86523
\(\int_{-1}^{3}\left[\tan ^{-1}\left(\frac{x}{x^{2}+1}\right)+\tan ^{-1}\left(\frac{x^{2}+1}{x}\right)\right] d x\) is equal to
1 \(2 \pi\)
2 \(\pi\)
3 \(\frac{\pi}{2}\)
4 \(\frac{\pi}{4}\)
Explanation:
(A) : Given, \(I =\int_{-1}^{3}\left[\tan ^{-1}\left(\frac{x}{x^{2}+1}\right)+\tan ^{-1}\left(\frac{x^{2}+1}{x}\right)\right] d x\) \(=\int_{-1}^{3} \tan ^{-1}\left\{\frac{\frac{x}{x^{2}+1}+\frac{x^{2}+1}{x}}{1-\frac{x}{x^{2}+1} \times \frac{x^{2}+1}{x}}\right\} d x\) \(=\int_{-1}^{3} \tan ^{-1}(\infty) d x=\int_{-1}^{3} \tan ^{-1}\left(\tan \frac{\pi}{2}\right) d x\) \(= \int_{-1}^{3}\left(\frac{\pi}{2}\right) d x=\left[\frac{\pi x}{2}\right]_{-1}^{3}=\frac{\pi}{2}[3+1]=2 \pi\)
COMEDK-2020
Integral Calculus
86524
The value of the integral \(\int_{3}^{6} \frac{\sqrt{x}}{\sqrt{9-x}+\sqrt{x}} d x\) is
1 \(3 / 2\)
2 2
3 1
4 \(1 / 2\)
Explanation:
(A) : Given, \(I=\int_{3}^{6} \frac{\sqrt{x}}{\sqrt{9-x}+\sqrt{x}} d x \tag{i}\) Apply property- \(I=\int_{3}^{6} \frac{\sqrt{6+3-x}}{\sqrt{9-(6+3-x)}+\sqrt{6+3-x}} d x\) \(I=\int_{3}^{6} \frac{\sqrt{9-x}}{\sqrt{x}+\sqrt{9-x}} d x \tag{ii}\) On adding equation (i) and (ii), we get - \(2 \mathrm{I}=\int_{3}^{6} \frac{\sqrt{\mathrm{x}}+\sqrt{9-\mathrm{x}}}{\sqrt{\mathrm{x}}+\sqrt{9-\mathrm{x}}} \mathrm{dx}=\int_{3}^{6} 1 \mathrm{dx}=[\mathrm{x}]_{3}^{6}\) \(2 \mathrm{I} =[6-3]\) \(2 \mathrm{I} =3\) \(\mathrm{I} =\frac{3}{2}\)
SRM JEE-2019
Integral Calculus
86525
\(\int_{0}^{1} x(1-x)^{99} d x\) is equal to
86523
\(\int_{-1}^{3}\left[\tan ^{-1}\left(\frac{x}{x^{2}+1}\right)+\tan ^{-1}\left(\frac{x^{2}+1}{x}\right)\right] d x\) is equal to
1 \(2 \pi\)
2 \(\pi\)
3 \(\frac{\pi}{2}\)
4 \(\frac{\pi}{4}\)
Explanation:
(A) : Given, \(I =\int_{-1}^{3}\left[\tan ^{-1}\left(\frac{x}{x^{2}+1}\right)+\tan ^{-1}\left(\frac{x^{2}+1}{x}\right)\right] d x\) \(=\int_{-1}^{3} \tan ^{-1}\left\{\frac{\frac{x}{x^{2}+1}+\frac{x^{2}+1}{x}}{1-\frac{x}{x^{2}+1} \times \frac{x^{2}+1}{x}}\right\} d x\) \(=\int_{-1}^{3} \tan ^{-1}(\infty) d x=\int_{-1}^{3} \tan ^{-1}\left(\tan \frac{\pi}{2}\right) d x\) \(= \int_{-1}^{3}\left(\frac{\pi}{2}\right) d x=\left[\frac{\pi x}{2}\right]_{-1}^{3}=\frac{\pi}{2}[3+1]=2 \pi\)
COMEDK-2020
Integral Calculus
86524
The value of the integral \(\int_{3}^{6} \frac{\sqrt{x}}{\sqrt{9-x}+\sqrt{x}} d x\) is
1 \(3 / 2\)
2 2
3 1
4 \(1 / 2\)
Explanation:
