List of top Mathematics Questions on integral asked in COMEDK UGET

Evaluate: \[ \int \tan^{-1}\left( \sqrt{\frac{1-\sin x}{1+\sin x}} \right)\,dx \]
\[ \int \frac{dx}{x\sqrt{x^2+4}}= \]
\[ \int(\sin^6x+\cos^6x+3\sin^2x\cos^2x)\,dx= \]
\( \int \left(e^{x\log_e 6}\right)e^x dx = \phi(x) + c \) then \( \phi(x) = \)
If \( \int \frac{\sin 2x}{(1+\sin x)(2+\sin x)} \, dx = a \log |1+\sin x| - b \log |2+\sin x| + c \), then the value of \( a \) and \( b \) is __________.
Evaluate the integral:
\[ \int \frac{dx}{x \sqrt{4x^2 - 9}} \]
Evaluate the integral \( \int \frac{e^{\tan^{-1} x}}{1 + x^2} \, dx \)
If \[ \int \frac{\sin x + \cos x}{\sqrt{1 + 2 \sin x \cos x}} \, dx = \varphi(x) + C, \text{ then } \varphi(x) = \]
The integral \( \int \frac{2^x}{\sqrt{1 - 4^x}} \, dx \) is equal to:
The integral \( \int_{\pi/2}^{-\pi/2} \sin x \, dx \) is:
If $\int\limits \frac{\cos \, 8x + 1}{ \tan \, 2x - \cot \, 2x} dx = a \, \cos \, 8x + c , $ then $a$ =
$\int\limits \frac{x^3 -1}{x^3 + x} dx = $
$ \int\limits_1^e \log \, x \, dx = $
The interval I such that $\int^1_0\frac{dx}{\sqrt{1+x^4}}\in \,I$ is given by
The value of $\int^2_{-2}(ax^3+bx +c)dx$ depends on the
$\int^{\frac{\pi}{2}}_0 \log(\tan\,x)dx=$
The solution of $\left(x+y\right)^{2} \left(x \frac{dy}{dx}+y\right)=xy \left(1 +\frac{dy}{dx}\right) $ is
$\int e^{x}\left(\cos x -\sin x\right)dx $ is equal to
If $\int f\left(x\right) dx =\Psi\left(x\right)$ then $ \int x^{5} f\left(x^{3}\right)dx $ is equal to
The intercepts on $x-axis$ made by tangents to the curve, $y = \displaystyle\int_0^x |t| dt, x \in R$, which are parallel to the line $y = 2x$, are equal to
If $ \int \frac{xe^{x}}{\left(1+x\right)^{2}} dx = e^{x} f\left(x\right) +c, $ then $f(x)$ is equal to
Let $f (x)$ and $g(x)$ be differentiable functions on (0, 2] such that $f"(x) - g"(x) = 0, f'(1) = 2g'(1) = 4, f(2) = 3g(2) = 9.$ Then $f (x)- g(x)$ at $ x = 3/2$ is
$ \int\limits_0^{\pi /8} \tan^2 (2x) dx = $
$\int\limits_{-1/2}^{1/2} \cos x\log\left(\frac{1+x}{1-x}\right) dx = $
$\int\limits_{1}^{e^{\frac{17}{2}}} \frac{\, \pi \cos (\pi \log x)}{x} dx =$