Definite Integration

Let \(f:[-1,2] \rightarrow[0, \infty)\) be a continuous function such that \(\mathrm{f}(x)=\mathrm{f}(1-x), \forall x \in[-1,2]\)
Let \(\mathrm{R}_1=\int_{-1}^2 x \mathrm{f}(x) \mathrm{d} x\) and \(\mathrm{R}_2\) be the area of the region b...

If \(\int_\limits0^{\frac{1}{2}} \frac{x^2}{\left(1-x^2\right)^{\frac{3}{2}}} \mathrm{~d} x=\frac{\mathrm{k}}{6}\), then the value of \(\mathrm{k}\) is

\(\int_\limits0^1 \cos ^{-1} x d x=\)

\(\int_\limits 0^\pi \frac{x \tan x}{\sec x+\cos x} d x=\)

Let \(\mathrm{f}: \mathrm{R} \rightarrow \mathrm{R}\) and \(\mathrm{g}: \mathrm{R} \rightarrow \mathrm{R}\) be continuous functions. Then the value of the integral $$\int_\limits{\frac{-\pi}{2}}^{\frac{\pi}{2}}[\mathrm{f}(x)+\mathrm{f}(-x)]...

\(\int_\limits0^2[x] \mathrm{d} x+\int_\limits0^2|x-1| \mathrm{d} x=\)
(where \([x]\) denotes the greatest integer function.)

The value of the integral \(\int_\limits0^1 \sqrt{\frac{1-x}{1+x}} \mathrm{~d} x\) is

\(\int_\limits{\frac{\pi}{4}}^{\frac{3 \pi}{4}} \frac{\mathrm{d} x}{1+\cos x}\) is equal to

\(\int_\limits0^{\pi / 2} \log \left(\frac{4+3 \sin x}{4+3 \cos x}\right) d x=\)

\(\int_\limits0^4 x[x] d x=\)
(where \([\mathrm{x}]\) denotes greatest integer function not greater than \(\mathrm{x}]\)

\(\int_\limits2^5 2[\mathrm{x}] \mathrm{dx}=\{\text { where }[\mathrm{x}] \text { denotes the greatest integer function } \leq \mathrm{x}\}\)

\(\int_\limits0^\pi x \sin x \cos ^4 x d x=\)

\(\int_0^{\pi / 2} \frac{\cos x}{3 \cos x+\sin x} d x=\)

\(\int_\limits0^1|5 x-3| d x=\)

\(\int_\limits0^\pi \frac{1}{4+3 \cos x} d x=\)

\(\int_\limits1^3\left[\tan ^{-1}\left(\frac{x}{x^2-1}\right)+\tan ^{-1}\left(\frac{x^2-1}{x}\right)\right] d x=\)

The value of \(\int\limits_0^1 {{{\tan }^{ - 1}}\left( {{{2x - 1} \over {1 + x - {x^2}}}} \right)dx}\) is

\(\int\limits_{ - \pi }^\pi {{{x\sin x} \over {1 + {{\cos }^2}x}}dx = }\)

\(\int_\limits0^2|2 x-3| d x=\)

\(\int_\limits{\frac{\pi}{6}}^{\frac{\pi}{2}} \frac{\operatorname{cosec} x \cdot \cot x}{1+\operatorname{cosec}^2 x} d x=\)

If \(\int_\limits0^a \sqrt{\frac{a-x}{x}} d x=\frac{k}{2}\), then \(k=\)

If \(f(x)=|x-1|+|x-2|+|x-3|, \forall x \in[1,4]\), then \(\int_\limits1^4 f(x) d x=\)

\(\int_\limits0^{\frac{\pi}{2}} \frac{\sin x-\cos x}{1-\sin x \cos x} d x=\)

\(\int\limits_{{{ - \pi } \over 2}}^{{\pi \over 2}} {{{\cos x} \over {1 + {e^x}}}dx = }\)

\(\int\limits_5^{10} \frac{d x}{(x-1)(x-2)}=\)

If \(\int_\limits0^{\frac{\pi}{2}} \frac{d x}{5+4 \sin x}=A \tan ^{-1} B\), then A + B =

\(\int_\limits0^{\pi / 4} \log (1+\tan x) d x=\)

If \(\int_\limits2^e\left[\frac{1}{\log x}-\frac{1}{(\log x)^2}\right] d x=a+\frac{b}{\log 2}\), then

If \(2 f(x)-3 f\left(\frac{1}{x}\right)=x\), then \(\int_\limits1^e f(x) d x=\)

$\int_\limits0^1\left(1-\frac{x}{1!}+\frac{x^2}{2!}-\frac{x^3}{3!}+\ldots\right.$ upto $\left.\infty\right) e^{2 x} d x=$

\(\int_\limits0^{\frac{\pi}{2}} \frac{\sqrt[7]{\sin x}}{\sqrt[7]{\sin x}+\sqrt[7]{\cos x}} d x=\)

The c.d.f, $F(x)$ associated with p.d.f. $f(x)=3\left(1-2 x^2\right)$. If $0< x<1$ is $k\left(x-\frac{2 x^3}{k}\right)$, then value of $k$ is

$\int_\limits0^1 \tan ^{-1}\left(\frac{2 x-1}{1+x-x^2}\right) d x=$

\(\int_\limits0^{\frac{\pi}{2}} \log \left[\sqrt{\frac{1-\cos 2 x}{1+\cos 2 x}}\right] d x=\)

\(\int_\limits2^3 \frac{x}{x^2-1} d x=\)

\(\int_0^a \sqrt{\frac{x}{a-x}} d x=\)

\(\int_\limits0^1\left(\frac{x^2-2}{x^2+1}\right) d x=\)

\(\int_{\frac{\pi}{5}}^{\frac{3 \pi}{10}}\left[\frac{\tan x}{\tan x+\cot x}\right] d x=\)

\(\int_\limits{-5}^5 \log \left(\frac{7-x}{7+x}\right) d x=\)

\(\int_{\frac{\pi}{18}}^{\frac{4 \pi}{9}} \frac{2 \sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x=\ldots \ldots\)

\(\int_0^{\frac{\pi}{2}} \sqrt{\cos \theta} \cdot \sin ^3 \theta d \theta=\) ............

\(\int_0^4 \frac{1}{1+\sqrt{x}} d x=\ldots \ldots\)

The value of $\int_{-3}^3\left(a x^5+b x^3+c x+k\right) d x$, where $a, b, c, k$ are constants, depends only on

If $\int_0^a \sqrt{\frac{a-x}{x}} d x=\frac{K}{2}$, then $K=\ldots .$.

\(\int_0^1 x(1-x)^5 d x=\ldots \ldots\)

\(\int_\limits a^b \frac{\sqrt{x}}{\sqrt{x}+\sqrt{a+b-x}} d x=\ldots \ldots\)