How 9 Cheenta students ranked in top 100 in ISI and CMI Entrances?
Learn More

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) $

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

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

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

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

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


$\quad I= I_1-I_2-I_3$

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

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

Subscribe to Cheenta at Youtube

Knowledge Partner

Cheenta is a knowledge partner of Aditya Birla Education Academy

Cheenta Academy

Aditya Birla Education Academy

Aditya Birla Education Academy

Cheenta. Passion for Mathematics

Advanced Mathematical Science. Taught by olympians, researchers and true masters of the subject.