Definite Integration

Evaluate $\displaystyle \int \limits_{0}^{\frac{\pi}{2}} x^2 \; \cos 2x \; dx$

$\begin{aligned} \text{Let } I & = \int \limits_{0}^{\frac{\pi}{2}} x^2 \; \cos 2x \; dx \\\\ & \left[\begin{aligned} \text{Note: }& \int \limits_{a}^{b} u\; v \; dx = \left[u \int v \; dx\right]_{a}^{b} - \int \limits_{a}^{b} \left\{\int v \; dx \times \dfrac{d}{dx} \left(du\right) \right\} \; dx \\\\ & \text{Here } u = x^2; \; v = \cos 2x \end{aligned}\right] \\\\ & = \left[x^2 \int \cos 2x \; dx\right]_{0}^{\frac{\pi}{2}} - \int \limits_{0}^{\frac{\pi}{2}} \left[\int \cos 2x \; dx \times \dfrac{d}{dx} \left(x^2\right)\right] dx \\\\ & = \dfrac{1}{2} \left[x^2 \; \sin 2x\right]_{0}^{\pi / 2} - \int \limits_{0}^{\frac{\pi}{2}} \dfrac{\sin 2x}{2} \times 2 x \; dx \\\\ & = \dfrac{1}{2} \left[\dfrac{\pi^2}{4} \; \sin \left(\pi\right) - 0\right] - \int \limits_{0}^{\frac{\pi}{2}} x \; \sin \left(2x\right) \; dx \\\\ & \hspace{10em} \left[\text{Here } u = x; \; v = \sin \left(2x\right)\right] \\\\ & = 0 - \left[x \int \sin \left(2x\right) \; dx\right]_{0}^{\pi / 2} + \int \limits_{0}^{\frac{\pi}{2}} \left[\sin \left(2x\right) \; dx \times \dfrac{d}{dx} \left(x\right)\right] \; dx \\\\ & = \dfrac{1}{2} \left[x \; \cos \left(2x\right)\right]_{0}^{\pi / 2} - \int \limits_{0}^{\frac{\pi}{2}} \dfrac{\cos \left(2x\right)}{2} \; dx \\\\ & = \dfrac{1}{2} \left[\dfrac{\pi}{2} \; \cos \left(0\right) - 0\right] - \dfrac{1}{4} \left[\sin \left(2x\right)\right]_{0}^{\pi / 2} \\\\ & = \dfrac{- \pi}{4} - \dfrac{1}{4} \left[\sin \left(\pi\right) - \sin \left(0\right)\right] \\\\ & = \dfrac{- \pi}{4} \end{aligned}$

Definite Integration

Evaluate $\displaystyle \int \limits_{0}^{1} \tan^{-1}x \; dx$

Let $I = \displaystyle \int \limits_{0}^{1} \tan^{-1}x \; dx$ $\;\;\; \cdots \; (1)$

Put $\tan^{-1}x = \theta$ $\;\;\; \cdots \; (2)$

Differentiating equation $(2)$ gives

$dx = \sec^2 \theta \; d\theta$ $\;\;\; \cdots \; (2a)$

From equation $(2)$, $x = \tan \theta$

When $x = 0$, $\theta = \tan^{-1}\left(0\right) = 0$ $\;\;\; \cdots \; (2b)$

When $x = 1$, $\theta = \tan^{-1} \left(1\right) = \dfrac{\pi}{4}$ $\;\;\; \cdots \; (2c)$

$\therefore$ $\;$ In view of equations $(2)$, $(2a)$, $(2b)$ and $(2c)$, equation $(1)$ can be written as

