Mathematics

Match the correct pair.

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

ANSWER

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

Hence, this is the answer.

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

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

Hence, this is the answer.

$$(C)$$ 
$$I=\int \sqrt {a^2-x^2}dx$$

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

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

Hence, this is the answer.

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

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

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

Hence, this is the answer.

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

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

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

Hence, this is the answer.
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Matrix Medium Published on 17th 09, 2020
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