List of top Mathematics Questions on Definite Integral asked in KEAM

The value of the integral $\int_{0}^{\pi / 2} \sqrt{\cos x \sin 2x} dx$ is equal to

The value of $\int_{-1}^{1} \frac{\log_e(1 + |x|)}{1 + |x|} dx$ is equal to

The value of \(\displaystyle \int_{0}^{1} \frac{t}{(t+1)^3}\,dt\) is equal to:

The value of \(\displaystyle \int_{0}^{2\pi} \frac{\sin 2x}{x(2\pi - x)}\,dx\) is equal to:

The value of \(\int_0^3 |3x^2 - 3| dx\) is equal to

The value of $\int_{0}^{3}x^{2}[x]dx$ is equal to

The value of $\int_{0}^{\pi}\frac{\sin x}{1+\sin x}dx$ is equal to} \textit{Note: The upper limit in the integral from the original paper contains a typo ($x$ instead of $\pi$). It has been corrected here to yield the valid options provided.

The value of $\int_{-2}^{1}\frac{|x|}{x}dx$ is equal to

The value of $\displaystyle \int_{0}^{1} x(1-x)^4 \, dx$ is equal to:

$\displaystyle \int_{-6}^{0} \left[t^3 + 9t^2 + 27t + 29 + (t+3)\cos(t+3)\right] dt$ is equal to:

If $I = \displaystyle \int_{-1}^{1} \frac{x^4}{1 - x^4} \cos^{-1}\left(\frac{2x}{1+x^2}\right) dx$, then $2I$ is equal to:

$\displaystyle \int_{0}^{1} \left[\tan^{-1}\left(\frac{1}{1+x+x^2+x^3}\right) + \tan^{-1}(1+x+x^2+x^3)\right] dx$ is equal to:

The value of the integral \(\int_{0}^{\pi} \frac{\cos x}{1 + \sin^{2} x} \, dx\) is

$\int_{-\log 3}^{+\log 3} e^{|x|} dx = $ ________.

$\int_{-\pi/2}^{\pi/2}(x^{5}+x^{3}+x)\cos x dx = $ ________.

If $[x]$ is the greatest integer less than or equal to x, then $\int_{-3}^{3}[x]dx=$ ________.

$\int_{4}^{5}\frac{1}{x(1+x)}dx=$ ________.

\(\int_{0}^{\frac{\pi}{2}} \sin 2x\,e^{\sin x}\,dx\) is equal to

$\int_{0}^{1}\frac{\sin x}{\sin x+\sin(1-x)}\,dx$ is equal to:

$\int_{-2}^{2}|x+3|\,dx =$

$\int_{0}^{\frac{\pi}{2}}\frac{1}{1+\sin x}\,dx =$

\( \int_{0}^{1} \frac{3^{2x}}{3^x + 1}\,dx = \)

The value of \( \int_{\pi/10}^{2\pi/5} \frac{\cot^3 x}{1+\cot^3 x}\,dx \) is equal to

If \( \int_{-\sqrt{3}}^{1} (-6x^2 + 18)\,dx = \alpha + \beta\sqrt{3} \), then the value of \( \alpha + \beta \) is equal to

\(\int_{0}^{2} \frac{x^{4}}{x^{4} + (2 - x)^{4}} \, dx =\)