$\begin{aligned} I & = \int \limits_{0}^{\frac{\pi}{4}} \theta \; \sec^2 \theta \; d\theta \\\\ & \left[\begin{aligned} \text{Note: }& \int \limits_{a}^{b} u\; v \; dx = \left[u \int v \; dx\right]_{a}^{b} - \int \limits_{a}^{b} \left\{\int v \; dx \times \dfrac{d}{dx} \left(du\right) \right\} \; dx \\\\ & \text{Here } u = \theta; \; v = \sec^2 \theta \end{aligned}\right] \\\\ & = \left[\theta \int \sec^2 \; d\theta\right]_{0}^{\frac{\pi}{4}} - \int \limits_{0}^{\frac{\pi}{4}} \left[\int \sec^2 \theta \; d \theta \times \dfrac{d}{d \theta} \left(\theta\right)\right] d \theta \\\\ & = \left[\theta \; \tan \theta\right]_{0}^{\pi / 4} - \int \limits_{0}^{\frac{\pi}{4}} \tan \theta \; d \theta \\\\ & = \dfrac{\pi}{4} \tan \left(\dfrac{\pi}{4}\right) - 0 - \left[\log \left|\sec \theta\right|\right]_{0}^{\pi / 4} \\\\ & = \dfrac{\pi}{4} - \left[\log \left|\sec \left(\dfrac{\pi}{4}\right)\right| - \log \left|\sec \left(0\right)\right|\right] \\\\ & = \dfrac{\pi}{4} - \left[\log \sqrt{2} - \log 1\right] \\\\ & = \dfrac{\pi}{4} - \log \sqrt{2} = \dfrac{\pi}{4} - \dfrac{1}{2} \log 2 \end{aligned}$

Definite Integration

Evaluate $\displaystyle \int \limits_{0}^{\frac{\pi}{2}} \dfrac{\sin^2 x}{\left(1 + \cos x\right)^2} \; dx$

$\begin{aligned} \text{Let } I & = \int \limits_{0}^{\frac{\pi}{2}} \dfrac{\sin^2 x}{\left(1 + \cos x\right)^2} \; dx \\\\ & = \int \limits_{0}^{\frac{\pi}{2}} \dfrac{\left(1 - \cos^2 x\right) \; dx}{\left(1 + \cos x\right)^2} \\\\ & = \int \limits_{0}^{\frac{\pi}{2}} \dfrac{\left(1 + \cos x\right) \left(1 - \cos x\right) \; dx}{\left(1 + \cos x\right)^2} \\\\ & = \int \limits_{0}^{\frac{\pi}{2}} \dfrac{\left(1 - \cos x\right) \; dx}{1 + \cos x} \\\\ & = \int \limits_{0}^{\frac{\pi}{2}} \dfrac{2 \sin^2 \left(\dfrac{x}{2}\right)}{2 \cos^2 \left(\dfrac{x}{2}\right)} \; dx \\\\ & = \int \limits_{0}^{\frac{\pi}{2}} \tan^2 \left(\dfrac{x}{2}\right) \; dx \\\\ & = \int \limits_{0}^{\frac{\pi}{2}} \left[\sec^2 \left(\dfrac{x}{2}\right) - 1\right] \; dx \\\\ & = \int \limits_{0}^{\frac{\pi}{2}} \sec^2 \left(\dfrac{x}{2}\right) \; dx - \int \limits_{0}^{\frac{\pi}{2}} dx \\\\ & = 2 \left[\tan \left(\dfrac{x}{2}\right)\right]_{0}^{\pi / 2} - \left[x\right]_{0}^{\pi / 2} \\\\ & = 2 \left[\tan \left(\dfrac{\pi}{4}\right) - \tan \left(0\right)\right] - \left(\dfrac{\pi}{2} - 0\right) \\\\ & = 2 \left(1 - 0\right) - \dfrac{\pi}{2} = 2 - \dfrac{\pi}{2} \end{aligned}$

Definite Integration

Evaluate $\displaystyle \int \limits_{0}^{\sqrt{2}} \sqrt{2 - x^2} \; dx$

Let $I = \displaystyle \int \limits_{0}^{\sqrt{2}} \sqrt{2 - x^2} \; dx$ $\;\;\; \cdots \; (1)$

Let $x = \sqrt{2} \sin \theta$ $\;\;\; \cdots \; (2)$

Differentiating equation $(2)$ gives

$dx = \sqrt{2} \cos \theta \; d\theta$ $\;\;\; \cdots \; (2a)$

From equation $(2)$, $\theta = \sin^{-1} \left(\dfrac{x}{\sqrt{2}}\right)$

When $x = 0$, $\theta = \sin^{-1} \left(\dfrac{0}{\sqrt{2}}\right) = 0$ $\;\;\; \cdots \; (2b)$

