This set of Complex Analysis Multiple Choice Questions & Answers (MCQs) focuses on “Evaluation of Real Integrals”.
1. Which of the following values is obtained by evaluating \( \int_{0}^{\pi}{\frac{1}{a + \ bcos\ \theta\ }}, \) dθ , a > |b|?
a) \(\frac{\pi}{\sqrt{a^{2} + b^{2}}}\)
b) \(\frac{\pi}{\sqrt{a^{2} – b^{2}}}\)
c) \(\frac{\pi}{\sqrt{b^{2} – a^{2}}}\)
d) \(\frac{\pi}{\sqrt{a^{3}}}\)
View Answer
Explanation:\(\int_{0}^{\pi}{\frac{1}{a + b\ cos\theta}\text{dθ}} = \ \frac{1}{2}\int_{0}^{2\pi}{\frac{1}{a + bcos\ \theta}\text{dθ}}\)
\( = \frac{1}{2}\left\lbrack \frac{2\pi}{\sqrt{a^{2} – b^{2}}} \right\rbrack\) \( = \frac{\pi}{\sqrt{a^{2} – b^{2}}},\ \ \ a > |b|\)
[∴If f2a-x= fx, then , \( \int_{0}^{2a}{f(x)\ dx = 2\int_{0}^{a}{f(x) dx\rbrack}}\)
[Here,cos 2π-θ =cos θ]
2. Which of the following values is obtained by evaluating \( \int_{0}^{2\pi}{\frac{1}{2 + \cos\theta\ }\text{dθ}}\)?
a) \(\frac{2\pi}{\sqrt{3}}\)
b) \(\frac{\pi}{\sqrt{3}}\)
c) \(\frac{2\pi}{\sqrt{5}}\)
d) \(\frac{\pi}{\sqrt{5}}\)
View Answer
Explanation: We know that
\(\int_{0}^{2\pi}{\frac{1}{a + bcos \theta}d\theta = \frac{2\pi}{\sqrt{a^{2} – b^{2}}}},\ a > |b|\)
Here, a=2, b=1, 2>|1|
\(\sqrt{a^{2} – b^{2}} = \sqrt{4 – 1} = \sqrt{3}\)
Therefore \(\int_{0}^{2\pi}{\frac{1}{2 + \cos\theta}d\theta = \ \frac{2\pi}{\sqrt{3}}}\)
3. Which of the following values is obtained by evaluating \( \int_{0}^{2\pi}{\frac{1}{13 + 5\cos\theta\ }\text{dθ}}\)?
A) \(\frac{\pi}{6}\)
b) \(\frac{\pi}{3}\)
c) \(\frac{\pi}{2}\)
d) \(\frac{2\pi}{3}\)
View Answer
Explanation:
\(\int_{0}^{2\pi}{\frac{1}{a + bcos\ \theta}d\theta = \frac{2\pi}{\sqrt{a^{2} – b^{2}}}},\ a > |b|\)
Here, a=13, b=5, 13>|5|
\(\sqrt{a^{2} – b^{2}} = \sqrt{169 – 25} = 12\)
\(\int_{0}^{2\pi}{\frac{1}{13 + 5\cos\theta\ }\ d\theta = \ \frac{2\pi}{12} = \frac{\pi}{6}}\)
4. Which of the following values is obtained by evaluating \( \int_{0}^{2\pi}{\ \frac{1}{5 + 4\ \cos\theta}\text{dθ}}\)?
a) \(\frac{2π}{3}\)
b) \(\frac{π}{6}\)
C) \(\frac{π}{2}\)
d) \(\frac{π}{3}\)
View Answer
Explanation: \(\int_{0}^{2\pi}{\frac{1}{a + bcos\ \theta}d\theta = \frac{2\pi}{\sqrt{a^{2} – b^{2}}}},\ a > |b|\)
Here, a=5, b=4, 5>|4|
\(\sqrt{a^{2} – b^{2}} = \sqrt{25 – 16} = 3\)
\(\int_{0}^{2\pi}{\ \frac{1}{5 + 4\ \cos\theta}\text{dθ}} = \frac{2\pi}{3}\).
