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Question: If \(f(x) = \left\{ \begin{matrix} e^{\cos x}.\sin x,|x| \leq 2 \\ 2,\text{otherwise} \end{matrix} \...

If f(x)={ecos⁡x.sin⁡x,∣x∣≤22,otherwise f(x) = \left\{ \begin{matrix} e^{\cos x}.\sin x,|x| \leq 2 \\ 2,\text{otherwise} \end{matrix} \right.\ , then ∫−23f(x)dx=\int_{- 2}^{3}{f(x)dx} =

A

0

B

1

C

2

D

3

Answer

2

Explanation

Solution

∣x∣≤2⇒−2≤x≤2|x| \leq 2 \Rightarrow - 2 \leq x \leq 2 and f(x)=ecos⁡xsin⁡xf(x) = e^{\cos x}\sin x is an odd function.

∴ I=∫−23f(x)dx=∫−22f(x)dx+∫23f(x)dxI = \int_{- 2}^{3}{f(x)dx} = \int_{- 2}^{2}{f(x)dx} + \int_{2}^{3}{f(x)dx}

⇒ I=0+∫232dx=[2x]23=2I = 0 + \int_{2}^{3}{2dx} = \lbrack 2x\rbrack_{2}^{3} = 2 [∵∫−aaf(x)=0\because\int_{- a}^{a}{f(x) = 0} if f(x) is odd and in (2, 3) f(x) is 2]

(4) ∫0af(x)dx=∫0af(a−x)dx\int_{0}^{a}{f(x)dx = \int_{0}^{a}{f(a - x)dx}} : This property can be used only when lower limit is zero. It is generally used for those complicated integrals whose denominators are unchanged when x is replaced by (a – x).

Following integrals can be obtained with the help of above property.

(i) ∫0π⥂/⥂2sin⁡nxsin⁡nx+cos⁡nxdx=∫0π/2cos⁡nxcos⁡nx+sin⁡nxdx=π4\int_{0}^{\pi ⥂ / ⥂ 2}\frac{\sin^{n}x}{\sin^{n}x + \cos^{n}x}dx = \int_{0}^{\pi/2}{\frac{\cos^{n}x}{\cos^{n}x + \sin^{n}x}dx = \frac{\pi}{4}}

(ii) ∫0π/2tan⁡nx1+tan⁡nxdx=∫0π/2cot⁡nx1+cot⁡nxdx=π4\int_{0}^{\pi/2}{\frac{\tan^{n}x}{1 + \tan^{n}x}dx = \int_{0}^{\pi/2}{\frac{\cot^{n}x}{1 + \cot^{n}x}dx = \frac{\pi}{4}}}

(iii) ∫0π/211+tan⁡nxdx=∫0π/211+cot⁡nxdx=π4\int_{0}^{\pi/2}{\frac{1}{1 + \tan^{n}x}dx = \int_{0}^{\pi/2}\frac{1}{1 + \cot^{n}x}dx = \frac{\pi}{4}}

(iv) ∫0π/2sec⁡nxsec⁡nx+cos⁡⥂ecnxdx=∫0π/2cos⁡⥂ecnxcos⁡⥂ecnx+sec⁡nxdx=π4\int_{0}^{\pi/2}{\frac{\sec^{n}x}{\sec^{n}x + \cos ⥂ ec^{n}x}dx =}\int_{0}^{\pi/2}{}\frac{\cos ⥂ ec^{n}x}{\cos ⥂ ec^{n}x + \sec^{n}x}dx = \frac{\pi}{4}

(v) ∫0π/2f(sin⁡2x)sin⁡xdx=∫0π/2f(sin⁡2x)cos⁡xdx\int_{0}^{\pi/2}{f(\sin 2x)\sin xdx =}\int_{0}^{\pi/2}{f(\sin 2x)\cos xdx}

(vi) ∫0π/2f(sin⁡x)dx=∫0π/2f(cos⁡x)dx\int_{0}^{\pi/2}{f(\sin x)dx = \int_{0}^{\pi/2}{f(\cos x)dx}}

(vii) ∫0π/2f(tan⁡x)dx=∫0π/2f(cot⁡x)dx\int_{0}^{\pi/2}{f(\tan x)dx = \int_{0}^{\pi/2}{f(\cot x)dx}}

(viii) ∫01f(log⁡x)dx=∫01f[log⁡(1−x)]dx\int_{0}^{1}{f(\log x)dx = \int_{0}^{1}{f\lbrack\log(1 - x)\rbrack dx}}

(ix)∫0π/2logtanxdx=∫0π/2logcotxdx\int_{0}^{\pi/2}{{logtan}xdx =}\int_{0}^{\pi/2}{{logcot}xdx}

(x) ∫0π/4log⁡(1+tan⁡x)dx=π8log⁡2\int_{0}^{\pi/4}{\log(1 + \tan x)dx = \frac{\pi}{8}\log 2}

(xi) ∫0π/2logsinxdx=∫0π/2logcosxdx=−π2log⁡2=π2log⁡12\int_{0}^{\pi/2}{{logsin}xdx} = \int_{0}^{\pi/2}{{logcos}xdx} = \frac{- \pi}{2}\log 2 = \frac{\pi}{2}\log\frac{1}{2}

(xii) ∫0π/2asin⁡x+bcos⁡xsin⁡x+cos⁡xdx=∫0π/2asec⁡x+bcos⁡⥂ecxsec⁡x+cos⁡⥂ecxdx\int_{0}^{\pi/2}{}\frac{a\sin x + b\cos x}{\sin x + \cos x}dx = \int_{0}^{\pi/2}{}\frac{a\sec x + b\cos ⥂ ecx}{\sec x + \cos ⥂ ecx}dx =∫0π/2atan⁡x+bcot⁡xtan⁡x+cot⁡xdx=π4(a+b)= \int_{0}^{\pi/2}{}\frac{a\tan x + b\cot x}{\tan x + \cot x}dx = \frac{\pi}{4}(a + b)