When $x = \sqrt{2}$, $\theta = \sin^{-1} \left(\dfrac{\sqrt{2}}{\sqrt{2}}\right) = \sin^{-1} \left(1\right) = \dfrac{\pi}{2}$ $\;\;\; \cdots \; (2c)$

$\therefore$ $\;$ In view of equations $(2)$, $(2a)$, $(2b)$ and $(2c)$, equation $(1)$ becomes

$\begin{aligned} I & = \int \limits_{0}^{\frac{\pi}{2}} \sqrt{2 - 2 \sin^2 \theta} \times \sqrt{2} \cos \theta \; d\theta \\\\ & = \sqrt{2} \int \limits_{0}^{\frac{\pi}{2}} \sqrt{2 \left(1 - \sin^2 \theta\right)} \cos \theta \; d\theta \\\\ & = 2 \int \limits_{0}^{\frac{\pi}{2}} \cos \theta \times \cos \theta \; d\theta \\\\ & = 2 \int \limits_{0}^{\frac{\pi}{2}} \cos^2 \theta \; d\theta \\\\ & = 2 \int \limits_{0}^{\frac{\pi}{2}} \left(\dfrac{1 + \cos 2 \theta}{2}\right) \; d\theta \\\\ & = \int \limits_{0}^{\frac{\pi}{2}} d\theta + \int \limits_{0}^{\frac{\pi}{2}} \cos 2 \theta \; d\theta \\\\ & = \left[\theta\right]_{0}^{\pi / 2} + \dfrac{1}{2} \left[\sin 2 \theta\right]_{0}^{\pi / 2} \\\\ & = \dfrac{\pi}{2} - 0 + \dfrac{1}{2} \left(\sin \pi - \sin 0\right) \\\\ & = \dfrac{\pi}{2} + \dfrac{1}{2} \left(0 - 0\right) \\\\ & = \dfrac{\pi}{2} \end{aligned}$

Definite Integration

Evaluate $\displaystyle \int \limits_{0}^{\frac{\pi}{2}} \sin^2 x \; dx$

$\begin{aligned} \text{Let } I & = \int \limits_{0}^{\frac{\pi}{2}} \sin^2 x \; dx \\\\ & = \int \limits_{0}^{\frac{\pi}{2}} \left(\dfrac{1 - \cos 2x}{2}\right) \; dx \\\\ & = \dfrac{1}{2} \int \limits_{0}^{\frac{\pi}{2}} dx - \dfrac{1}{2} \int \limits_{0}^{\frac{\pi}{2}} \cos 2x \; dx \\\\ & = \dfrac{1}{2} \left[x\right]_{0}^{\pi / 2} - \dfrac{1}{2} \left[\dfrac{\sin 2x}{2}\right]_{0}^{\pi / 2} \\\\ & = \dfrac{1}{2} \left(\dfrac{\pi}{2} - 0\right) - \dfrac{1}{4} \left(\sin \pi - \sin 0\right) \\\\ & = \dfrac{\pi}{4} - \dfrac{1}{4} \left(0 - 0\right) \\\\ & = \dfrac{\pi}{4} \end{aligned}$

Definite Integration

Obtain $\displaystyle \int \limits_{1}^{3} x^3 \; dx$ as the limits of a sum.

Let $f\left(x\right) = x^3$ $\;\;\; \cdots \; (1)$

$f\left(x\right)$ is continuous on $\left[1, 3\right]$.

Lower limit $= a = 1$; Upper limit $= b = 3$

Divide $\left[1, 3\right]$ into n congruent sub-intervals.