5. Which of the following values is obtained by evaluating \( \int_{0}^{2\pi}{\ \frac{1}{13 + 12\ \cos\theta}\text{dθ}}\)?
a) \(\frac{2π}{5}\)
b) \(\frac{π}{6}\)
c) \(\frac{π}{4}\)
d) \(\frac{π}{3}\)
View Answer
Explanation:
\(\int_{0}^{2\pi}{\frac{1}{a + bcos\ \theta}d\theta = \frac{2\pi}{\sqrt{a^{2} – b^{2}}}},\ a > |b|\)
Here, a=13, b=12, 13>|12|
\(\sqrt{a^{2} – b^{2}} = \sqrt{169 – 144} = 5\)
\(\int_{0}^{2\pi}{\frac{1}{13 + 12\cos\theta}d\theta = \ \frac{2\pi}{5}}\)
6. Which of the following values is obtained by evaluating \(\int_{0}^{2\pi}{\ \frac{\cos 2\theta}{5 + 4\ \cos\theta}\text{dθ}}\)?
a) \(\frac{π}{3}\)
b) \(\frac{π}{6}\)
c) \(\frac{π}{4}\)
d) 0
View Answer
Explanation:
We know that
\( \int_{0}^{2\pi}{\frac{\cos\text{mθ}}{a + b\cos\theta}d\theta = \frac{4\pi}{b}\left\lbrack \frac{\alpha^{m}}{\alpha – \beta} \right\rbrack\ }\)
\(\int_{0}^{2\pi}{\frac{\cos{2\theta}}{a + b\cos\theta}d\theta = \frac{4\pi}{b}\left\lbrack \frac{\alpha^{2}}{\alpha – \beta} \right\rbrack\ }\)
\(a = 5,\ b = 4,\ 5 > |4|,\ m = 2\)
\(\alpha = \frac{- a + \sqrt{a^{2} – b^{2}}}{b} = \ \frac{- 5 + \sqrt{25 – 16}}{4} = \frac{- 1}{2}\)
\(\alpha – \beta = \frac{2\sqrt{a^{2} – b^{2}}}{b} = \ \frac{2(3)}{4} = \frac{3}{2}\)
\(\int_{0}^{2\pi}{\ \frac{\cos 2\theta}{5 + 4\ \cos\theta}d\theta = \ \frac{4\pi}{4}\left\lbrack \frac{\left( – \frac{1}{2} \right)^{2}}{\frac{3}{2}} \right\rbrack = \ \frac{\pi}{6}}\)
7. Which of the following values is obtained by evaluating \( \int_{0}^{2\pi}{\ \frac{\cos 2\theta}{5 – 4\ \cos\theta}\text{dθ}}\)?
a) \(\frac{π}{6}\)
b) \(\frac{π}{3}\)
c) \(\frac{π}{2}\)
d) \(\frac{π}{4}\)
View Answer
Explanation: We know that
\(\int_{0}^{2\pi}{\frac{\cos\text{mθ}}{a + b\cos\theta}d\theta = \frac{4\pi}{b}\left\lbrack \frac{\alpha^{m}}{\alpha – \beta} \right\rbrack\ }\)
Here, a=5, b=-4, 5>4, m=2
\(\int_{0}^{2\pi}{\ \frac{\cos 2\theta}{5 – 4\ \cos\theta}\text{dθ}} = \ \frac{4\pi}{b}\left\lbrack \frac{\alpha^{2}}{\alpha – \beta} \right\rbrack\ \)
\(\alpha = \frac{- a + \sqrt{a^{2} – b^{2}}}{b} = \ \frac{- 5 + \sqrt{25 – 16}}{- 4} = \frac{1}{2}\)
\(\alpha – \beta = \ \frac{2\sqrt{a^{2} – b^{2}}}{b} = \ \frac{2(3)}{- 4} = \frac{- 3}{2}\)
\(\int_{0}^{2\pi}{\ \frac{\cos 2\theta}{5 – 4\ \cos\theta}\text{dθ}} = \ \frac{4\pi}{- 4}\left\lbrack \frac{\left( \frac{1}{2} \right)^{2}}{\left( – \frac{3}{2} \right)} \right\rbrack = \ \frac{\pi}{6}\)
8. Which of the following values is obtained by evaluating \(\int_{0}^{2\pi}{\ \frac{\cos 3\theta}{5 – 4\ \cos\theta}\text{dθ}}\)?
