If _0^(/2) √( x) dx = ()/√2, then the value of is Option 1: Option 2: -1 Option 3: ()/2 Option 4: 3/√2

Solution: Let I = _0^(/2) √( x) dx. By the property _0^a f(x)dx = _0^a f(a-x)dx: I = _0^(/2) √( x) dx. Adding: 2I = _0^(/2) ( x + x)/(√( x x)) dx. Substituting t = x - x, dt = ( x + x)dx: 2I = √2_(-1)^1 dt/(√(1-t^2)) = √2 . So I = /√2, giving = .

Calculus

Integrals

Hard

If $\int_0^{\pi/2} \sqrt{\tan x} \, dx = \frac{\lambda}{\sqrt{2}}$ , then the value of $\lambda$ is

$\pi$

$-1$

$\frac{\pi}{2}$

$\frac{3}{\sqrt{2}}$

✅ Correct Option: 1

Let $I = \int_0^{\pi/2} \sqrt{\tan x}\, dx$ . By the property $\int_0^a f(x)dx = \int_0^a f(a-x)dx$ : $I = \int_0^{\pi/2} \sqrt{\cot x}\, dx$ . Adding: $2I = \int_0^{\pi/2} \frac{\sin x + \cos x}{\sqrt{\sin x \cos x}} dx$ . Substituting $t = \sin x - \cos x$ , $dt = (\cos x + \sin x)dx$ : $2I = \sqrt{2}\int_{-1}^{1} \frac{dt}{\sqrt{1-t^2}} = \sqrt{2}\,\pi$ . So $I = \pi/\sqrt{2}$ , giving $\lambda = \pi$ .

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(Here, c is an arbitrary constant)

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If ∫0π/2tan⁡x dx=λ2\int_0^{\pi/2} \sqrt{\tan x} \, dx = \frac{\lambda}{\sqrt{2}}∫0π/2​tanx​dx=2​λ​, then the value of λ\lambdaλ is

Solution

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CUET Mathematics Past Year Questions (Topic-Wise)

If $\int_0^{\pi/2} \sqrt{\tan x} \, dx = \frac{\lambda}{\sqrt{2}}$ , then the value of $\lambda$ is