Integral Calculus-1 Questions (52)

If $f(x) = \lim_{n \to \infty} \frac{\tan(1/n)\log(1/n)}{n}$, and $\int \frac{f(x)}{\sqrt{\sin^{11} x \cos x}} dx = g(x) + C$ (C being the constant of integration). Then:

If $f'(x) = \frac{1}{-x + \sqrt{x^2 + 1}}$ and $f(0) = -\frac{1 + \sqrt{2}}{2}$, then the value of $f(5)$ will be :

The integral $\int \frac{\sec^{3/2}\theta - \sec^{1/2}\theta}{2 + \tan^2\theta} \tan\theta d\theta$ is:

If $f(x) = \lim_{n \to \infty} \frac{x^n - x^{-n}}{x^n + x^{-n}}$, $0 < x < 1$, $n \in \mathbb{N}$ then $\int (\sin^{-1} x) f'(x) dx$ is equal to:

The derivative of $x^4 + x^{-5}$ is $-\left(4x^{-5} + 5x^{-6}\right)$. So, $$\int \frac{5x^3 + 4x^5}{\left(x^5 + x + 1\right)^2} dx =$$

If the anti-derivative of $\frac{x^3}{\sqrt{4+2x^2}}$ which passes through $(1, 2)$ is $\frac{1}{m}\left(1+2x^2\right)^{1/2}\left(x^2-1\right)+c$. Then:

Let $f(xy) = f(x) \cdot f(y)$, $\forall x > 0, y > 0$ and $f(1 + x) = 1 + x[1 + g(x)]$, where $\lim_{x \to 0} g(x) = 0$, then $$\int \frac{f(x)}{f'(x)} dx$$ is:

If $$\int \frac{dx}{x^2(x^n + 1)^{(n-1)/n}} = -[f(x)]^{1/n} + C$$, then $f(x)$ is

Let $f(x) = \frac{1}{4 - 3\cos^2 x + 5\sin^2 x}$ and its anti-derivative $F(x) = \frac{1}{3}\tan^{-1}(g(x)) + c$, then :

$I_1 = \int f(x) dx$ and $I_2 = \int_0^1 f(x) dx$ where $f(x) = x^2 \ln\left(1-x^2\right)$, then:

$f(x) = \int e^{\tan^{-1}x}\left(1+x+x^2\right)d\left(\cot^{-1}x\right)$ is equal to:

If $\int\left[\frac{1}{1-x^8}\left\{\cos^{-1}\left(\frac{2x}{1+x^2}\right) + \tan^{-1}\left(\frac{2x}{1-x^2}\right)\right\}\right]dx$ has $p\tan^{-1}f(x)$ & $q\tan^{-1}g(x)$ terms and $x \in (-1, 1)$, then: (where $p$ and $q$ are constant)

If $A = \int e^{ax}\cos bx dx$ and $B = \int e^{ax}\sin bx dx$, then which of the following may be correct?

If $\int \sqrt{\frac{\cos x - \cos^3 x}{\left(1-\cos^3 x\right)}} dx = f(x) + c$, then $f(x)$ is equal to:

Let $f$ is a differentiable function such that $f(x) = x^2 + \int_0^x e^{-t}f(x-t)dt$, then:

Let $f$ is a differentiable function such that $f'(x) = f(x) + \int_0^2 f(x)dx, f(0) = \frac{4 - e^2}{3}$, then:

The integral $\int \frac{\sec^{3/2}\theta - \sec^{1/2}\theta}{2 + \tan^2\theta} \tan\theta d\theta$ is:

If $f(x) = \lim_{n \to \infty} \frac{x^n - x^{-n}}{x^n + x^{-n}}$, $0 < x < 1$, $n \in \mathbb{N}$ then $\int (\sin^{-1} x) f'(x) dx$ is equal to:

$$\int \frac{x^4 - 2}{x^2\sqrt{x^4 + x^2 + 2}} dx =$$

Anti-derivative of $$\frac{x - 1}{(x + 1)\sqrt{x^3 + x^2 + x}}$$ is:

The derivative of $x^4 + x^{-5}$ is $-\left(4x^{-5} + 5x^{-6}\right)$. So, $$\int \frac{5x^3 + 4x^5}{\left(x^5 + x + 1\right)^2} dx =$$

If $f'(x) = \frac{1}{-x + \sqrt{x^2 + 1}}$ and $f(0) = -\frac{1 + \sqrt{2}}{2}$, then the value of $f(5)$ will be :

Let $f(xy) = f(x) \cdot f(y)$, $\forall x > 0, y > 0$ and $f(1 + x) = 1 + x[1 + g(x)]$, where $\lim_{x \to 0} g(x) = 0$, then $$\int \frac{f(x)}{f'(x)} dx$$ is:

If $$\int \frac{dx}{x^2(x^n + 1)^{(n-1)/n}} = -[f(x)]^{1/n} + C$$, then $f(x)$ is

If $I_{m,n} = \int \cos^m x \sin nx dx$, then $7I_{4,3} - 4I_{3,2} =$$

$I = \int \frac{dx}{(\sin x - 2\cos x)(2\cos x + \sin x)}$ is equal to

If $\int \sqrt{\cos ecx + 1}\,dx = kfog(x) + c$, where $k$ is a real constant, then :

Let $f(x) = \frac{1}{4 - 3\cos^2 x + 5\sin^2 x}$ and its anti-derivative $F(x) = \frac{1}{3}\tan^{-1}(g(x)) + c$, then :

A function $f(x)$ continuous on $\mathbb{R}$ and periodic with $2\pi$ satisfies $f(x) + (\sin x) f(x + \pi) = \sin^2 x$ then,

$$\int \frac{dx}{\prod_{i=0}^{n}(x+r)} \text{ is equal to:}$$

$I_1 = \int f(x) dx$ and $I_2 = \int_0^1 f(x) dx$ where $f(x) = x^2 \ln\left(1-x^2\right)$, then:

$f(x) = \int e^{\tan^{-1}x}\left(1+x+x^2\right)d\left(\cot^{-1}x\right)$ is equal to:

Let $f: \mathbb{R} \to \mathbb{R}$ be a function satisfying $f(x+2y) = f(x)e^{2y} + f(2y)e^x + x^2\left(1-e^{2y}\right) + 4y^2\left(1-e^x\right) + 4xyy$ for all $x, y \in \mathbb{R}$ and $f'(0) = 1$, then:

If $\int\left[\frac{1}{1-x^8}\left\{\cos^{-1}\left(\frac{2x}{1+x^2}\right) + \tan^{-1}\left(\frac{2x}{1-x^2}\right)\right\}\right]dx$ has $p\tan^{-1}f(x)$ & $q\tan^{-1}g(x)$ terms and $x \in (-1, 1)$, then: (where $p$ and $q$ are constant)

If $A = \int e^{ax}\cos bx dx$ and $B = \int e^{ax}\sin bx dx$, then which of the following may be correct?

If $\int \sqrt{\frac{\cos x - \cos^3 x}{\left(1-\cos^3 x\right)}} dx = f(x) + c$, then $f(x)$ is equal to:

If $\int \frac{dx}{x^2\left(x^7 - 6\right)} = A\left[\ln\left(p^9 + 9p^2 - 2p^3 - 18p\right)\right] + c$, then:

If the anti-derivative of $\frac{x^3}{\sqrt{4+2x^2}}$ which passes through $(1, 2)$ is $\frac{1}{m}\left(1+2x^2\right)^{1/2}\left(x^2-1\right)+c$. Then:

If $I = \int \frac{\sin x + \sin^3 x}{\cos 2x} dx = P\cos x + Q\ln |f(x)| + R$, then:

Let $f$ is a differentiable function such that $f(x) = x^2 + \int_0^x e^{-t}f(x-t)dt$, then:

Let $f$ is a differentiable function such that $f'(x) = f(x) + \int_0^2 f(x)dx, f(0) = \frac{4 - e^2}{3}$, then:

If $f(x) = \lim_{n \to \infty} \left[2x + 4x^3 + ... + 2nx^{2n-1}\right]$ $(0 < x < 1)$ then $\int f(x)dx$ is equal to:

$$\int \frac{x^4 - 2}{x^2\sqrt{x^4 + x^2 + 2}} dx =$$

Let $f: \mathbb{R} \to \mathbb{R}$ be a function satisfying $f(x+2y) = f(x)e^{2y} + f(2y)e^x + x^2\left(1-e^{2y}\right) + 4y^2\left(1-e^x\right) + 4xyy$ for all $x, y \in \mathbb{R}$ and $f'(0) = 1$, then:

If $I_{m,n} = \int \cos^m x \sin nx dx$, then $7I_{4,3} - 4I_{3,2} =$$

A function $f(x)$ continuous on $\mathbb{R}$ and periodic with $2\pi$ satisfies $f(x) + (\sin x) f(x + \pi) = \sin^2 x$ then,

If $\int \frac{dx}{x^2\left(x^7 - 6\right)} = A\left[\ln\left(p^9 + 9p^2 - 2p^3 - 18p\right)\right] + c$, then:

If $I = \int \frac{\sin x + \sin^3 x}{\cos 2x} dx = P\cos x + Q\ln |f(x)| + R$, then:

If $\int \sqrt{\cos ecx + 1}\,dx = kfog(x) + c$, where $k$ is a real constant, then :

Anti-derivative of $$\frac{x - 1}{(x + 1)\sqrt{x^3 + x^2 + x}}$$ is: