### Definite Integrals

• Find the definite integral of algebraic, trigonometric/inverse trigonometric, exponential and logarithmic functions.

• Use properties of definite integrals.

### Solved Example:

#### 2-8-01

$\int_{-a}^{a}(\sin^6 x + \sin^7 x)dx = ?$

Solution:
A function is said to be even function if: $f(-x) = f(x)$
A function is said to be odd function if: $f(-x) = -f(x)$
Here, $\sin(x)$ is an odd function because: $\sin(-x) = - \sin x$
$\sin^6 x$, being an even power of $\sin x$ is an even function.
$\sin^7 x$, being an odd power of $\sin x$ is an odd function. \begin{align*} \int_{-a}^{a}(\sin^6 x + \sin^7 x)dx &= \int_{-a}^{a}(\sin^6 x) dx + \int_{-a}^{a}(\sin^7 x)dx\\ &= \int_{-a}^{a}(\sin^6 x) dx + 0\\ &= 2 \int_{0}^{a}(\sin^6 x) dx \end{align*}

### Solved Example:

#### 2-8-02

The value of the integral: $\int ^{\infty }_{-\infty }\dfrac {dx}{1+x^{2}}$ is:

Solution:
\begin{align*} I &=\int ^{\infty }_{-\infty }\dfrac {dx}{1+x^{2}}\\ &=\left[ \tan ^{-1}x\right] ^{\infty }_{-\infty }\\ &=\left[ \tan ^{-1}\left(\infty \right) -\tan ^{-1}\left( -\infty \right) \right]\\ &=\dfrac {\pi }{2}-\left( -\dfrac {\pi }{2}\right)\\ &=\pi \end{align*}

### Solved Example:

#### 2-8-03

If f (x) is an even function and a is a positive real number, then $\int ^{a}_{-a}f\left( x\right) dx$ equals:

Solution:
For a function, whose limits bounded between -a to a and a is a positive real number. The solution is given by:
If f(x) is an even function, which means if f(-x) = f(x) $\int ^{a}_{-a}f\left( x\right) dx=2\int ^{a}_{0}f\left( x\right) dx$
If f(x) is an odd function, which means if f(-x) = - f(x) $\int ^{a}_{-a}f\left( x\right) dx=0$

### Solved Example:

#### 2-8-04

Let f be a continuous and +ve real valued function on [0, 1]. Then $\int_0^\pi f(\sin x) \cos x\ dx = ?$

Solution:
Let t = $\sin x$, So, dt = $\cos x\ dx$
When x = 0, t = 0, When x = $\pi$, t = 0
$\int_0^\pi f(\sin x) \cos x\ dx$ becomes $\int_0^0 f(t)\ dt$
Since, both upper and lower limits are identical, the value of the integral is zero.

### Solved Example:

#### 2-8-05

The value of: $\int_{-1}^1 x e^{\lvert x \rvert} dx$ is: (GATE Civil 2021)

Solution: $f(x) = x e^{\lvert x \rvert}$ $f(-x) = -x e^{\lvert -x \rvert} = -x e^{\lvert x \rvert} = -f(x)$

$x e^{\lvert x \rvert}$ is an odd function because f(-x) = -f(x), also the limits are symmetric about zero. Hence, the value of the integral between -1 to 0 cancels out the one from 0 to +1.