a) \(\frac{π}{12}\)
b) \(\frac{π}{3}\)
c) \(\frac{π}{4}\)
d) \(\frac{π}{5}\)
View Answer
Explanation:
Here, a=5, b=-4, 5>-4, m=3
\(\int_{0}^{2\pi}{\frac{\cos 3\theta}{5 – 4\ \cos\theta}d\theta = \ }\frac{4\pi}{b}\left\lbrack \frac{\alpha^{3}}{\alpha – \beta} \right\rbrack\)
\(\alpha = \frac{- a + \sqrt{a^{2} – b^{2}}}{b} = \ \frac{- 5 + \sqrt{25 – 16}}{- 4} = \frac{1}{2}\)
\(\alpha – \beta = \ \frac{2\sqrt{a^{2} – b^{2}}}{b} = \ \frac{2(3)}{- 4} = \frac{- 3}{2}\)
\(\int_{0}^{2\pi}{\ \frac{\cos 3\theta}{5 – 4\ \cos\theta}\text{dθ}} = \frac{4\pi}{- 4}\ \left\lbrack \frac{\left( \frac{1}{2} \right)^{3}}{\left( – \frac{3}{2} \right)} \right\rbrack = \ \frac{\pi}{12}\)
9. Which of the following values is obtained by evaluating \(\int_{0}^{2\pi}{\ \frac{\sin^{2}\theta}{5 + 4\cos\theta}\text{dθ}}\)?
a) \(\frac{π}{4}\)
b) \(\frac{π}{2}\)
c) \(\frac{π}{3}\)
d) \(\frac{π}{6}\)
View Answer
Explanation:
\(\int_{0}^{2\pi}{\ \frac{\sin^{2}\theta}{5 + 4\cos\theta}\text{dθ}} = \ \frac{1}{2}\int_{0}^{2\pi}{\frac{1 – \cos{2\theta}}{5 + 4\cos\theta}\text{dθ}}\)
\( = \frac{1}{2}\ \frac{4\pi}{b}\ \left\lbrack \frac{1 – \alpha^{2}}{\alpha – \beta} \right\rbrack = \ \frac{2\pi}{b}\left\lbrack \frac{1 – \alpha^{2}}{\alpha – \beta} \right\rbrack\)
Here, a=5, b=4, 5>4, m=2
\(\alpha = \frac{- a + \sqrt{a^{2} – b^{2}}}{b} = \ \frac{- 5 + \sqrt{25 – 16}}{4} = \frac{- 1}{2}\ \)
\(\alpha – \beta = \ \frac{2\sqrt{a^{2} – b^{2}}}{b} = \ \frac{2(3)}{4} = \frac{3}{2}\)
\(\int_{0}^{2\pi}{\ \frac{\sin^{2}\theta}{5 + 4\cos\theta}\text{dθ}} = \frac{2\pi}{4}\left\lbrack \frac{1 – \left( – \frac{1}{2} \right)^{2}}{\left( \frac{3}{2} \right)} \right\rbrack = \ \frac{\pi}{4}\ \)
10. Which of the following values is obtained by evaluating \(\int_{0}^{\pi}{\ \frac{1 + 2\cos\theta}{5 + 4\cos\theta}\text{dθ}}\)?
a) 0
b) \(\frac{1}{3}\)
c) \(\frac{11}{3}\)
d) \(\frac{7}{3}\)
View Answer
Explanation:
\(\int_{0}^{\pi}{\ \frac{1 + 2\cos\theta}{5 + 4\cos\theta}\text{dθ}} = \ \frac{1}{2}\int_{0}^{2\pi}{\ \frac{1 + 2\cos\theta}{5 + 4\cos\theta}\text{dθ}}\)
\(\ \frac{1}{2}\ \frac{4\pi}{b}\ \left\lbrack \frac{1 + 2\alpha}{\alpha – \beta} \right\rbrack = \ \frac{2\pi}{b}\left\lbrack \frac{1 + 2\alpha}{\alpha – \beta} \right\rbrack\ \)
Here, a=5, b=4, 5>4, m=1
\(\alpha = \frac{- a + \sqrt{a^{2} – b^{2}}}{b} = \ \frac{- 5 + \sqrt{25 – 16}}{4} = \frac{- 1}{2}\ \)
\(\alpha – \beta = \ \frac{2\sqrt{a^{2} – b^{2}}}{b} = \ \frac{2(3)}{4} = \frac{3}{2}\)
\(\int_{0}^{\pi}{\ \frac{1 + 2\cos\theta}{5 + 4\cos\theta}\text{dθ}} = \ \frac{2\pi}{4}\left\lbrack \frac{1 + 2\left( – \frac{1}{2} \right)}{\left( \frac{3}{2} \right)} \right\rbrack = 0\ \)