(A) : Given, \(I=\int_{3}^{6} \frac{\sqrt{x}}{\sqrt{9-x}+\sqrt{x}} d x \tag{i}\) Apply property- \(I=\int_{3}^{6} \frac{\sqrt{6+3-x}}{\sqrt{9-(6+3-x)}+\sqrt{6+3-x}} d x\) \(I=\int_{3}^{6} \frac{\sqrt{9-x}}{\sqrt{x}+\sqrt{9-x}} d x \tag{ii}\) On adding equation (i) and (ii), we get - \(2 \mathrm{I}=\int_{3}^{6} \frac{\sqrt{\mathrm{x}}+\sqrt{9-\mathrm{x}}}{\sqrt{\mathrm{x}}+\sqrt{9-\mathrm{x}}} \mathrm{dx}=\int_{3}^{6} 1 \mathrm{dx}=[\mathrm{x}]_{3}^{6}\) \(2 \mathrm{I} =[6-3]\) \(2 \mathrm{I} =3\) \(\mathrm{I} =\frac{3}{2}\)
SRM JEE-2019
Integral Calculus
86525
\(\int_{0}^{1} x(1-x)^{99} d x\) is equal to
86523
\(\int_{-1}^{3}\left[\tan ^{-1}\left(\frac{x}{x^{2}+1}\right)+\tan ^{-1}\left(\frac{x^{2}+1}{x}\right)\right] d x\) is equal to
1 \(2 \pi\)
2 \(\pi\)
3 \(\frac{\pi}{2}\)
4 \(\frac{\pi}{4}\)
Explanation:
(A) : Given, \(I =\int_{-1}^{3}\left[\tan ^{-1}\left(\frac{x}{x^{2}+1}\right)+\tan ^{-1}\left(\frac{x^{2}+1}{x}\right)\right] d x\) \(=\int_{-1}^{3} \tan ^{-1}\left\{\frac{\frac{x}{x^{2}+1}+\frac{x^{2}+1}{x}}{1-\frac{x}{x^{2}+1} \times \frac{x^{2}+1}{x}}\right\} d x\) \(=\int_{-1}^{3} \tan ^{-1}(\infty) d x=\int_{-1}^{3} \tan ^{-1}\left(\tan \frac{\pi}{2}\right) d x\) \(= \int_{-1}^{3}\left(\frac{\pi}{2}\right) d x=\left[\frac{\pi x}{2}\right]_{-1}^{3}=\frac{\pi}{2}[3+1]=2 \pi\)
COMEDK-2020
Integral Calculus
86524
The value of the integral \(\int_{3}^{6} \frac{\sqrt{x}}{\sqrt{9-x}+\sqrt{x}} d x\) is
1 \(3 / 2\)
2 2
3 1
4 \(1 / 2\)
Explanation:
(A) : Given, \(I=\int_{3}^{6} \frac{\sqrt{x}}{\sqrt{9-x}+\sqrt{x}} d x \tag{i}\) Apply property- \(I=\int_{3}^{6} \frac{\sqrt{6+3-x}}{\sqrt{9-(6+3-x)}+\sqrt{6+3-x}} d x\) \(I=\int_{3}^{6} \frac{\sqrt{9-x}}{\sqrt{x}+\sqrt{9-x}} d x \tag{ii}\) On adding equation (i) and (ii), we get - \(2 \mathrm{I}=\int_{3}^{6} \frac{\sqrt{\mathrm{x}}+\sqrt{9-\mathrm{x}}}{\sqrt{\mathrm{x}}+\sqrt{9-\mathrm{x}}} \mathrm{dx}=\int_{3}^{6} 1 \mathrm{dx}=[\mathrm{x}]_{3}^{6}\) \(2 \mathrm{I} =[6-3]\) \(2 \mathrm{I} =3\) \(\mathrm{I} =\frac{3}{2}\)
SRM JEE-2019
Integral Calculus
86525
\(\int_{0}^{1} x(1-x)^{99} d x\) is equal to
86523
\(\int_{-1}^{3}\left[\tan ^{-1}\left(\frac{x}{x^{2}+1}\right)+\tan ^{-1}\left(\frac{x^{2}+1}{x}\right)\right] d x\) is equal to
1 \(2 \pi\)
2 \(\pi\)
3 \(\frac{\pi}{2}\)
4 \(\frac{\pi}{4}\)
Explanation:
(A) : Given, \(I =\int_{-1}^{3}\left[\tan ^{-1}\left(\frac{x}{x^{2}+1}\right)+\tan ^{-1}\left(\frac{x^{2}+1}{x}\right)\right] d x\) \(=\int_{-1}^{3} \tan ^{-1}\left\{\frac{\frac{x}{x^{2}+1}+\frac{x^{2}+1}{x}}{1-\frac{x}{x^{2}+1} \times \frac{x^{2}+1}{x}}\right\} d x\) \(=\int_{-1}^{3} \tan ^{-1}(\infty) d x=\int_{-1}^{3} \tan ^{-1}\left(\tan \frac{\pi}{2}\right) d x\) \(= \int_{-1}^{3}\left(\frac{\pi}{2}\right) d x=\left[\frac{\pi x}{2}\right]_{-1}^{3}=\frac{\pi}{2}[3+1]=2 \pi\)
