Mathematics

Evaluate: $$\displaystyle \int_{0}^{\pi} \dfrac {x}{a^{2} \cos^{2}x + b^{2}\sin^{2}x} dx$$


SOLUTION
We have, $$I = \displaystyle \int_{0}^{\pi} \dfrac {x}{a^{2} \cos^{2}x + b^{2} \sin^{2}x}dx$$  .... (i)
$$I =\displaystyle  \int_{0}^{\pi} \dfrac {\pi - x}{a^{2}\cos^{2} (\pi - x) + b^{2}\sin^{2} (\pi - x)}dx$$
$$\left (\because \displaystyle \int_{0}^{a} f(x) dx = \int_{0}^{a} f(a - x)dx\right )$$
$$I = \displaystyle \int_{0}^{\pi} \dfrac {\pi - x}{a^{2}\cos^{2}x + b^{2}\sin x}dx$$  .... (ii)
On adding equations (i) and (ii), we get
$$2I = \displaystyle \int_{0}^{\pi} \dfrac {x + \pi - x}{a^{2}\cos^{2} x + b^{2} \sin^{2}x}dx$$
$$2I = \pi \displaystyle \int_{0}^{\pi} \dfrac {1}{a^{2} \cos^{2}x + b^{2}\sin^{2} x}dx$$
Using the property, $$\int_{0}^{2a} f(x) \cdot d = \int_{0}^{a} [f(x) + f(2a - x)]dx$$
$$\therefore 2I = \pi \left [\displaystyle \int_{0}^{\pi /2} \dfrac {1}{a^{2}\cos^{2} x + b^{2} \sin^{2} x} dx + \int_{0}^{\pi/2} \dfrac {1}{a^{2}\cos^{2} (\pi - x) + b^{2}\sin^{2} (\pi - x)}dx\right ]$$
$$I = \pi \displaystyle \int_{0}^{\pi/2} \dfrac {1}{a^{2}\cos^{2} x + b^{2}\sin^{2} x}dx = \pi \int_{0}^{\pi/2} \dfrac {dx}{a^{2}\cos^{2}x\left (1 + \dfrac {b^{2}}{a^{2}}\tan^{2}x\right )}$$
$$= \dfrac {\pi}{a^{2}} \displaystyle \int_{0}^{\pi/2} \dfrac {\sec^{2}xdx}{1 + \dfrac {b^{2}}{a^{2}}\tan^{2}x}$$ [By putting $$u = \dfrac {b}{a}\tan x, du = \dfrac {b}{a}\sec^{2}x dx$$ when $$x = 0, u = 0$$ when $$x = \dfrac {\pi}{2}, u = \infty]$$
$$= \dfrac {\pi}{a^{2}} \displaystyle \int_{0}^{\infty} \dfrac {\dfrac {a}{b}du}{1 + u^{2}}$$
$$= \dfrac {\pi}{ab} \displaystyle \int_{0}^{\infty} \dfrac {du}{1 + u^{2}} = \dfrac {\pi}{ab} [\tan^{-1} (u)]_{0}^{\infty} + c$$
$$= \dfrac {\pi}{ab} [\tan^{-1} \infty - \tan^{-1} 0] + c$$
$$= \dfrac {\pi}{ab} \left [\dfrac {\pi}{2} - 0\right ] + c$$
$$= \dfrac {\pi^{2}}{2ab}$$
View Full Answer

Its FREE, you're just one step away


Subjective Hard Published on 17th 09, 2020
Next Question
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 114
Enroll Now For FREE

Realted Questions

Q1 Subjective Medium
$$\int_{0}^{4}x+e^{x} dx $$

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer
Q2 Single Correct Medium
Value of $${\int}_{-\pi}^{\dfrac{17\pi}{2}}\left(\left|\sin x\right|\right)dx$$
  • A. $$20$$
  • B. $$\dfrac{19}{2}$$
  • C. $$38$$
  • D. $$19$$

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer
Q3 Subjective Medium

Integrate $$\displaystyle \int \frac {\sec^2 x }{\sqrt {\tan^2x+4}}dx$$

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer
Q4 Subjective Medium
Solve  $$\displaystyle\int \dfrac {1}{2x+3}dx$$

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer
Q5 Single Correct Medium
$$\displaystyle \int\frac{e^{x}(1+x)}{sin^{2}(xe^{x})}dx=$$
  • A. $$ \tan(xe^{x})+c$$
  • B. $$\sin(xe^{x})+c $$
  • C. $$\sin e^{x}+c$$
  • D. $$-\cot(xe^{x})+c$$

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer