Indefinite Integration 3 Question 17
17. $f(x)=\left|\begin{array}{ccc}\sec x & \cos x & \sec ^{2} x+\cot x \operatorname{cosec} x \ \cos ^{2} x & \cos ^{2} x & \operatorname{cosec}^{2} x \ 1 & \cos ^{2} x & \cos ^{2} x\end{array}\right|$.
Then, $\int _0^{\pi / 2} f(x) d x=\ldots$.
$(1987,2 M)$
Analytical & Descriptive Questions
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Answer:
Correct Answer: 17. $-\frac{15 \pi+32}{60}$
Solution:
- Given, $f(x)=\left|\begin{array}{ccc}\sec x & \cos x & \operatorname{cosec} x \cdot \cot x+\sec ^{2} x \ \cos ^{2} x & \cos ^{2} x & \operatorname{cosec}^{2} x \ 1 & \cos ^{2} x & \cos ^{2} x\end{array}\right|$ Applying $R _3 \rightarrow \frac{1}{\cos x} R _3$,
$f(x)=\cos x\left|\begin{array}{ccc}\sec x & \cos x & \operatorname{cosec} x \cdot \cot x+\sec ^{2} x \ \cos ^{2} x & \cos ^{2} x & \operatorname{cosec}^{2} x \ \sec x & \cos x & \cos x\end{array}\right|$
Applying $R _1 \rightarrow R _1-R _3 \Rightarrow f(x)$
$$ =\cos x\left|\begin{array}{ccc} 0 & 0 & \operatorname{cosec} x \cdot \cot x+\sec ^{2} x-\cos x \\ \cos ^{2} x & \cos ^{2} x & \operatorname{cosec}^{2} x \\ \sec x & \cos x & \cos x \end{array}\right| $$
$$ \begin{aligned} & =\left(\operatorname{cosec} x \cdot \cot x+\sec ^{2} x-\cos x\right) \cdot\left(\cos ^{3} x-\cos x\right) \cdot \cos x \\ & =-\frac{\sin ^{2} x+\cos ^{3} x-\cos ^{3} x \cdot \sin ^{2} x}{\sin ^{2} x \cdot \cos ^{2} x} \cdot \cos ^{2} x \cdot \sin ^{2} x \\ & =-\sin ^{2} x-\cos ^{3} x\left(1-\sin ^{2} x\right)=-\sin ^{2} x-\cos ^{5} x \\ & \therefore \quad \int _0^{\pi / 2} f(x) d x=-\int _0^{\pi / 2}\left(\sin ^{2} x+\cos ^{5} x\right) d x \\ & \because \int _0^{\pi / 2} \sin ^{m} x \cdot \cos ^{n} x d x=\frac{\sqrt{\frac{m+1}{2}} \sqrt{\frac{n+2}{2}}}{2 \sqrt{\frac{m+n+2}{2}}} \\ & \int _0^{\pi / 2} f(x) d x=-\frac{\sqrt{\frac{3}{2}} \cdot \sqrt{\frac{1}{2}}}{2 \sqrt{2}}+\frac{\sqrt{\frac{6}{2}} \cdot \sqrt{\frac{1}{2}}}{2 \sqrt{\frac{7}{2}}} \\ & =-\frac{1 / 2 \cdot \pi}{2}+\frac{2 \sqrt{\pi}}{2 \cdot \frac{5}{2} \cdot \frac{3}{2} \cdot \frac{1}{2} \cdot \sqrt{\pi}}=-\frac{\pi}{4}+\frac{8}{15}=-\frac{15 \pi+32}{60} \end{aligned} $$
