{2} \int d \theta=\frac{1}{2} \theta+C \\ & =\frac{1}{2} \sec ^{-1}\left(x^2\right)+C \end{aligned}$$

Let $$I=\int_0^2\left(x^2+3\right) d x$$

$$\begin{array}{ll} \text { Here, } & a=0, b=2 \text { and } h=\frac{b-a}{n}=\frac{2-0}{n} \\ \Rightarrow & h=\frac{2}{n} \Rightarrow n h=2 \Rightarrow f(x)=\left(x^2+3\right) \end{array}$$

Now, $\quad \int_0^2\left(x^2+3\right) d x=\lim _{h \rightarrow 0} h[f(0)+f(0+h)+f(0+2 h)+\ldots+f\{0+(n-1) h\}]\quad\text{.... (i)}$

$\because\quad f(0)=3$

$$\begin{aligned} \Rightarrow \quad f(0+h)=h^2+3, f(0+2 h) & =4 h^2+3=2^2 h^2+3 \\ f[0+(n-1) h] & =\left(n^2-2 n+1\right) h+3=(n-1)^2 h+3 \end{aligned}$$

From Eq. (i),

$$\int_0^2\left(x^2+3\right) d x=\lim _{h \rightarrow 0} h\left[3+h^2+3+2^2 h^2+3+3^2 h^2+3+\ldots+(n-1)^2 h^2+3\right]$$

$$\begin{aligned} & =\lim _{h \rightarrow 0} h\left[3 n+h^2\left\{1^2+2^2+\ldots+(n-1)^2\right\}\right] \\ & =\lim _{h \rightarrow 0} h\left[3 n+h^2\left(\frac{(n-1)(2 n-2+1)(n-1+)}{6}\right)\right]\left[\because \Sigma n^2=\frac{n(n+1)(2 n+1)}{6}\right] \\ & =\lim _{h \rightarrow 0} h\left[3 n+h^2\left(\frac{\left(n^2-n\right)(2 n-1)}{6}\right)\right] \end{aligned}$$

$$\begin{aligned} & =\lim _{h \rightarrow 0} h\left[3 n+\frac{h^2}{6}\left(2 n^3-n^2-2 n^2+n\right)\right] \\ & =\lim _{h \rightarrow 0}\left[3 n h+\frac{2 n^3 h^3-3 n^2 h^2 \cdot h+n h \cdot h^2}{6}\right] \\ & =\lim _{h \rightarrow 0}\left[3 \cdot 2+\frac{2 \cdot 8-3 \cdot 2^2 \cdot h+2 \cdot h^2}{6}\right]=\lim _{h \rightarrow 0}\left[6+\frac{16-12 h+2 h^2}{6}\right] \\ & =6+\frac{16}{6}=6+\frac{8}{3}=\frac{26}{3} \end{aligned}$$

$$\begin{aligned} &\begin{aligned} \text { Let }\quad & I=\int_0^2 e^x d x \\ \text{Here,}\quad & a=0 \text { and } b=2 \end{aligned} \end{aligned}$$

$$\begin{array}{ll} \therefore & h=\frac{b-a}{n} \\ \Rightarrow & n h=2 \text { and } f(x)=e^x \end{array}$$

Now, $$\int_0^2 e^x d x=\lim _{h \rightarrow 0} h[f(0)+f(0+h)+f(0+2 h)+\ldots+f\{0+(n-1) h\}]$$

$\therefore \quad I=\lim _{h \rightarrow 0} h\left[1+e^h+e^{2 h}+\ldots+e^{(n-1) h}\right]$

$$ \begin{aligned} & =\lim _{h \rightarrow 0} h\left[\frac{1 \cdot\left(e^h\right)^n-1}{e^h-1}\right]=\lim _{h \rightarrow 0} h\left(\frac{e^{n h}-1}{e^h-1}\right) \\ & =\lim _{h \rightarrow 0} h\left(\frac{e^2-1}{e^h-1}\right) \\ & =e^2 \lim _{h \rightarrow 0} \frac{h}{e^h-1}-\lim _{h \rightarrow 0} \frac{h}{e^h-1} \quad \left[\because \lim _{h \rightarrow 0} \frac{h}{\mathrm{e}^h-1}=1\right]\\ & =e^2-1=e^2-1 \end{aligned}$$