11. Which of the following values is obtained by evaluating \( \int_{0}^{2\pi}{\ \frac{1}{5 – 4\sin\theta}\text{dθ}}\)?
a) \(\frac{2π}{3}\)
b) \(\frac{π}{3}\)
c) \(\frac{2π}{5}\)
d) \(\frac{π}{4}\)
View Answer
Explanation:
\(\int_{0}^{2\pi}{\frac{1}{a + bsin\ \theta}d\theta = \frac{2\pi}{\sqrt{a^{2} – b^{2}}}},\ a > |b|\)
Here, a=5, b=-4, 5>|-4|
\(\int_{0}^{2\pi}{\frac{1}{5 – 4\sin\theta}d\theta = \ \frac{2\pi}{\sqrt{25 – 16}} = \ \frac{2\pi}{3}}\)
12. Which of the following values is obtained by evaluating \(\int_{0}^{2\pi}{\frac{1}{13 + 5\sin\theta}\text{dθ}}\)?
a) \(\frac{π}{4}\)
b) \(\frac{π}{2}\)
c) \(\frac{π}{3}\)
d) \(\frac{π}{6}\)
View Answer
Explanation:
\(\int_{0}^{2\pi}{\frac{1}{a + bsin\ \theta}d\theta = \frac{2\pi}{\sqrt{a^{2} – b^{2}}}},\ a > |b|\)
\(\sqrt{a^{2} – b^{2}} = \ \sqrt{169 – 25} = 12\)
\(\int_{0}^{2\pi}{\frac{1}{13 + 5\sin\theta}d\theta = \ \frac{2\pi}{12} = \ \frac{\pi}{6}}\)
13. Which of the following values is obtained by evaluating \(\int_{- \infty}^{\infty}{\frac{x^{2}}{\left( x^{2} + 1 \right)(x^{2} + 4)}\text{dx}}\)?
a) \(\frac{π}{4}\)
b) \(\frac{π}{2}\)
c) \(\frac{π}{3}\)
d) \(\frac{π}{6}\)
View Answer
Explanation:
\(\ \int_{- \infty}^{\infty}{\frac{x^{2}}{\left( x^{2} + a^{2} \right)(x^{2} + b^{2})} dx = \frac{\pi}{a + b\ }}\)
Here, a2=1→ a=1, b2=4→b=2
\(\ \int_{- \infty}^{\infty}{\frac{x^{2}}{\left( x^{2} + 1 \right)(x^{2} + 4)}\text{dx}} = \frac{\pi}{1 + 2} = \ \frac{\pi}{3}\)
14. Which of the following values is obtained by evaluating \(\int_{0}^{\infty}{\frac{x^{2}}{\left( x^{2} + 9 \right)(x^{2} + 4)}\text{dx}}\)?
a) \(\frac{π}{10}\)
b) \(\frac{π}{4}\)
c) \(\frac{π}{9}\)
d) \(\frac{π}{8}\)
View Answer
Explanation:
\(\int_{- \infty}^{\infty}{\frac{x^{2}}{\left( x^{2} + a^{2} \right)(x^{2} + b^{2})}\ dx = \ \frac{\pi}{a + b\ }}\)
Here, a2=9 →a = 3 , b2 = 4 → b = 2
\(\int_{0}^{\infty}{\frac{x^{2}}{\left( x^{2} + 9 \right)(x^{2} + 4)}\text{dx}} = \frac{1}{2}\left( \frac{\pi}{3 + 2} \right) = \ \frac{\pi}{10}\)
15. Which of the following values is obtained by evaluating \( \int_{- \infty}^{\infty}{\frac{1}{\left( x^{2} + 1 \right)(x^{2} + 4)}\text{dx}}\)?
a) \(\frac{π}{10}\)
b) \(\frac{π}{4}\)
c) \(\frac{π}{9}\)
d) \(\frac{π}{8}\)
View Answer
Explanation:
\(\int_{- \infty}^{\infty}{\frac{1}{\left( x^{2} + a^{2} \right)(x^{2} + b^{2})}\ dx = \frac{\pi}{2ab(a + b)\ }}\)
Here, a2=1 →a = 1 , b2 = 4 → b = 2
\(\int_{- \infty}^{\infty}{\frac{1}{\left( x^{2} + 1 \right)(x^{2} + 4)}\ dx = \frac{\pi}{(1)(2)(1 + 2)\ } = \ \frac{\pi}{6}\ }\)
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