Recent questions in Integrals

$\large \int \frac {\cos x}{\cos(x-a)} \normalsize dx=A\;x+B\;log \cos(x-\alpha),then $\[(a)\;A=\cos \alpha \qquad (b)\;B=\sin \alpha \qquad(c)\;A=\sin \alpha \qquad(d)\;B=\cos \alpha\]

asked Feb 1, 2013 by meena.p

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If $\Large \int \normalsize \tan^4 xdx = a\tan^3x+b\tan x+cx,then $\[(a)\;a=\frac{1}{3} \qquad (b)\;b=-1 \qquad(c)a=1\qquad(d)\;c=1\]

asked Feb 1, 2013 by meena.p

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If $\Large \int \frac {dx}{(x+1)(x-2)} \normalsize = A\; log(x+1)+B\; log(x-2)+c,then $\[(a)\;A+B=0 \qquad (b)\;AB=- \qquad(c)AB=\frac{1}{9}\qquad(d)\;AB=-9\]

asked Feb 1, 2013 by meena.p

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$S\sin 2x\;log\;\cos x\;dx=$\[(a)\;\frac{1}{2}\cos^2x-cos^2x log\cos x \qquad (b)\;\frac{1}{2}\cos^2x+\cos^2log\cos x \qquad(c)\;\cos^2x.log\cos x \qquad(d)\;\cos^2x (1-log\cos x)\]

asked Feb 1, 2013 by meena.p

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$\Large \int \frac {dx}{x+\sqrt x}=$\[(a)\;\log (1+\sqrt x) \qquad (b)\;\frac{1}{2}\log (1+\sqrt x) \qquad(c)\;2\log (1+\sqrt x) \qquad(d)\;\log (x+\sqrt x)\]

asked Feb 1, 2013 by meena.p

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$\Large\int \frac {dx}{(2-x)\sqrt{1+x}}=$\[(a)\;2\tan^{-1}\sqrt {1+x}\qquad (b)\;\frac{1}{2}\tan^{-1}\sqrt{1+x}\qquad(c)\;\tan^{-1}\sqrt {1+x}\qquad(d)\;log\bigg\{ \normalsize (2+x)\sqrt{1+x}\bigg\}\]

asked Feb 1, 2013 by meena.p

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$\Large \int \frac{e^{log}\bigg(1+\frac{1}{x^2}\bigg)dx}{x^2+\frac{1}{x^2}}=$\[(a)\;\frac{1}{\sqrt 2} \tan^{-1}\bigg(x+\frac{1}{x}\bigg)\qquad (b)\;\frac{1}{\sqrt 2}\tan^{-1}\bigg(\frac{x-\frac{1}{x}}{\sqrt 2}\bigg)\qquad(c)\;\sqrt 2 \tan^{-1}\bigg(x+\frac{1}{x}\bigg)\qquad(d)\;\frac{1}{\sqrt 2}\tan^{-1}\bigg(\frac{1}{x}-x\bigg)\]

asked Jan 31, 2013 by meena.p

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$\Large \frac {1}{x(x^4-1)^4}=$\[(a)\;\log \bigg[\frac{x^4}{x^4-1}\bigg]\qquad (b)\;\frac{1}{2}\log \bigg[\frac{x^2+1}{x^2+1}\bigg]\qquad(c)\;\frac{1}{4} log \bigg[\frac{x^4-1}{x^4}\bigg]\qquad(d)\;\log \frac{x(x^2-1}{x^2+1}\bigg]\]

asked Jan 31, 2013 by meena.p

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$\Large \frac {3^xdx}{\sqrt{1-9^x}} =$\[(a)\;\log _e 3\sin^{-1}(3^x) \qquad (b)\;\frac{1}{log_e^3} \sin^{-1}(3^x)\qquad(c)\;\log_e3\sin^{-1}(3^{x/2})\qquad(d)\;\frac{1}{log _e3} \sin^{-1}(3^{x/2}\]

asked Jan 31, 2013 by meena.p

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$\Large\int \frac {2xdx}{1+x^4}=$\[(a)\;\tan^{-1} (x^2) \qquad (b)\;\frac{1}{2}\tan^{-1} x^2 \qquad(c)\;log (1+x^4)\qquad(d)\;\tan^{-1} \bigg(\frac{1}{x^2}\bigg)\]

asked Jan 31, 2013 by meena.p

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$\int \sqrt {1+x^2} d(x^2)=$\[(a)\;\frac{2}{3x}(1+x^2)^{3/2}\qquad (b)\;\frac{2}{3}(1+x^2)^{3/2}\qquad(c)\;\frac{2x}{3}(1+x^2)^{3/2}\qquad(d)\;\frac{2x}{3}(1+x^2)\]

asked Jan 31, 2013 by meena.p

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$\Large\int \frac {dx}{x-X^3}=$\[(a)\;\frac{1}{2}log \frac{x^2}{1-x^2}\qquad (b)\;\frac{1}{2}log \frac{1-x^2}{x^2}\qquad(c)\;log \frac{(1-x)}{x(1+x)}\qquad(d)\;log x(1-x^2)\]

asked Jan 31, 2013 by meena.p

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$\Large\int \frac {dx}{x\sqrt{1-x^3}} =$\[\begin {array} {1 1} (a)\;\frac{1}{3} log \bigg | \frac {\sqrt {1-x^2}+1}{\sqrt {1-x^2}-1}\bigg | & \quad (b)\;\frac{1}{3} log \bigg |\frac{\sqrt {1-x^2}-1}{\sqrt {1-x^2}+1}\bigg| & \quad \\(c)\;\frac{2}{3} log \bigg |\frac{1}{\sqrt {1-x^3}}\bigg| & \quad(d)\;\frac{1}{3} log | 1-x^3 | \end {array}\]

