# Complex Analysis Questions and Answers – Evaluation of Real Integrals

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) $$\frac{\pi}{\sqrt{a^{3}}}$$

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) $$\frac{\pi}{\sqrt{5}}$$

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) $$\frac{2\pi}{3}$$

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) $$\frac{π}{3}$$

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) $$\frac{π}{3}$$

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

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) $$\frac{π}{4}$$

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) $$\frac{π}{5}$$

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) $$\frac{π}{6}$$

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) $$\frac{7}{3}$$

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) $$\frac{π}{4}$$

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) $$\frac{π}{6}$$

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) $$\frac{π}{6}$$

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) $$\frac{π}{8}$$

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) $$\frac{π}{8}$$

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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