Definite Integration Questions (1340)

If \(I = \displaystyle\int_{-\frac{\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}} \sqrt{\frac{1-x}{1+x}}\,\sin^{-1}x\,dx = \frac{\pi}{M} - \sqrt{N}\), find the value of \((M+N)\).
Evaluate: \( L = \lim_{n \to \infty} \sum_{k=0}^{n-1} \frac{k}{n} \left[ \left(\frac{k+1}{n}\right)^{\frac{1}{m}} - \left(\frac{k}{n}\right)^{\frac{1}{m}} \right] \). Find the value of \(m\) if \(L = \dfrac{1}{10}\).
The value of \(\displaystyle\int_0^{100\pi} \left(\left[\cot^{-1} x\right] + \left[\tan^{-1} x\right]\right) dx\) equals ________. (where \([\cdot]\) denotes the greatest integer function)
Let \( f(x) \) be a continuous function \( \forall\, x \in R \) such that \[ \lim_{x \to \pi/4} \frac{\displaystyle\int_{2}^{\sec^2 x} f(t)\,dt}{x^2 - \dfrac{\pi^2}{16}} = \frac{k}{\pi} f(a) \] where \( a, k \in N \), then the value of \( k^a \) is equal to:
\(\lim_{n \to \infty} \left[\dfrac{1}{n^2} \sec^2 \dfrac{1}{n^2} + \dfrac{2}{n^2} \sec^2 \dfrac{4}{n^2} + \cdots + \dfrac{1}{n^2} \sec^2 1\right]\) equals
The integral \(\int\left(1+x-\frac{1}{x}\right)e^{x+\frac{1}{x}}dx\) is equal to
If \(f(x)\) and \(g(x)\) are both continuous functions then the value of \[\displaystyle\int_{\ln \lambda}^{\ln(1/\lambda)} \dfrac{f\!\left(\dfrac{x^2}{4}\right)(f(x) - f(-x))}{g\!\left(\dfrac{x^2}{4}\right)(g(x) + g(-x))} \, dx\] is equal to:
Find the following limit: 16. \(\lim_{n \to \infty} \left[\left(1+\frac{1}{n^2}\right)\left(1+\frac{2^2}{n^2}\right)\left(1+\frac{3^2}{n^2}\right)\cdots\left(1+\frac{n^2}{n^2}\right)\right]^{1/n}\)
Find $\int \frac{dx}{x^2 + a^2}$
[JEE Main 2019] \(\displaystyle\int\frac{dx}{x(x^n+1)}\) equals
Evaluate $$\int \text{cosec} x \ln |\cot x - \text{cosec} x| dx$$
Evaluate the integral: \[ I = \int \frac{3(\tan x - 1)\sec^2 x}{(\tan x + 1)\sqrt{\tan^3 x + \tan^2 x + \tan x}}\, dx \]
If \(L = \lim_{n \to \infty}\left(\dfrac{1}{\sqrt{n}\sqrt{n+1}} + \dfrac{1}{\sqrt{n}\sqrt{n+2}} + \cdots + \dfrac{1}{\sqrt{n}\sqrt{n+n}}\right) = a\sqrt{b} - c\) and \(a^4 + b^3 + c^2 + d = 29\), find \(d\).
Evaluate $\int \frac{dx}{\sqrt{9 - 16x^2}}$
If \(\int e^{\sec x}(\sec x \tan x + (\sec x \tan x + \sec^2 x))\,dx = e^{\sec x} f(x) + C\), then a possible choice of \(f(x)\) is
Given\[ \int \frac{dx}{x^3(1+x^6)^{2/3}} = x\, f(x)\cdot(1+x^6)^{1/3} + C \]find \(f(x)\).
If In = ∫(sinx)ⁿ dx where n ∈ ℕ, then 5I4 - 6I6 is equal to:
The value of definite integral \(\displaystyle\int_{1}^{\sqrt{3}} \left(x^{2x^2+1} + \ln\left(x^{x^{(2x^2+1)}}\right)\right)dx\) is equal to:
If \(f'(x) + g'(x) = (f(x) + g(x))^2 + 1\), then adding two given equations gives \(\displaystyle\int \dfrac{f'(x)+g'(x)}{(f(x)+g(x))^2+1}\,dx = \displaystyle\int 1\,dx\). What is the result?
