Indefinite Integration

\(\int \frac{\log (\cot x)}{\sin 2 x} d x=\)

If \(\int \frac{\mathrm{d} x}{x \sqrt{1-x^3}}=\mathrm{k} \log \left(\frac{\sqrt{1-x^3}-1}{\sqrt{1-x^3}+1}\right)+\mathrm{c}\), (where \(\mathrm{c}\) is a constant of integration), then value of \(\mathrm{k}\) is

\(\int \frac{\mathrm{e}^x(1+x)}{\cos ^2\left(\mathrm{e}^x \cdot x\right)} \mathrm{d} x=\)

If \(\mathrm{I}=\int \sin (\log (x)) \mathrm{d} x\), then \(\mathrm{I}\) is given by

The value of \(\int(1-\cos x) \cdot \operatorname{cosec}^2 x d x\) is

If \(\int \frac{\cos 8 x+1}{\cot 2 x-\tan 2 x} \mathrm{~d} x=\mathrm{A} \cos 8 x+\mathrm{c}\), where \(\mathrm{c}\) is an arbitrary constant, then the value of \(\mathrm{A}\) is

\(\int x \sqrt{\frac{2 \sin \left(x^2+1\right)-\sin 2\left(x^2+1\right)}{2 \sin \left(x^2+1\right)+\sin 2\left(x^2+1\right)}} d x=\)

If \(\mathrm{I}=\int \frac{\mathrm{d} x}{x^2\left(x^4+1\right)^{\frac{3}{4}}}\), then \(\mathrm{I}\) is

\(\int \frac{d x}{\sin x+\cos x}=\)

\(\int \frac{1}{7-6 x-x^2} d x=\)

If \(\int \sqrt{\frac{x-7}{x-9}} d x=A \sqrt{x^2-16 x+63}+\log \left|(x-8)+\sqrt{x^2-16 x+63}\right|+c,\)
(where \(\mathrm{c}\) is a constant of integration) then \(\mathrm{A}\) is

\(\int \mathrm{e}^x\left(1-\cot x+\cot ^2 x\right) \mathrm{d} x=\)

The value of \(\int \frac{\left(x^2-1\right) d x}{x^3 \sqrt{2 x^4-2 x^2+1}}\) is

\(\int \frac{5 \tan x}{\tan x-2} \mathrm{~d} x=x+\mathrm{a} \log |\sin x-2 \cos x|+\mathrm{c},\)
(where \(c\) is a constant of integration), then the value of \(a\) is

The value of \(\int \frac{\mathrm{d} x}{x^2\left(x^4+1\right)^{\frac{3}{4}}}\) is

\(\int \frac{3 x-2}{(x+1)(x-2)^2} \mathrm{~d} x=\)
(where \(C\) is a constant of integration)

\(\text { If } \int e^{x^2} \cdot x^3 \mathrm{~d} x=e^{x^2} \cdot[f(x)+C]\)
(where \(C\) is a constant of integration.) and \(f(1)=0\), then value of \(f(2)\) will be

If \(f(x)=\sqrt{\tan x}\) and \(g(x)=\sin x \cdot \cos x\) then \(\int \frac{f(x)}{g(x)} \mathrm{d} x\) is equal to (where \(C\) is a constant of integration)

\(\int \frac{\sin \frac{5 x}{2}}{\sin \frac{x}{2}} d x=\)
(where \(C\) is a constant of integration.)

\(\int \frac{\mathrm{dx}}{32-2 \mathrm{x}^2}=\mathrm{A} \log (4-\mathrm{x})+\mathrm{B} \log (4+\mathrm{x})+\mathrm{c}\), then the values of \(\mathrm{A}\) and \(\mathrm{B}\) are respectively (where c is a constant of integration)

\(\int \cos ^3 x \cdot e^{\log (\sin x)} d x=\)

If \(\int \frac{(\cos x-\sin x)}{8-\sin 2 x} d x=\frac{1}{p} \log \left[\frac{3+\sin x+\cos x}{3-\sin x-\cos x}\right]+c\), then \(p=\) (where \(\mathrm{c}\) is a constant of integration)

\(\int e^x\left(\frac{1+\sin x}{1+\cos x}\right) d x=\)

\(\int \frac{d x}{e^x+e^{-x}+2}=\)

\(\int \cos ^3 x e^{\log (\sin x)^2} d x=\)

\(\int e^{\left(e^x+x\right)} d x=\)

If \(\int \frac{\sin x}{\sin (x-\alpha)} d x=A x+B \log \sin (x-\alpha)+c\), then the value of A and B are respectively (where \(\mathrm{c}\) is a constant of integration)

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

If \(\int \frac{\sqrt{x}}{x(x+1)} d x=k \tan ^{-1} m+c\), (where c is constant of integration), then

\(\int \frac{d x}{\cos x \sqrt{\cos 2 x}}=\)

\(\int \sec ^{-1} x d x=\)

\(\int \frac{\sec ^8 x}{\operatorname{cosec} x} d x=\)

\(\int \sin ^{-1}\left(\frac{2 x}{1+x^2}\right) d x=\quad(\text { where }|x| < 1)\)

\(\int {{e^x}\left( {{{x - 1} \over {{x^2}}}} \right)dx = }\)

\(\int \frac{1}{\cos x+\sqrt{3} \sin x} d x=\)

\(\int \frac{\tan ^4 \sqrt{x} \cdot \sec ^2 \sqrt{x}}{\sqrt{x}} d x=\)

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

\(\int e^{\tan x}\left(\sec ^2 x+\sec ^3 x \sin x\right) d x=\)

If \(\int \frac{x^3}{\sqrt{1+x^2}} d x=a\left(1+x^2\right)^{\frac{3}{2}}+b \sqrt{1+x^2}+c\), then \(a+b=\), (where \(c\) is constant of integration)

\(\int[1+2 \tan x(\tan x+\sec x)]^{\frac{1}{2}} d x=\)

If \(\int {{{5\tan x} \over {\tan x - 2}}dx = x + a\log |\sin x - 2\cos x| + c}\), then a = (Where c is constant of integration)

\(\int[\sin |\log x|+\cos |\log x|] d x=\)

\(\int \frac{1}{x^{\frac{1}{2}}+x^{\frac{1}{3}}} d x=\)

\(\int \frac{x+\sin x}{1+\cos x} d x=\)

If \(\int \frac{1+x^2}{1+x^4} d x=\frac{1}{\sqrt{2}} \tan ^{-1}\left[\frac{f(x)}{\sqrt{2}}\right]+c\), then \(f(x)=\)

\(\int \tan ^{-1}(\sec x+\tan x) d x=\)

\(\int \frac{2 x^2-1}{x^4-x^2-20} d x=\)

\(\int \operatorname{cosec}(x-a) \operatorname{cosec} x d x=\)

\(\int \sec ^4 x \cdot \tan ^4 x d x=\frac{\tan ^m x}{m}+\frac{\tan ^n x}{n}+c\) (where c is constant of integration), then m + n =

\(\int \frac{d x}{x^2+4 x+13}=\)