Let $a \in \left(0, \dfrac{\pi}{2}\right)$ be fixed.
If
$\displaystyle \int \dfrac{\tan x + \tan a}{\tan x - \tan a} , dx = A(x)\cos 2a + B(x)\sin 2a + C,$
where $C$ is a constant of integration,
then the functions $A(x)$ and $B(x)$ are respectively:
If $I=\displaystyle\int_{0}^{\pi/2}\frac{\sin^{3/2}x}{\sin^{3/2}x+\cos^{3/2}x}\,dx$, then $\displaystyle\int_{0}^{2I}\frac{x\sin x\cos x}{\sin^{4}x+\cos^{4}x}\,dx$ equals:
The value of the definite integral$ \int\limits_{ - {\pi \over 4}}^{{\pi \over 4}} {{{dx} \over {(1 + {e^{x\cos x}})({{\sin }^4}x + {{\cos }^4}x)}}} $ is equal to :
The value of $\displaystyle \int_{-\pi/2}^{\pi/2} \dfrac{dx}{[x] + [\sin x] + 4}$,
where $[t]$ denotes the greatest integer less than or equal to $t$, is :
If the area of the bounded region $R = \left\{ {(x,y):\max \{ 0,{{\log }_e}x\} \le y \le {2^x},{1 \over 2} \le x \le 2} \right\}$ is , $\alpha {({\log _e}2)^{ - 1}} + \beta ({\log _e}2) + \gamma $, then the value of ${(\alpha + \beta - 2\lambda )^2}$ is equal to :
Let $f(x)+2f!\left(\frac{1}{x}\right)=x^{2}+5$ and $2g(x)-3g!\left(\frac{1}{x}\right)=x$, $x>0$. If $\alpha=\displaystyle\int_{1}^{2} f(x),dx$ and $\beta=\displaystyle\int_{1}^{2} g(x),dx$, then the value of $9\alpha+\beta$ is:
Statement-1 : The value of the integral
$\displaystyle \int_{\pi/6}^{\pi/3}\frac{dx}{1+\sqrt{\tan x}}$ is equal to $\pi/6$
Statement-2 : $\displaystyle \int_{a}^{b}f(x)\,dx=\int_{a}^{b}f(a+b-x)\,dx$.
Let f(x) be a differentiable function defined on [0, 2] such that f'(x) = f'(2 $-$ x) for all x$ \in $ (0, 2), f(0) = 1 and f(2) = e2. Then the value of $\int\limits_0^2 {f(x)} dx$ is :
The value of $\displaystyle \int_{0}^{2}!\left(,|2x^{2}-3x|+\big[x-\tfrac{1}{2}\big]\right),dx$, where $[\cdot]$ is the greatest integer function, is equal to:
Let \(f(x)\) be a function satisfying \(f(x)+f(\pi-x)=\pi^{2}\), \(\forall x\in\mathbb{R}\).
Then \(\displaystyle \int_{0}^{\pi} f(x)\sin x\,dx\) is equal to:
If the value of the integral $\int\limits_0^5 {{{x + [x]} \over {{e^{x - [x]}}}}dx = \alpha {e^{ - 1}} + \beta } $, where $\alpha$, $\beta$ $\in$ R, 5$\alpha$ + 6$\beta$ = 0, and [x] denotes the greatest integer less than or equal to x; then the value of ($\alpha$ + $\beta$)2 is equal to :
Let $\beta(m,n)=\displaystyle\int_{0}^{1}x^{m-1}(1-x)^{,n-1},dx,; m,n>0$. If $\displaystyle\int_{0}^{1}(1-x^{10})^{20},dx=a\times \beta(b,c)$, then $100(a+b+c)$ equals:
If $I_1=\displaystyle\int_{0}^{1} e^{-x}\cos^{2}x,dx$;
$I_2=\displaystyle\int_{0}^{1} e^{-x^{2}}\cos^{2}x,dx$ and
$I_3=\displaystyle\int_{0}^{1} e^{-x^{3}},dx$; then
$ \text{Let } \alpha,\beta,\gamma \text{ be the three roots of } x^{3}+bx+c=0. \text{ If } \beta\gamma=1=-\alpha,\ \text{then } b^{3}+2c^{3}-3\alpha^{3}-6\beta^{3}-8\gamma^{3} \text{ is equal to:} $
Let $f$ and $g$ be continuous functions on $[0,a]$ such that $f(x)=f(a-x)$ and $g(x)+g(a-x)=4$. Then $\displaystyle \int_{0}^{a} f(x)\,g(x)\,dx$ is equal to :
If $f(\alpha)=\int\limits_{1}^{\alpha} \frac{\log _{10} \mathrm{t}}{1+\mathrm{t}} \mathrm{dt}, \alpha>0$ then $f\left(\mathrm{e}^{3}\right)+f\left(\mathrm{e}^{-3}\right)$ is equal to :
