Chapter 7: Integrals

Integration is the inverse process of differentiation. Instead of differentiating a function, we are given the derivative of a function and asked to find its primitive, i.e., the original function. Such a process is called integration or anti-differentiation.

Indefinite Integrals

If $\frac{d}{dx}[F(x)] = f(x)$, then we write $\int f(x) dx = F(x) + C$. Here $C$ is an arbitrary constant known as the constant of integration.

Some Standard Integrals

1. $\int x^n dx = \frac{x^{n+1}}{n+1} + C, \quad n \neq -1$
2. $\int \cos x dx = \sin x + C$
3. $\int \sin x dx = -\cos x + C$
4. $\int e^x dx = e^x + C$
5. $\int \frac{1}{x} dx = \log|x| + C$

Methods of Integration

1. Integration by Substitution: To evaluate $\int f(\phi(x))\cdot\phi’(x) dx$, put $\phi(x) = t \implies \phi’(x)dx = dt$. The integral becomes $\int f(t) dt$.

2. Integration by Parts: For the product of two functions $u$ and $v$: $$ \int u v dx = u \int v dx - \int \left( u’ \int v dx \right) dx $$ (The choice of first function $u$ and second function $v$ is typically guided by the rule “ILATE”, standing for Inverse Trigonometric, Logarithmic, Algebraic, Trigonometric, Exponential).

3. Integration by Partial Fractions: Used for rational functions of the form $\frac{P(x)}{Q(x)}$ where $P(x)$ and $Q(x)$ are polynomials.

Definite Integrals

The definite integral $\int_a^b f(x) dx$ represents the area bounded by the curve $y = f(x)$, the ordinates $x = a$, $x = b$, and the $x$-axis. Fundamental Theorem of Calculus: Let $f$ be continuous function defined on the closed interval $[a, b]$ and $F$ be an anti-derivative of $f$. Then $\int_a^b f(x) dx = [F(x)]_a^b = F(b) - F(a)$.

There are several properties of definite integrals, such as: $\int_a^b f(x) dx = \int_a^b f(a + b - x) dx$ $\int_0^a f(x) dx = \int_0^a f(a - x) dx$

Competency-Based Questions

1. [CBSE 2022] Evaluate: $\int \frac{x \sin^{-1} x}{\sqrt{1 - x^2}} dx$

Solution:

Let $I = \int \frac{x \sin^{-1} x}{\sqrt{1 - x^2}} dx$

Let $\sin^{-1} x = t \implies \frac{1}{\sqrt{1 - x^2}} dx = dt$. Also, $x = \sin t$.

So the integral becomes:

$I = \int (\sin t) \cdot t dt = \int t \cdot \sin t dt$

Now use Integration by Parts (taking $t$ as first function and $\sin t$ as second function):

$I = t \int \sin t dt - \int (\frac{d}{dt}(t) \cdot \int \sin t dt) dt$

$I = t (-\cos t) - \int 1 \cdot (-\cos t) dt$

$I = -t \cos t + \int \cos t dt$

$I = -t \cos t + \sin t + C$

Substitute back $t = \sin^{-1} x$.

Also $\sin t = x \implies \cos t = \sqrt{1 - \sin^2 t} = \sqrt{1 - x^2}$.

$I = -(\sin^{-1} x)\sqrt{1 - x^2} + x + C$.

2. [Sample Paper 2021] Evaluate: $\int_0^{\pi/2} \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} dx$

Solution:

Let $I = \int_0^{\pi/2} \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} dx \quad \dots \text{(Eq i)}$

Using the property $\int_0^a f(x) dx = \int_0^a f(a - x) dx$:

$I = \int_0^{\pi/2} \frac{\sqrt{\sin(\pi/2 - x)}}{\sqrt{\sin(\pi/2 - x)} + \sqrt{\cos(\pi/2 - x)}} dx$

$I = \int_0^{\pi/2} \frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}} dx \quad \dots \text{(Eq ii)}$

Adding (Eq i) and (Eq ii):

$2I = \int_0^{\pi/2} \frac{\sqrt{\sin x} + \sqrt{\cos x}}{\sqrt{\sin x} + \sqrt{\cos x}} dx$

$2I = \int_0^{\pi/2} 1 dx = [x]_0^{\pi/2} = \pi/2$

Therefore, $I = \pi/4$.

3. [CBSE 2019] Evaluate: $\int e^x \left(\frac{1 + \sin x}{1 + \cos x}\right) dx$

Solution:

Let $I = \int e^x \left(\frac{1 + \sin x}{1 + \cos x}\right) dx$

$I = \int e^x \left(\frac{1 + 2\sin(x/2)\cos(x/2)}{2\cos^2(x/2)}\right) dx$

$I = \int e^x \left( \frac{1}{2\cos^2(x/2)} + \frac{2\sin(x/2)\cos(x/2)}{2\cos^2(x/2)} \right) dx$

$I = \int e^x \left( \frac{1}{2} \sec^2(x/2) + \tan(x/2) \right) dx$

Let $f(x) = \tan(x/2)$. Then $f’(x) = \frac{1}{2}\sec^2(x/2)$.

We know the standard integral form: $\int e^x [f(x) + f’(x)] dx = e^x f(x) + C$

Therefore, $I = e^x \tan(x/2) + C$.