COMEDK-2020
Integral Calculus
86524
The value of the integral \(\int_{3}^{6} \frac{\sqrt{x}}{\sqrt{9-x}+\sqrt{x}} d x\) is
1 \(3 / 2\)
2 2
3 1
4 \(1 / 2\)
Explanation:
(A) : Given, \(I=\int_{3}^{6} \frac{\sqrt{x}}{\sqrt{9-x}+\sqrt{x}} d x \tag{i}\) Apply property- \(I=\int_{3}^{6} \frac{\sqrt{6+3-x}}{\sqrt{9-(6+3-x)}+\sqrt{6+3-x}} d x\) \(I=\int_{3}^{6} \frac{\sqrt{9-x}}{\sqrt{x}+\sqrt{9-x}} d x \tag{ii}\) On adding equation (i) and (ii), we get - \(2 \mathrm{I}=\int_{3}^{6} \frac{\sqrt{\mathrm{x}}+\sqrt{9-\mathrm{x}}}{\sqrt{\mathrm{x}}+\sqrt{9-\mathrm{x}}} \mathrm{dx}=\int_{3}^{6} 1 \mathrm{dx}=[\mathrm{x}]_{3}^{6}\) \(2 \mathrm{I} =[6-3]\) \(2 \mathrm{I} =3\) \(\mathrm{I} =\frac{3}{2}\)
SRM JEE-2019
Integral Calculus
86525
\(\int_{0}^{1} x(1-x)^{99} d x\) is equal to
86523
\(\int_{-1}^{3}\left[\tan ^{-1}\left(\frac{x}{x^{2}+1}\right)+\tan ^{-1}\left(\frac{x^{2}+1}{x}\right)\right] d x\) is equal to
1 \(2 \pi\)
2 \(\pi\)
3 \(\frac{\pi}{2}\)
4 \(\frac{\pi}{4}\)
Explanation:
(A) : Given, \(I =\int_{-1}^{3}\left[\tan ^{-1}\left(\frac{x}{x^{2}+1}\right)+\tan ^{-1}\left(\frac{x^{2}+1}{x}\right)\right] d x\) \(=\int_{-1}^{3} \tan ^{-1}\left\{\frac{\frac{x}{x^{2}+1}+\frac{x^{2}+1}{x}}{1-\frac{x}{x^{2}+1} \times \frac{x^{2}+1}{x}}\right\} d x\) \(=\int_{-1}^{3} \tan ^{-1}(\infty) d x=\int_{-1}^{3} \tan ^{-1}\left(\tan \frac{\pi}{2}\right) d x\) \(= \int_{-1}^{3}\left(\frac{\pi}{2}\right) d x=\left[\frac{\pi x}{2}\right]_{-1}^{3}=\frac{\pi}{2}[3+1]=2 \pi\)
COMEDK-2020
Integral Calculus
86524
The value of the integral \(\int_{3}^{6} \frac{\sqrt{x}}{\sqrt{9-x}+\sqrt{x}} d x\) is
1 \(3 / 2\)
2 2
3 1
4 \(1 / 2\)
Explanation:
(A) : Given, \(I=\int_{3}^{6} \frac{\sqrt{x}}{\sqrt{9-x}+\sqrt{x}} d x \tag{i}\) Apply property- \(I=\int_{3}^{6} \frac{\sqrt{6+3-x}}{\sqrt{9-(6+3-x)}+\sqrt{6+3-x}} d x\) \(I=\int_{3}^{6} \frac{\sqrt{9-x}}{\sqrt{x}+\sqrt{9-x}} d x \tag{ii}\) On adding equation (i) and (ii), we get - \(2 \mathrm{I}=\int_{3}^{6} \frac{\sqrt{\mathrm{x}}+\sqrt{9-\mathrm{x}}}{\sqrt{\mathrm{x}}+\sqrt{9-\mathrm{x}}} \mathrm{dx}=\int_{3}^{6} 1 \mathrm{dx}=[\mathrm{x}]_{3}^{6}\) \(2 \mathrm{I} =[6-3]\) \(2 \mathrm{I} =3\) \(\mathrm{I} =\frac{3}{2}\)
SRM JEE-2019
Integral Calculus
86525
\(\int_{0}^{1} x(1-x)^{99} d x\) is equal to
