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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 determinant and definite integral. Sequential hints are given here.

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]

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