_0^(4) x/(1 +, x, ) dx = Option 1: 4 Option 2: 16 Option 3: 32 Option 4: 64

Answer: 16. Solution: Let I = _0^(4) x/(1+, x, ) dx. Using _0^a f(x)dx = _0^a f(a-x)dx: I = _0^(4) (4 - x)/(1+, x, ) dx. Adding: 2I = 4 _0^(4) dx/(1+, x, ). Since, x, has period, _0^(4) = 4 _0^() = 4 * 2 = 8. So 2I = 32, giving I = 16.

Calculus

Integrals

Hard

$\int_0^{4\pi} \frac{x}{1 + |\cos x|} dx =$

$4\pi$

$16\pi$

$32\pi$

$64\pi$

✅ Correct Option: 2

Let $I = \int_0^{4\pi} \frac{x}{1+|\cos x|} dx$ . Using $\int_0^a f(x)dx = \int_0^a f(a-x)dx$ : $I = \int_0^{4\pi} \frac{4\pi - x}{1+|\cos x|} dx$ . Adding: $2I = 4\pi \int_0^{4\pi} \frac{dx}{1+|\cos x|}$ . Since $|\cos x|$ has period $\pi$ , $\int_0^{4\pi} = 4 \int_0^{\pi} = 4 \cdot 2 = 8$ . So $2I = 32\pi$ , giving $I = 16\pi$ .

2026: 18 May Shift 1

Match List-I with List-II

List-I	List-II
Integral	Value of the integral
(A) $\displaystyle\int_{1}^{2} e^{2\log x} \, dx$	(I) $\log 2$
(B) $\displaystyle\int_{-1}^{1} \left\vert x \right\vert \, dx$	(II) $\dfrac{1}{2}$
(C) $\displaystyle\int_{0}^{1} \log e^x \, dx$	(III) $\dfrac{7}{3}$
(D) $\displaystyle\int_{1}^{2} \dfrac{1}{x} \, dx$	(IV) $1$

Choose the correct answer from the options given below:

2026: 30 May Shift 1

The value of $\int_{-\pi/2}^{\pi/2}(\sin|x|+\cos|x|)dx$ equals:

2026: 25 May Shift 1

$\displaystyle\int \dfrac{1}{(x-1)^{3/4}(x+2)^{5/4}}\, dx$ is equal to :

[where C is an arbitrary constant]

2022: 23 Aug Shift 1

$\int \frac{dx}{(e^x - 1)} =$

2026: 31 May Shift 2

In which of the following cases the value of the integral is more than 10?

A. $\displaystyle\int_0^2 |x|\,dx$

B. $\displaystyle\int_0^2 (x+3)^2\,dx$

C. $\displaystyle\int_0^2 e^x\,dx$

D. $\displaystyle\int_0^2 (2x+5)\,dx$

Choose the correct answer from the options given below:

2026: 29 May Shift 1

$\displaystyle\int_1^3 \dfrac{\sqrt{x}}{\sqrt{x}+\sqrt{4-x}}dx$ is equal to:

∫04πx1+∣cos⁡x∣dx=\int_0^{4\pi} \frac{x}{1 + |\cos x|} dx =∫04π​1+∣cosx∣x​dx=

Solution

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CUET Mathematics Past Year Questions (Topic-Wise)

$\int_0^{4\pi} \frac{x}{1 + |\cos x|} dx =$