# Definite Integral & Expansion of a Determinant |ISI QMS 2019 |QMB Problem 7(a)

Try this beautiful problem from ISI QMS 2019 exam. This problem requires knowledge of expansion of determinant and definite integral.

## Definite Integral and Expansion of Determinant - ISI QMS 2019 (QMB Problem 7a)

Let $f(x)= \begin{vmatrix}\sec x & \cos x & \sec^2 x+\cot x\textrm{cosec}^2 x\\ \cos^2x & \cos^2 x & \textrm{cosec}^2x \\ 1 & \cos^2 x & \textrm{cosec}^2 x \end{vmatrix}$. Then find $\displaystyle\int\limits_{\frac{\pi}{4}}^{\frac{\pi}{2}} f(x) \mathrm d x$

### Key Concepts

Expansion of Determinant

Evaluation of Definite Integral

## Check the Answer

Answer: $1-\frac{1}{\sqrt{2}}-\frac{\pi}{8}-\frac12\log 2$

ISI QMS 2019 (QMB Problem 7a)

Secrets in Inequalities.

## Try with Hints

Way to proceed with this problem is obtaining a simpler form of $f(x)$ by expanding the given determinant and then integrate it within the given limit.

You can do this ... Give it a try !!!!

Expanding the given determinant :

$f(x)= \begin{vmatrix}\sec x & \cos x & \sec^2 x+\cot x\textrm{cosec}^2 x\\ \cos^2x & \cos^2 x & \textrm{cosec}^2x \\ 1 & \cos^2 x & \textrm{cosec}^2 x \end{vmatrix}$

$=\textrm{sec} x[\cos^2 x \textrm{coesc}^2 x - \cos^x \textrm{coesc}^2 x ]-\\ \quad \cos x [\cos^2 x-\textrm{cosec}^2 x- \textrm{coesc}^2 x]+ \\ \quad (\textrm{sec}^2 x+\textrm{cot} x \textrm{coesc}^2 x)[\cos^4 x-cos^2 x]$

$=0+\cos x[ \textrm{coesc}^2 x (1-cos^2 x)]- \\ \quad (\textrm{sec}^2 x+\textrm{cot} x \textrm{coesc}^2 x)[\cos^2 x(1-\cos^2 x)]$

$=\cos x[\textrm{cosec}^2 x . \sin^2 x]- \\ \quad (\textrm{sec}^2 x+\textrm{cot} x \textrm{coesc}^2 x)[\cos^2x . \sin^2 x]$

$=\cos x-[\sin^2 x+\textrm{cot}x\cos^2 x]$

Therefore $f(x)= \cos x-\sin^2 x-\textrm{cot}x\cos^2 x$

Now integrating $f(x)$ within the given limits we will get our answer.

Integrating $f(x)$ within the given limits :

$I=\displaystyle\int\limits_{\frac{\pi}{4}}^{\frac{\pi}{2}} f(x)\quad \mathrm d x\\= \displaystyle\int\limits_{\frac{\pi}{4}}^{\frac{\pi}{2}} \cos x \quad\mathrm d x - \displaystyle\int\limits_{\frac{\pi}{4}}^{\frac{\pi}{2}} \sin^2 x \quad\mathrm d x - \displaystyle\int\limits_{\frac{\pi}{4}}^{\frac{\pi}{2}} \textrm{cot}x\cos^2 x \quad\mathrm d x \\= I_1-I_2-I_3 \text{[say]}$

Evaluating $I_1$ :

Now $I_1 = \displaystyle\int \limits_{\frac{\pi}{4}}^{\frac{\pi}{2}} \cos x \quad \mathrm d x = \sin x \bigg|_{{\pi}/{4}}^{{\pi}/{2}}= [\sin (\frac{\pi}{2})-\sin (\frac{\pi}{4})]=[1-\frac{1}{\sqrt 2}]$

Now can you find the values of $I_2 \text{and} I_3$ ?

Evaluating $I_2$ :

$I_2= \displaystyle\int\limits_{\frac{\pi}{4}}^{\frac{\pi}{2}} \sin^2 x \quad \mathrm d x$

$\quad = \frac12\displaystyle\int\limits_{\frac{\pi}{4}}^{\frac{\pi}{2}} 2\sin^2 x \quad \mathrm d x$

$\quad = \frac12\displaystyle\int\limits_{\frac{\pi}{4}}^{\frac{\pi}{2}}[1-\cos 2x]\quad\mathrm d x$

$\quad= \frac12\bigg[x-\frac {\sin2x}{2}\bigg]_{ \frac{\pi}{4}}^{\frac{\pi}{2}}$

$\quad =\frac12[\frac{\pi}{2}-\frac{\sin \pi}{2}-\frac{\pi}{4}+\frac{\sin \frac{\pi}{2}}{2}]$

$\quad= \frac{\pi}{8}+\frac14$

Evaluating $I_3$ :

$I_3= \displaystyle\int\limits_{\frac{\pi}{4}}^{\frac{\pi}{2}} \textrm{cot} x\cos^2 x \mathrm d x$

$\quad= \displaystyle\int\limits_{\frac{\pi}{4}}^{\frac{\pi}{2}} \frac{\cos x(1-\sin^2 x)}{\sin x} \mathrm d x$

