dx/((a^2-x^2)^(3/2)) is equal to: (where C is an arbitrary constant) Option 1: 1/a^2√(a^2-x^2) + C Option 2: x/(a^2√(a^2-x^2)) + C Option 3: x/(a^2√(a^2+x^2)) + C Option 4: 2/5a^2(a^2-x^2)^(5/2) + C

Answer: x/(a^2√(a^2-x^2)) + C. Solution: Put x = a, dx = a d. The integral becomes 1/a^2^2 d = ()/a^2. Back-substituting, = x/√(a^2-x^2), so the answer is x/(a^2√(a^2-x^2)) + C.

Calculus

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$\int \frac{dx}{(a^2-x^2)^{3/2}}$ is equal to: (where C is an arbitrary constant)

$\frac{1}{a^2}\sqrt{a^2-x^2} + C$

$\frac{x}{a^2\sqrt{a^2-x^2}} + C$

$\frac{x}{a^2\sqrt{a^2+x^2}} + C$

$\frac{2}{5a^2}(a^2-x^2)^{5/2} + C$

✅ Correct Option: 2

Put $x = a\sin\theta$ , $dx = a\cos\theta\,d\theta$ . The integral becomes $\frac{1}{a^2}\int\sec^2\theta\,d\theta = \frac{\tan\theta}{a^2}$ . Back-substituting, $\tan\theta = x/\sqrt{a^2-x^2}$ , so the answer is $\frac{x}{a^2\sqrt{a^2-x^2}} + C$ .

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$\int_1^4 |x - 2|dx$ is equal to

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$\int_0^1 tan^{-1}\left(\frac{2x-1}{1+x-x^2}\right)dx$ is equal to

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$\int \sqrt{1 - 49x^2} \, dx$ is equal to

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$\int \left(\frac{\cos 2x - \cos 2\alpha}{\cos x - \cos \alpha}\right) dx =$

(Given that $c$ is an arbitrary constant)

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$\int e^x \left(\frac{x-1}{2x^2}\right) dx$ is equal to ( where $C$ is an arbitrary constant)

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Match the LIST-I with LIST-II

	LIST-I (Definite integrals)		LIST-II (Value)
A.	$\displaystyle\int_{-6}^{6} \vert x \vert \, dx$	I.	$1$
B.	$\displaystyle\int_{-\pi/2}^{\pi/2} \sin^{2017} x \, dx$	II.	$36$
C.	$\displaystyle\int_{0}^{\pi/4} \cos x \, dx$	III.	$0$
D.	$\displaystyle\int_{0}^{2} \dfrac{\sqrt{x}}{\sqrt{x}+\sqrt{2-x}} \, dx$	IV.	$\dfrac{1}{\sqrt{2}}$

Choose the correct answer from the options given below:

∫dx(a2−x2)3/2\int \frac{dx}{(a^2-x^2)^{3/2}}∫(a2−x2)3/2dx​ is equal to: (where C is an arbitrary constant)

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CUET Mathematics Past Year Questions (Topic-Wise)

$\int \frac{dx}{(a^2-x^2)^{3/2}}$ is equal to: (where C is an arbitrary constant)