Length of each sub-interval $= h = \dfrac{b - a}{n} = \dfrac{3 - 1}{n} = \dfrac{2}{n}$ $\;\;\; \cdots \; (2)$

$\begin{aligned} \text{Now, } f\left(a + kh\right) & = f\left(1 + kh\right) \\\\ & = \left(1 + k h\right)^3 \;\;\; \left[\text{by equation } (1)\right] \\\\ & = 1 + 3 k h + 3 k^2 h^2 + k^3 h^3 \;\;\; \cdots \; (3) \end{aligned}$

By definition, $\displaystyle \int \limits_{1}^{3} x^3 \; dx = \lim\limits_{n \to \infty} \left[h \sum \limits_{k = 1}^{n} f \left(a + k h\right)\right]$

$\begin{aligned} \therefore \; \int \limits_{1}^{3} x^3 \; dx & = \lim \limits_{n \to \infty} \left[h \sum_{k = 1}^{n} \left(1 + 3 k h + 3 k^2 h^2 + k^3 h^3\right)\right] \;\;\; \left[\text{by equation } (3)\right] \\\\ & = \lim\limits_{n \to \infty} \left[h \sum \limits_{k = 1}^{n} 1 + 3 h \sum \limits_{k = 1}^{n} k + 3 h^2 \sum \limits_{k = 1}^{n} k^2 + h^3 \sum \limits_{k = 1}^{n} k^3\right] \\\\ & = \lim\limits_{n \to \infty} \dfrac{2}{n} \left[n + 3 \times \dfrac{2}{n} \times \dfrac{n \left(n + 1\right)}{2} + 3 \times \left(\dfrac{2}{n}\right)^2 \times \dfrac{n \left(n + 1\right) \left(2 n + 1\right)}{6} \right. \\\\ & \hspace{11em} \left. + \left(\dfrac{2}{n}\right)^3 \times \dfrac{n^2 \left(n + 1\right)^2}{4}\right] \;\;\; \left[\text{by equation } (2)\right] \\\\ & = \lim\limits_{n \to \infty} \dfrac{2}{n} \left[n + 3 \left(n + 1\right) + \dfrac{2 \left(n + 1\right) \left(2 n + 1\right)}{n} + \dfrac{2 \left(n + 1\right)^2}{n}\right] \\\\ & = \lim\limits_{n \to \infty} \dfrac{2}{n} \left[4 n + 3 + \dfrac{2 \left(2 n^2 + 3 n + 1\right)}{n} + \dfrac{2 \left(n^2 + 2 n + 1\right)}{n}\right] \\\\ & = \lim\limits_{n \to \infty} \left[8 + \dfrac{6}{n} + 8 + \dfrac{12}{n} + \dfrac{4}{n^2} + 4 + \dfrac{8}{n} + \dfrac{4}{n^2}\right] \\\\ & \left[\text{As } n \to \infty, \; \dfrac{1}{n} \to 0 \right] \\\\ & = 8 + 8 + 4 = 20 \end{aligned}$

Definite Integration

Obtain $\displaystyle \int \limits_{0}^{\pi} \sin x \; dx$ $\;$ as the limit of a sum.

Let $f\left(x\right) = \sin x$ $\;\;\; \cdots \; (1)$

$f\left(x\right)$ is continuous on $\left[0, \pi\right]$.

Lower limit $= a = 0$; Upper limit $= b = \pi$

Divide $\left[0, \pi\right]$ into n congruent sub-intervals.

Length of each sub-interval $= h = \dfrac{b - a}{n} = \dfrac{\pi - 0}{n} = \dfrac{\pi}{n}$

$\implies$ $n h = \pi$ $\;\;\; \cdots \; (2)$

Now, $f\left(a + kh\right) = f\left(kh\right) = \sin \left(k h\right)$ $\;\;\; \cdots \; (3)$ $\;\;\;$ [from equation $(1)$]

Since $h = \dfrac{1}{n}$ $\implies$ when $n \rightarrow \infty, \; h \rightarrow 0$ $\;\;\; \cdots \; (4)$

By definition, $\displaystyle \int \limits_{0}^{\pi} \sin x \; dx = \lim\limits_{n \to \infty} \left[h \sum \limits_{k = 1}^{n} f \left(a + k h\right)\right]$

$\begin{aligned} \therefore \; \int \limits_{0}^{\pi} \sin x \; dx & = \lim\limits_{h \to 0} \left[h \sum \limits_{k = 1}^{n} \sin\left(k h\right)\right] \;\;\; \left[\text{by equations }(3) \text{ and }(4)\right] \\\\ & = \lim\limits_{h \to 0} h \left[\sin \left(h\right) + \sin \left(2 h\right) + \sin \left(3 h\right) + \cdots + \sin \left(n h\right)\right] \;\;\; \cdots \; (5) \end{aligned}$