For $x\in\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)$, if
$y(x)=\displaystyle\int \frac{\csc x+\sin x}{\csc x\sec x+\tan x\sin^2 x}\,dx$,
and $\displaystyle\lim_{x\to \left(\frac{\pi}{2}\right)} y(x)=0$, then $y\!\left(\dfrac{\pi}{4}\right)$ is equal to:
The integral $\int\limits_1^e {\left\{ {{{\left( {{x \over e}} \right)}^{2x}} - {{\left( {{e \over x}} \right)}^x}} \right\}} \,$ loge x dx is equal to :
If the value of the integral
$\displaystyle \int_{-\pi/2}^{\pi/2}
\left(
\dfrac{x^{2}\cos x}{1+x^{2}}
+\dfrac{1+\sin^{2}x}{1+e^{\sin(2\tan^{-1}x)}}
\right)\,dx
= \dfrac{\pi}{4}\,(\pi+a)-2,$
then the value of $a$ is:
Let $f:\mathbb{R}\to\mathbb{R}$ be a twice differentiable function such that $f(2)=1$. If $F(x)=x f(x)$ for all $x\in\mathbb{R}$, $\displaystyle\int_{0}^{2} x F''(x),dx=6$ and $\displaystyle\int_{0}^{2} x^{2} F''(x),dx=40$, then $F'(2)+\displaystyle\int_{0}^{2} F(x),dx$ is equal to:
Let A1 be the area of the region bounded by the curves y = sinx, y = cosx and y-axis in the first quadrant. Also, let A2 be the area of the region bounded by the curves y = sinx, y = cosx, x-axis and x = ${\pi \over 2}$ in the first quadrant. Then,
Let $f$ be a continuous function satisfying
$\displaystyle \int_{0}^{t^2} \big(f(x) + x^2\big)\,dx = \dfrac{4}{3}t^3, \; \forall t > 0.$
Then $f\!\left(\dfrac{\pi^2}{4}\right)$ is equal to:
If $\sin\!\left(\dfrac{y}{x}\right)=\log_e|x|+\dfrac{\alpha}{2}$
is the solution of the differential equation
$x\cos\!\left(\dfrac{y}{x}\right)\dfrac{dy}{dx}=y\cos\!\left(\dfrac{y}{x}\right)+x$
and $y(1)=\dfrac{\pi}{3}$, then $\alpha^{2}$ is equal to:
If $\displaystyle \int \frac{1}{a^{2}\sin^{2}x+b^{2}\cos^{2}x},dx=\frac{1}{12}\tan^{-1}(3\tan x)+\text{constant}$, then the maximum value of $a\sin x+b\cos x$ is:
The value of $k\in\mathbb{N}$ for which the integral $I_n=\displaystyle\int_{0}^{1}(1-x^{k})^{n},dx,\ n\in\mathbb{N}$, satisfies $147I_{20}=148I_{21}$ is:
If $f(x)=\dfrac{2-x\cos x}{2+x\cos x}$ and $g(x)=\log_e x,\ (x>0)$, then the value of the integral $\displaystyle \int_{-\pi/4}^{\pi/4} g\big(f(x)\big),dx$ is:
Consider the integral $I = \int_0^{10} {{{[x]{e^{[x]}}} \over {{e^{x - 1}}}}dx} $, where [x] denotes the greatest integer less than or equal to x. Then the value of I is equal to :
Let $f(x)$ be a positive function and $I_{1}=\int_{-\tfrac{1}{2}}^{1} 2x,f\left(2x(1-2x)\right),dx$ and $I_{2}=\int_{-1}^{2} f\left(x(1-x)\right),dx$. Then the value of $\dfrac{I_{2}}{I_{1}}$ is equal to
If $f:\mathbb{R}\to\mathbb{R}$ is a continuous function satisfying
\[
\int_{0}^{\pi/2} f(\sin 2x)\,\sin x\,dx \;+\; \alpha \int_{0}^{\pi/4} f(\cos 2x)\,\cos x\,dx \;=\; 0,
\]
then the value of $\alpha$ is:
Which of the following statements is correct for the function g($\alpha$) for $\alpha$ $\in$ R such that $g(\alpha ) = \int\limits_{{\pi \over 6}}^{{\pi \over 3}} {{{{{\sin }^\alpha }x} \over {{{\cos }^\alpha }x + {{\sin }^\alpha }x}}dx} $
Let the function $f:[0,2]\to\mathbb{R}$ be defined as
\[
f(x)=
\begin{cases}
e^{\min\{x^2,\; x-[x]\}}, & x\in[0,1),\\[4pt]
e^{[\,x-\log_e x\,]}, & x\in[1,2],
\end{cases}
\]
where $[t]$ denotes the greatest integer less than or equal to $t$.