$\quad = \displaystyle\int\limits_{\frac{\pi}{4}}^{\frac{\pi}{2}} \frac{\cos x}{\sin x} \mathrm d x \quad- \displaystyle\int\limits_{\frac{\pi}{4}}^{\frac{\pi}{2}} \sin x . \cos x \quad \mathrm d x$

$\quad = \displaystyle\int\limits_{\frac{\pi}{4}}^{\frac{\pi}{2}} \frac{\mathrm d (\sin x)}{\sin x}\quad- \displaystyle\int\limits_{\frac{\pi}{4}}^{\frac{\pi}{2}} \sin x\quad \mathrm d (\sin x)$

$\quad = \bigg[\log (\sin x)\quad-\quad \frac{\sin^2 x}{2}\bigg]_{\frac{\pi}{2}}^{\frac{\pi}{2}}$

$\quad= \bigg[\log (\sin \frac{\pi}{2})-\frac{\sin^2 \frac{\pi}{2}}{2}-\log (\sin \frac{\pi}{4}) + \frac{\sin^2 \frac{\pi}{4}}{2}\bigg]$

$\quad =\bigg[0-\frac12-\log \frac{1}{\sqrt 2}+\frac14\bigg]$

$\quad = \bigg[-\frac12 -\log (1) + \log (\sqrt 2) +\frac14 \bigg]$

$\quad = \bigg[\frac12\log 2- \frac14 \bigg]$

Therefore,

$\quad I= I_1-I_2-I_3$

$\Rightarrow I = 1-\frac{1}{\sqrt 2}-\frac{\pi}{8}-\frac14-\frac12\log(2) +\frac14$

$\Rightarrow I = 1 -\frac{\pi}{8}-\frac{1}{\sqrt 2}-\frac12\log(2)$ [ANS]

# Definite Integral | IIT JAM 2018 | Problem 4

Try this beautiful problem from IIT JAM 2018 which requires knowledge of the properties of Definite integral.

## Properties of Definite Integral -IIT JAM2018 (Problem 4)

Let $a$ be positive real number. If $f$ is a continuous and even function defined on the interval $[-a,a]$, then $\displaystyle\int_{-a}^a \frac{f(x)}{1+e^x} \mathrm d x$ is equal to :-

• $\displaystyle\int_0^a f(x) \mathrm d x$
• $2\displaystyle\int_0^a \frac{f(x)}{1+e^x}\mathrm d x$
• $2\displaystyle\int_0^a f(x) \mathrm d x$
• $2a\displaystyle\int_0^a \frac{f(x)}{1+e^x}\mathrm d x$

### Key Concepts

Definite Integral

Properties of definite Integral

Even function / Odd function

## Check the Answer

Answer: $\displaystyle\int_0^a f(x) \mathrm d x$

IIT JAM 2018, Problem 4

Definite and Integral calculus : R Courant

## Try with Hints

In this first I will give you the properties we need to solve this problem :

Property 1 : $\displaystyle\int_a^b f(x) \mathrm d x = \displaystyle\int_a^b f(a+b-x) \mathrm d x$

[Where $f$ is continuous on $[a,b]$]

Property 2 : If $f$ is an even function i.e., $f(x)=f(-x)$ then

$\displaystyle\int_{-a}^{a} f(x) \mathrm d x = 2 \displaystyle\int_{0}^{a} f(x) \mathrm d x$

Can you drive it from here !!!! Give it a try !!!

Let $I=\displaystyle\int_{-a}^a \frac{f(x)}{1+e^x} \mathrm d x \quad \ldots (i)$

$\Rightarrow I= \displaystyle\int_{-a}^a \frac{f(a-a-x)}{1+e^{(a-a-x)}} \mathrm d x$

[Since, $f$ is continuous then $\displaystyle\int_{a}^b f(x) \mathrm{d}x = \displaystyle\int_{a}^b f(a+b-x) \mathrm{d} x$]

$\Rightarrow I= \displaystyle\int_{-a}^a \frac{f(-x)}{1+e^{-x}} \mathrm d x$

$\Rightarrow I= \displaystyle\int_{-a}^a \frac{f(x)}{1+\frac{1}{e^x}} \mathrm d x$ [Since $f(x)$ is even]

$\Rightarrow I= \displaystyle\int_{-a}^a \frac{e^x.f(x)}{1+e^{x}} \mathrm d x \quad \ldots (ii)$

Adding $(i)$ and $(ii)$ we can get some interesting result !!!

Adding $(i)$ and $(ii)$ we get ,

$2I= \displaystyle\int_{-a}^a \frac{f(x)}{1+e^{x}} \mathrm d x + \displaystyle\int_{-a}^a \frac{e^x .f(x)}{1+e^{x}} \mathrm d x$

$\Rightarrow 2I = \displaystyle\int_{-a}^a \frac{[f(x)+e^x.f(x)]}{1+e^{x}} \mathrm d x$

$\Rightarrow 2I = \displaystyle\int_{-a}^a \frac{f(x)[1+e^x]}{[1+e^{x}]}$

$\Rightarrow 2I= \displaystyle\int_{-a}^a f(x) \mathrm d x$

$\Rightarrow 2I = 2\displaystyle\int_0^a f(x) \mathrm d x$ [Since $f(x)$ is even ]

$\Rightarrow I = \displaystyle\int_0^a f(x) \mathrm d x$ [ANS]