Let $S_n = \sin \left(h\right) + \sin \left(2 h\right) + \sin \left(3 h\right) + \cdots + \sin \left(n h\right)$ $\;\;\; \cdots \; (6)$

Multiplying equation $(6)$ with $2 \sin \left(\dfrac{h}{2}\right)$ gives

$\begin{aligned} 2 \sin \left(\dfrac{h}{2}\right) S_n & = 2 \sin \left(h\right) \sin \left(\dfrac{h}{2}\right) + 2 \sin \left(2 h\right) \sin \left(\dfrac{h}{2}\right) \\\\ & \hspace{3em} + 2 \sin \left(3 h\right) \sin \left(\dfrac{h}{2}\right) + \cdots + 2 \sin \left(n h\right) \sin \left(\dfrac{h}{2}\right) \\\\ & \left[\text{Note: } 2 \sin A \; \sin B = \cos \left(A - B\right) - \cos \left(A + B\right)\right] \\\\ & = \cos \left(h - \dfrac{h}{2}\right) - \cos \left(h + \dfrac{h}{2}\right) + \cos \left(2 h - \dfrac{h}{2}\right) - \cos \left(2 h + \dfrac{h}{2}\right) \\\\ & \hspace{3em}+ \cos \left(3 h - \dfrac{h}{2}\right) - \cos \left(3 h + \dfrac{h}{2}\right) + \\\\ & \hspace{7em} \cdots + \cos \left(n h - \dfrac{h}{2}\right) - \cos \left(n h + \dfrac{h}{2}\right) \\\\ & = \cos \left(\dfrac{h}{2}\right) - \cos \left(\dfrac{3 h}{2}\right) + \cos \left(\dfrac{3 h}{2}\right) - \cos \left(\dfrac{5 h}{2}\right) \\\\ & \hspace{3em} + \cos \left(\dfrac{5 h}{2}\right) + \cdots + \cos \left(n h - \dfrac{h}{2}\right) - \cos \left(n h + \dfrac{h}{2}\right) \end{aligned}$

$\begin{aligned} \therefore \; S_n & = \dfrac{\cos \left(\dfrac{h}{2}\right) - \cos \left(n h + \dfrac{h}{2}\right)}{2 \sin \left(\dfrac{h}{2}\right)} \\\\ & = \dfrac{\cos \left(\dfrac{h}{2}\right) - \cos \left(\pi + \dfrac{h}{2}\right)}{2 \sin \left(\dfrac{h}{2}\right)} \;\;\; \left[\text{from equation }(2)\right] \\\\ & = \dfrac{\cos \left(\dfrac{h}{2}\right) + \cos \left(\dfrac{h}{2}\right)}{2 \sin \left(\dfrac{h}{2}\right)} \;\;\; \left[\text{Note: } \cos \left(\pi + \theta\right) = - \cos \theta\right] \\\\ & = \dfrac{2 \cos \left(h / 2\right)}{2 \sin \left(h / 2\right)} \\\\ & = \dfrac{\cos \left(h / 2\right)}{\sin \left(h / 2\right)} \;\;\; \cdots \; (7) \end{aligned}$

$\therefore$ $\;$ In view of equation $(7)$, equation $(5)$ becomes

$\begin{aligned} \int \limits_{0}^{\pi} \sin x \; dx & = \lim\limits_{h \to 0} \; h \times \dfrac{\cos \left(h / 2\right)}{\sin \left(h / 2\right)} \\\\ & = \lim\limits_{\frac{h}{2} \to 0} \; \cos \left(\dfrac{h}{2}\right) \times 2 \times \dfrac{1}{\lim\limits_{\frac{h}{2} \to 0} \; \dfrac{\sin \left(h / 2\right)}{h / 2}} \;\;\; \left[\text{Note: As } h \to 0, \; \dfrac{h}{2} \to 0\right] \\\\ & = 1 \times 2 \times 1 = 2 \end{aligned}$