asked Jan 31, 2013 by meena.p

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$\Large\int \frac {\sin^2x-\cos^2x}{\sin^2x-\cos^2x} \normalsize dx=$\[(a)\;\tan x+\cot x \qquad (b)\;\tan x-\cot x \qquad(c)\;\tan x+\sec x \qquad(d)\;\tan x+cosec x\]

asked Jan 31, 2013 by meena.p

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$\Large\int \frac {x+\sin x}{1+\cos x} \normalsize dx=$\[(a)\;\tan \frac{x}{2}\qquad (b)\;x\tan \frac{x}{2}\qquad(c)\;\cot \frac{x}{2}\qquad(d)\;x \cot \frac{x}{2}\]

asked Jan 31, 2013 by meena.p

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$\int x\tan^{-1}x dx=$ \[(a)\;\bigg(\frac{x^2+1}{2}\bigg) \tan^{-1}x- \frac{x}{2} \qquad (b)\;\bigg(\frac{x^2+1}{2}\bigg)+\tan^{-1}x-x \qquad(c)\;(x^2+1)\tan^{-1}x-x \qquad(d)\;(x^2+1) \tan^{-1}x+x \]

asked Jan 30, 2013 by meena.p

0 answers

$\Large\int \frac {x \sin^{-1}x}{\sqrt{1-x^2}} \normalsize dx=$\[(a)\;\sqrt{1-x^2}\sin^{-1}x\qquad (b)\;x\sin^{-1}x\qquad(c)\;x-\sqrt{1-x^2}\sin^{-1}x\qquad(d)\;(\sin^{-1}x)^2\]

asked Jan 30, 2013 by meena.p

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$\Large\int x^3 e^{x^2} dx=$ \[(a)\;x^2(e^{x^2}-1) \qquad (b)\;\frac{1}{2}x^2(e^{x^2}-1)\qquad(c)\;\frac{1}{2}e^{x^2}(x^2-1) \qquad(d)\;\frac{1}{2}(e^x-1)\]

asked Jan 30, 2013 by meena.p

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$\Large\int \frac {\sqrt x}{\sqrt{a^3-x^3}} dx=$\[(a)\;\frac{2}{3}\bigg(\frac{x}{a}\bigg)^{3/2} \qquad (b)\;\frac{2}{3}\sin^{-1}\bigg(\frac{x}{a}\bigg)^{3/2} \qquad(c)\;\frac{2}{3}\cos {-1}\bigg(\frac{x}{a}\bigg)^{3/2} \qquad(d)\;\frac{2}{3}\sin^{-1}\bigg(\frac{a}{x}\bigg)^{3/2}\]

asked Jan 30, 2013 by meena.p

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$\Large\int \frac {e^xdx} {e^{2x}+1}=$\[(a)\;log (e^x+e^{-x}) \qquad (b)\;log (e^{2x}+1)\qquad(c)\;\tan^{-1}(e^x) \qquad(d)\;\tan^{-1}(2e^x)\]

asked Jan 30, 2013 by meena.p

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$\Large\int \frac {\sin^2xdx}{a^2\cos ^2x+b^2\sin^2x \sin^2 x} =$\begin{array}{1 1}(a)\;(b-a)log(a^2\cos ^2x+b^2\sin^2x) & \quad (b)\;\frac{1}{b-a}log(a^2\cos ^2x+b^2\sin^2x) & \\(c)\;\frac{1}{b^2-a^2}log(a^2\cos ^2x+b^2\sin^2x) & \quad(d)\;\frac{1}{a^2+b^2}log(a^2\cos ^2x+b^2\sin^2x)\end{array}

asked Jan 30, 2013 by meena.p

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$\Large\int \frac {dx}{\sin x+\sqrt 3 \cos x} =$ \begin{array}{1 1}(a)\;\frac{1}{2} log \tan \bigg [\frac{\pi}{2}+\frac{\pi}{6}\bigg] & \quad (b)\;\frac{1}{2} log \bigg\{cosec \bigg(x+\frac{\pi}{3}\bigg)-\cot \bigg(x+\frac{\pi}{3}\bigg)\bigg\} & \\ (c)\;\frac{1}{2} log \bigg\{\sec \bigg(x-\frac{\pi}{6}\bigg)+\tan \bigg(x-\frac{\pi}{6}\bigg)\bigg\} & \quad (d)\;-\frac{1}{2} log \bigg\{cosec \bigg(x+\frac{\pi}{3}\bigg)+\cot \bigg(x+\frac{\pi}{3}\bigg)\bigg\}\end{array}

asked Jan 30, 2013 by meena.p

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$\Large\int \frac {\sqrt {\tan x}}{\sin x \cos x} \normalsize dx=$\[(a)\;2\sqrt {\tan x} \qquad (b)\;2\sqrt {\cot x} \qquad(c)\;\sqrt {\cot x} \qquad(d)\;\sqrt {\tan x}\]

asked Jan 30, 2013 by meena.p

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$\Large\int \frac {\tan (log)x} {x} dx=$\[(a)\;log\cos (log x) \qquad (b)\;log\sec (log x)\qquad(c)\;log\sin (log x) \qquad(d)\;-log\cos (log x)\]

asked Jan 30, 2013 by meena.p

0 answers