List of top Mathematics Questions on integral asked in KEAM

$\displaystyle \int \frac{\sin(\cot^{-1}x)}{1+x^2} \, dx$ is equal to:

$\displaystyle \int \left(\frac{1}{(1+x)^2} - \frac{2}{(1+x)^3}\right)e^x \, dx$ is equal to:

$\displaystyle \int \sqrt{1 + \sin\left(\frac{x}{8}\right)} \, dx =$

$\displaystyle \int \frac{x^4 - 1}{x + 1} \, dx$ is equal to:

$\displaystyle \int \frac{\sin t + \cos t}{13 + 36\sin^2 t} \, dt$ is equal to:

\( \int \left(\tan^{2}(2x) - \cot^{2}(2x)\right)\,dx \)

Evaluate the integral: \( \int \frac{1}{x^3} \sqrt{1 - \frac{1}{x^2}} \text{dx} = \)

If $\frac{dy}{dx} = \frac{1}{8\left(\sqrt{16+\sqrt{25+\sqrt{x}}}\right)\left(\sqrt{25+\sqrt{x}}\right)\sqrt{x}}$, then $y =$

$\int e^{\left(x+\frac{1}{x}\right)}\left(\frac{x^{2}-1}{x^{2}}\right)dx =$

$\int e^{2\theta}(2\cos^{2}\theta-\sin 2\theta)\, d\theta =$

$\int \frac{9e^{x}+4e^{-x}}{9e^{x}-4e^{-x}}dx =$

$\int \frac{\sec x}{(\sec x+\tan x)^{9}}dx =$

If $\int \frac{1}{x^{7}\left(\frac{1}{x^{8}}+1\right)^{p}}dx = -\frac{1}{2}\left(\frac{1}{\frac{1}{x^{8}}+1}\right)^{2} + c$, then $p =$

\( \int e^x(x^2-2)\cos(e^x(x^2-2x)) dx = \)

\( \int e^x \sec x (1+\tan x) dx = \)

\( \int x^7 (x^8 + 1)^{-3/4} \, dx = \)

\( \int \frac{\sin^{-1}x}{\sqrt{1-x^2}} \, dx = \)

\( \int \frac{2x^2 + 4x + 3}{x^2 + x + 1} \, dx = \)

\(\int \frac{1}{x(1 + x^{4})} \, dx\) is equal to

\(\int \frac{\sec^2(\sqrt{2x+5})}{\sqrt{2x+5}} \, dx =\)

\(\int \frac{\sin 4\theta}{\sin 2\theta} \, d\theta =\)

\(\int \frac{x}{5-x^2} \, dx =\)

\(\int \frac{e^{x}}{e^{-x} + 3e^{x}} dx =\)

If \( \int f(x)\cos x \, dx = \frac{1}{2}\{f(x)\}^2 + c \), then \( f\left(\frac{\pi}{2}\right) \) is

\( \int \frac{x e^x}{(1+x)^2} \, dx = \)