If \(\int x^{26}(x-1)^{17}(5x-3)\, dx = \dfrac{x^{27}(x-1)^{18}}{k} + C\), where \(C\) is constant of integration, then the value of \(k\) is:
Let $f(x) = \displaystyle\int \dfrac{2x}{(x^2+1)(x^2+3)}\,dx$. If $f(3) = \dfrac{1}{2}(\log_e 5 - \log_e 6)$, then $f(4)$ is equal to
If $I_{m,n} = \int \cos^m x \sin nx dx$, then $7I_{4,3} - 4I_{3,2} =$$
Evaluate \(\int \frac{dx}{e^x \sec x(1 + \tan x)}\)
If \(\int \dfrac{2x+5}{\sqrt{7-6x-x^2}}\, dx = A\sqrt{7-6x-x^2} + B\sin^{-1}\left(\dfrac{x+3}{4}\right) + C\), where \(C\) is a constant of integration, then the values of \(A\) and \(B\) are:
Evaluate the integral: \[I = \int \frac{\sin^2 x \cos^2 x}{(\sin^3 x + \cos^3 x)^2} dx\]
If \( f(x) = \displaystyle\int \frac{3x^2 - x^{-2}}{\left(x^3 + 1 + \dfrac{1}{x}\right)^2}\, dx \) and \( f(0) = 0 \), then \( f(-1) \) equals:
\(\displaystyle\int\frac{3x+1}{(x+1)^2(x^2+1)}\,dx\) equals (where \(C\) is the constant of integration)
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:
[JEE Main 2023] \(\displaystyle\int\frac{3x^2+2}{(x^2+x+1)^2}\,dx\) equals (where \(C\) is a constant)
If \(f(x) = \displaystyle\int \dfrac{(3x^4 - 1)}{(x^4 + x + 1)^2}\, dx\) and \(f(0) = 0\), then \(f(-1)\) is equal to:
For \(x > 0\), let \(f(x) = \int_{1}^{x} \frac{\log t}{1+t} dt\). Then \(f(x) + f\left(\frac{1}{x}\right)\) is equal to:
The value of the definite integral \(\displaystyle\int_{1}^{\sqrt{3}}\left(x^{2x^2+1}+\ln\left(x^{x^{\left(2x^2+1\right)}}\right)\right)dx\) is equal to:
\(\displaystyle\int \frac{x^2-1}{(x+1)\sqrt{x^2+x}}\,dx\) equals (where \(C\) is constant of integration)
Find $\int \frac{\sin x}{1 + \sin x \cos x} dx$
Evaluate: \(\displaystyle\int_{-4}^{4} \frac{x^2}{(x^2+16)(1+e^x)}\,dx\) (up to four decimal places).
If $\displaystyle\int(\sin x)^{-11/2}(\cos x)^{-5/2}\,dx=-\dfrac{p_1}{q_1}(\cot x)^{5/2}-\dfrac{p_2}{q_2}(\cot x)^{-5/2}-\dfrac{p_3}{q_3}(\cot x)^{1/2}+\dfrac{p_4}{q_4}(\cot x)^{-3/2}+C$, where $p_i$ and $q_i$ are positive integers with $\gcd(p_i,q_i)=1$ for $i=1,2,3,4$ and $C$ is the constant of integration, then $\dfrac{15p_1p_2p_3p_4}{q_1q_2q_3q_4}$ is equal to
Evaluate $\int (2x-4)\sqrt{4+3x-x^2} dx$
Let I_n = ∫₁^e (ln x)^n d(x²). Then the value of 2I_n + nI_{n-1} equals:
77. \(\int \cos(\log_e x) dx\) is equal to
Evaluate the integral: \[I = \int \frac{x^{5m-1} + 2x^{4m-1}}{(x^{2m} + x^m + 1)^3} dx\]
Evaluate $\int \frac{2x - \sqrt{\arcsin x}}{\sqrt{1-x^2}} dx$
The value of \(\sqrt{2}\int \frac{\sin x\, dx}{\sin\left(x - \frac{\pi}{4}\right)}\) is
If \(I_1 = \int_{1}^{3} f(x^3 - 2x^2 - 5x + 2020)\,dx\) and \(I_2 = \int_{1}^{3} f(x^3 - 2x^2 - 5x + 2020)\,dx\) are related such that \(\dfrac{2I_1}{3I_2} = \dfrac{a}{b}\), find the value of \(a + b\).
Given \[I = \int \frac{(\sin^n\theta - \sin\theta)^{\frac{1}{n}}\cos\theta}{\sin^{n+1}\theta}\,d\theta\] Evaluate the integral.
Let \(I = \displaystyle\int (\sin 4x)\, e^{\tan^2 x}\,dx\). If \(I = A^{10}\) when evaluated appropriately, find the value of \(\dfrac{A^{10}}{10}\) (in decimal). Given that the answer is of the form \(A = -2\cos^4 x \cdot e^{\tan^2 x} + C\), and \(\dfrac{A^{10}}{10} = \dfrac{1024}{100}\).
\(\displaystyle\int \cos(100x)\cdot\sin^{95}x\,dx\) equals
[JEE Main 2023] If \(\displaystyle\int\frac{x}{1+x^3}\,dx = p\ln|1+x|+q\ln|1-x+x^2|+r\tan^{-1}\!\dfrac{2x-1}{\sqrt3}+C\), then the value of \(18(p+q+r)\) is
Evaluate \(\displaystyle\int_{-1}^{1.5} [x^2]\, dx\), where \([x]\) is the greatest integral (floor) function.
Evaluate $$\int \frac{\cos^4 x \, dx}{\sin^3 x \{\sin^5 x + \cos^5 x\}^{3/5}}$$
If f_n(x) = \frac{n+1}{(n+1)!} \cdot x^n, then the value of \int_1^n \frac{1}{f_n(x)} \, dx is