Then the value of the integral $\displaystyle \int_{0}^{2} x f(x)\,dx$ is:
Let $f:\mathbb{R}\to\mathbb{R}$ be defined by $f(x)=\dfrac{x}{(1+2x^{4})^{1/4}}$, and $g(x)=f(f(f(f(x))))$. Then $18\displaystyle\int_{0}^{\sqrt{2\sqrt{5}}} x^{2}g(x),dx$ is equal to:
For $x \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$, if $y(x)=\int \frac{\operatorname{cosec} x+\sin x}{\operatorname{cosec} x \sec x+\tan x \sin ^2 x} d x$, and $\lim _\limits{x \rightarrow\left(\frac{\pi}{2}\right)^{-}} y(x)=0$ then $y\left(\frac{\pi}{4}\right)$ is equal to
Let $y=f(x)$ be a thrice differentiable function in $(-5,5)$.
Let the tangents to the curve $y=f(x)$ at $(1,f(1))$ and $(3,f(3))$ make angles $\dfrac{\pi}{6}$ and $\dfrac{\pi}{4}$ respectively with the positive $x$-axis.
If $27\displaystyle\int_{1}^{3}\big((f'(t))^{2}+1\big)f'''(t),dt=\alpha+\beta\sqrt{3}$,
where $\alpha,\beta$ are integers, then the value of $\alpha+\beta$ equals:
Let f be a differentiable function in $\left( {0,{\pi \over 2}} \right)$. If $\int\limits_{\cos x}^1 {{t^2}\,f(t)dt = {{\sin }^3}x + \cos x} $, then ${1 \over {\sqrt 3 }}f'\left( {{1 \over {\sqrt 3 }}} \right)$ is equal to
Let $\int_\limits\alpha^{\log _e 4} \frac{\mathrm{d} x}{\sqrt{\mathrm{e}^x-1}}=\frac{\pi}{6}$. Then $\mathrm{e}^\alpha$ and $\mathrm{e}^{-\alpha}$ are the roots of the equation :
If
$\displaystyle \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\left(\dfrac{x^2\cos x}{1+x^2}+\dfrac{1+\sin^2 x}{1+e^{\sin 2x}}\right)dx = \dfrac{\pi}{4}(\pi+a)-2$,
then the value of $a$ is:
$a$ and $b$ be real constants such that the function $f$ defined by $f(x)=\left\{\begin{array}{ll}x^2+3 x+a & , x \leq 1 \\ b x+2 & , x>1\end{array}\right.$ be differentiable on $\mathbb{R}$. Then, the value of $\int_\limits{-2}^2 f(x) d x$ equals
Let $\mathrm{f}: \mathbb{R} \rightarrow \mathbb{R}$ be defined as $f(x)=a e^{2 x}+b e^x+c x$. If $f(0)=-1, f^{\prime}\left(\log _e 2\right)=21$ and $\int_0^{\log _e 4}(f(x)-c x) d x=\frac{39}{2}$, then the value of $|a+b+c|$
Let f : R $ \to $ R be defined as f(x) = e$-$xsinx. If F : [0, 1] $ \to $ R is a differentiable function with that F(x) = $\int_0^x {f(t)dt} $, then the value of $\int_0^1 {(F'(x) + f(x)){e^x}dx} $ lies in the interval
$ \text{If } \displaystyle \int_{0}^{10}\frac{[\sin 2\pi x]}{e^{,x-[x]}},dx ;=; \alpha e^{-1}+\beta e^{-1/2}+\gamma,\ \text{ where } \alpha,\beta,\gamma \text{ are integers and } [x] \text{ is the greatest integer } \le x,\ \text{then the value of } \alpha+\beta+\gamma \text{ is:} $
The integral $\int {{{(2x - 1)\cos \sqrt {{{(2x - 1)}^2} + 5} } \over {\sqrt {4{x^2} - 4x + 6} }}} dx$ is equal to (where c is a constant of integration)
Let f : R $\to$ R be a differentiable function such that $f\left( {{\pi \over 4}} \right) = \sqrt 2 ,\,f\left( {{\pi \over 2}} \right) = 0$ and $f'\left( {{\pi \over 2}} \right) = 1$ and let $g(x) = \int_x^{\pi /4} {(f'(t)\sec t + \tan t\sec t\,f(t))\,dt} $ for $x \in \left[ {{\pi \over 4},{\pi \over 2}} \right)$. Then $\mathop {\lim }\limits_{x \to {{\left( {{\pi \over 2}} \right)}^ - }} g(x)$ is equal to :
Let f : R $\to$ R be a continuous function satisfying f(x) + f(x + k) = n, for all x $\in$ R where k > 0 and n is a positive integer. If ${I_1} = \int\limits_0^{4nk} {f(x)dx} $ and ${I_2} = \int\limits_{ - k}^{3k} {f(x)dx} $, then :
Let f be a non-negative function in [0, 1] and twice differentiable in (0, 1). If $\int_0^x {\sqrt {1 - {{(f'(t))}^2}} dt = \int_0^x {f(t)dt} } $, $0 \le x \le 1$ and f(0) = 0, then $\mathop {\lim }\limits_{x \to 0} {1 \over {{x^2}}}\int_0^x {f(t)dt} $ :
Let $f(x)=x+\dfrac{a}{\pi^{2}-4}\sin x+\dfrac{b}{\pi^{2}-4}\cos x,\ x\in\mathbb{R}$ be a function which satisfies
$\displaystyle f(x)=x+\int_{0}^{\pi/2}\sin(x+y)\,f(y)\,dy.$
Then $(a+b)$ is equal to:
The value of $\dfrac{e^{-\pi/4}+\displaystyle\int_{0}^{\pi/4} e^{-x}\tan^{50}x\,dx}{\displaystyle\int_{0}^{\pi/4} e^{-x}\big(\tan^{49}x+\tan^{51}x\big)\,dx}$ is:
Let $[x]$ denote the greatest integer $\le x$. Consider the function
$$f(x)=\max\{x^{2},\,1+[x]\}.$$
Then the value of the integral $\displaystyle \int_{0}^{2} f(x)\,dx$ i
If $\displaystyle \int_{0}^{1} \frac{1}{(5+2x-2x^2)\,(1+e^{\,2-4x})}\,dx=\frac{1}{\alpha}\log_e\!\left(\frac{\alpha+1}{\beta}\right),\ \alpha,\beta>0,$ then $\alpha^4-\beta^4$ is equal to:
If the value of the integral $\int\limits_0^{{1 \over 2}} {{{{x^2}} \over {{{\left( {1 - {x^2}} \right)}^{{3 \over 2}}}}}} dx$ is ${k \over 6}$, then k is equal to :
Let g(x) = $\int_0^x {f(t)dt} $, where f is continuous function in [ 0, 3 ] such that ${1 \over 3}$ $ \le $ f(t) $ \le $ 1 for all t$\in$ [0, 1] and 0 $ \le $ f(t) $ \le $ ${1 \over 2}$ for all t$\in$ (1, 3]. The largest possible interval in which g(3) lies is :
Let a be a positive real number such that $\int_0^a {{e^{x - [x]}}} dx = 10e - 9$ where [ x ] is the greatest integer less than or equal to x. Then a is equal to:
Let $f:\mathbb{R}\to(0,\infty)$ be a strictly increasing function such that
$\displaystyle \lim_{x\to\infty}\frac{f(7x)}{f(x)}=1$.
Then the value of $\displaystyle \lim_{x\to\infty}\Big[\frac{f(5x)}{f(x)}-1\Big]$ is:
Let $(a, b)$ be the point of intersection of the curve $x^2 = 2y$ and the straight line $y - 2x - 6 = 0$ in the second quadrant.
Then the integral $I = \int_a^b \dfrac{9x^2}{1 + 5x^4},dx$ is equal to:
$ \text{If [t] denotes the greatest integer } \le t, \text{ then the value of } \frac{3(e-1)}{e} \int_{1}^{2} x^2 e^{\lfloor x \rfloor + \lfloor x^3 \rfloor} dx \text{ is:} $
If [x] denotes the greatest integer less than or equal to x, then the value of the integral $\int_{ - \pi /2}^{\pi /2} {[[x] - \sin x]dx} $ is equal to :
Let the domain of the function $f(x) = \log_2 \log_4 \log_6 (3 + 4x - x^2)$ be $(a, b)$.
If $\int_0^{b - a} [x^2] , dx = p - \sqrt{q - \sqrt{r}}, ; p, q, r \in \mathbb{N}, ; \gcd(p, q, r) = 1$,
where $[,]$ is the greatest integer function, then $p + q + r$ is equal to
If
$2\displaystyle\int_{0}^{1} \tan^{-1} x , dx = \displaystyle\int_{0}^{1} \cot^{-1} (1 - x + x^{2}) , dx,$
then
$\displaystyle\int_{0}^{1} \tan^{-1} (1 - x + x^{2}) , dx$ is equal to :
Let $g(t) = \int_{ - \pi /2}^{\pi /2} {\cos \left( {{\pi \over 4}t + f(x)} \right)} dx$, where $f(x) = {\log _e}\left( {x + \sqrt {{x^2} + 1} } \right),x \in R$. Then which one of the following is correct?
If $\int\limits_0^{100\pi } {{{{{\sin }^2}x} \over {{e^{\left( {{x \over \pi } - \left[ {{x \over \pi }} \right]} \right)}}}}dx = {{\alpha {\pi ^3}} \over {1 + 4{\pi ^2}}},\alpha \in R} $ where [x] is the greatest integer less than or equal to x, then the value of $\alpha$ is :
Let $y=y(x)$ be the solution of the differential equation $\dfrac{dy}{dx}+3\tan^2 x,y+3y=\sec^2 x$, $y(0)=\dfrac{1}{3}+e^3$. Then $y!\left(\dfrac{\pi}{4}\right)$ is equal to:
Let $\alpha\in(0,1)$ and $\beta=\log_{e}(1-\alpha)$. Let $P_{n}(x)=x+\dfrac{x^{2}}{2}+\dfrac{x^{3}}{3}+\cdots+\dfrac{x^{n}}{n},\ x\in(0,1)$. Then the integral $\displaystyle \int_{0}^{\alpha}\frac{t^{50}}{1-t}\,dt$ is equal to
Let
$f(x)=
\begin{cases}
-2, & -2 \le x \le 0,\\[4pt]
x-2, & 0 < x \le 2,
\end{cases}$
and $h(x)=f(|x|)+|f(x)|.$
Then $\displaystyle \int_{-2}^{2} h(x)\,dx$ is equal to:
Let f : R $ \to $R be a continuously differentiable function such that f(2) = 6 and f'(2) = ${1 \over {48}}$. If $\int\limits_6^{f\left( x \right)} {4{t^3}} dt$ = (x - 2)g(x), then $\mathop {\lim }\limits_{x \to 2} g\left( x \right)$ is equal to :
The value of $\int_{e^2}^{e^4} \frac{1}{x}\left(\frac{e^{\left(\left(\log _e x\right)^2+1\right)^{-1}}}{e^{\left(\left(\log _e x\right)^2+1\right)^{-1}}+e^{\left(\left(6-\log _e x\right)^2+1\right)^{-1}}}\right) d x$ is