Mathematics

# Match the correct pair.

 $\int \text{cosec}\, xdx$ $-\log(\text{cosec} x+\cot x)+c$ $\displaystyle \int \dfrac{dx}{x\sqrt{x^2-1}}$ $\dfrac{a^2}{2}\left[\dfrac{x}{a}\sqrt{a^2 - x^2} + \sin^{-1} \dfrac{x}{a}\right] +c$ $\displaystyle \int \sqrt{a^2 -x^2dx}$ $\log [x + \sqrt{x^2 - a^2}]+c$ $\displaystyle \int \dfrac{dx}{\sqrt{a^2-x^2}}$ $\tan^{-1} \sqrt{x^2-1} + c$ $\displaystyle \int \frac{dx}{a^2 + x^2}$ $\sin^{-1} \dfrac{x}{a} + c$

0,9,13,22,29

##### SOLUTION
$(A)$
$I=\int \text{cosec} xdx$
$I=\int \dfrac{\csc x(\csc x+\cot x)}{\csc x+\cot s}dx$
$I=\int \dfrac{\csc^2 x+\csc x.\cot x}{\csc x+\cot s}dx$

Let $t=\csc x +\cot x$
$dx={\csc ^2 x+\csc x\cot x}dt$

Therefore,
$I=-\int \dfrac{1}{t}dt$
$I=-\log(t)+C$
$I=-\log(\csc x+\cot x)+C$

$(B)$
$I=\int \dfrac{1}{x\sqrt {x^2-1}}dx$

Let $t=\sqrt {x^2-1}$
$\dfrac{dt}{dx}=\dfrac{1}{2\sqrt {x^2-1}}\times 2x$
$\dfrac{dt}{dx}=\dfrac{x}{\sqrt {x^2-1}}$

Therefore,
$I=\int \dfrac{1}{t^2+1}$
$I=\tan^{-1} t+C$
$I=\tan^{-1} {\sqrt {x^2-1}}+C$

$(C)$
$I=\int \sqrt {a^2-x^2}dx$

Let $x=a\sin t$
$t=\sin^{-1} \dfrac{x}{a}$
$dx=a\cos t dt$

Therefore,
$I=a^2\int \cos^2 t dt$
$I=\dfrac{a^2}{2}\int (1+\cos 2t) dt$
$I=\dfrac{a^2}{2}\left[t+\dfrac{\sin 2t}{2}\right]+C$
$I=\dfrac{a^2}{2}\left[t+\cos t\sin t\right]+C$
$I=\dfrac{a^2}{2}\left[\sin^{-1}\dfrac{x}{a}+\cos \sin^{-1} \dfrac{x}{a}\sin \sin ^{-1}\dfrac{x}{a}\right]+C$
$I=\dfrac{a^2}{2}\left[\sin^{-1}\dfrac{x}{a}+\dfrac{x}{a}\sqrt{{1-\dfrac{x^2}{a^2}}}\right]+C$

$(D)$
$I=\int \dfrac{dx}{ \sqrt {a^2-x^2}}$

Let $t=\dfrac{x}{a}$
$dt=\dfrac{dx}{da}$

Therefore,
$I=\int \dfrac{dt}{ \sqrt {1-t^2}}$
$I=\sin^{-1} t+C$
$I=\sin^{-1}\dfrac{x}{a}+C$

$(E)$
$I=\int \dfrac{dx}{ {a^2+x^2}}$

Let $t=\dfrac{x}{a}$
$dt=\dfrac{dx}{da}$

Therefore,
$I=\int \dfrac{dt}{ a{1+t^2}}$
$I=\dfrac{1}{a}\tan^{-1} t+C$
$I=\dfrac{1}{a}\tan^{-1}\dfrac{x}{a}+C$

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Matrix Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

#### Realted Questions

Q1 Single Correct Medium
Solve $\displaystyle\int \dfrac {(e^{2x}-1)}{e^{2x}+1}dx$
• A. $I=2\log |e^x+e^{-x}|+C$
• B. $I=\log |e^x-e^{-x}|+C$
• C. None of these
• D. $I=\log |e^x+e^{-x}|+C$

1 Verified Answer | Published on 17th 09, 2020

Q2 Matrix Hard
The value of $\displaystyle \int_{a}^{b}f\left ( x \right )dx$
 $a= 0,\:b= \pi /3,\:f\left ( x \right )= \displaystyle \frac{x}{1+\sin x}$ $\displaystyle \frac{16}{3}\pi -2\sqrt{3}$ $a= 1,\:b= 16,\:f\left ( x \right )= \tan^{-1}\sqrt{\sqrt{x}-1}$ $\displaystyle \frac{\pi ^{2}}{16}+\displaystyle \frac{\pi }{4} \log 2$ $a= 0,\:b= \pi /2,\:f\left ( x \right )= \displaystyle \frac{x\sin 2x}{\left ( \sin x+\cos x \right )^{2}}$ $-\displaystyle \frac{\pi }{3} \left ( 2-\sqrt{3} \right )+\log \left ( \dfrac{1}{2}+\displaystyle \frac{\sqrt{3}}{2} \right )$ $a= 0,\:b= \pi /4,\:f\left ( x \right )= \displaystyle \frac{x^{2}\sec ^{2}x\left ( \cos 2x-\sin 2x \right )}{\left ( \sin x+\cos x \right )^{2}}$ $\displaystyle \frac{\pi ^{2}}{8}-\displaystyle \frac{\pi }{4}$

1 Verified Answer | Published on 17th 09, 2020

Q3 Subjective Medium
$\int e^{\ 10x} dx$

1 Verified Answer | Published on 17th 09, 2020

Q4 Subjective Medium
Solve
$\int (4x^{3}-\dfrac{3}{x^{4}}) dx$

Let $n \space\epsilon \space N$ & the A.M., G.M., H.M. & the root mean square of $n$ numbers $2n+1, 2n+2, ...,$ up to $n^{th}$ number are $A_{n}$, $G_{n}$, $H_{n}$ and $R_{n}$ respectively.