(C) : Let,\(I=\int_{0}^{1} \sqrt{\frac{1+x}{1-x}} d x=\int_{0}^{1} \sqrt{\frac{(1+x)^{2}}{\left(1^{2}-x^{2}\right)}} d x\) \(=\int_{0}^{1} \frac{1+\mathrm{x}}{\sqrt{1-\mathrm{x}^{2}}} \mathrm{dx}=\int_{0}^{1} \frac{1}{\sqrt{1-\mathrm{x}^{2}}} \mathrm{dx}+\int_{0}^{1} \frac{\mathrm{x}}{\sqrt{1-\mathrm{x}^{2}}} \mathrm{dx}\) Put, \(1-\mathrm{x}^{2}=\mathrm{t}\) \(-2 \mathrm{xdx}=\mathrm{dt}\) Now, when \(\mathrm{x}=0, \mathrm{t}=1\) and \(\mathrm{x}=1, \mathrm{t}=0\) \(=\left[\sin ^{-1} \mathrm{x}\right]_{0}^{1}-\frac{1}{2} \int_{1}^{0} \frac{\mathrm{dt}}{\sqrt{\mathrm{t}}}=\left(\sin ^{-1} 1-\sin ^{-1} 0\right)+\frac{1}{2} \int_{0}^{1} \frac{\mathrm{dt}}{\sqrt{\mathrm{t}}}\) \(=\frac{\pi}{2}+[\sqrt{\mathrm{t}}]_{0}^{1}=\frac{\pi}{2}+1\)
Karnataka CET-2019
Integral Calculus
86692
The value of \(\int_{2}^{8} \frac{\sqrt{10-x}}{\sqrt{x}+\sqrt{10-x}} d x\) is
1 10
2 0
3 8
4 3
Explanation:
(D) : Let, \(I=\int_{2}^{8} \frac{\sqrt{10-x}}{\sqrt{x}+\sqrt{10-x}} d x\) \(=\int_{2}^{8} \frac{\sqrt{10-10+x}}{\sqrt{10-x}+\sqrt{10-10+x}} d x\) \(I=\int_{2}^{8} \frac{\sqrt{x}}{\sqrt{10-x}+\sqrt{x}} d x\) Adding equation (i) and (ii) \(2 \mathrm{I}=\int_{2}^{8} \mathrm{dx}\) \(2 \mathrm{I}=[\mathrm{x}]_{2}^{8}\) \(2 \mathrm{I}=[8-2]=6\) \(\therefore \mathrm{I}=3\)
Karnataka CET-2016
Integral Calculus
86693
\(\int_{0}^{\pi / 4} \log \left(\frac{\sin x+\cos x}{\cos x}\right) d x\) is equal to
1 \(\frac{\pi}{2} \log 2\)
2 \(\log 2\)
3 \(\frac{\pi}{4} \log 2\)
4 \(\frac{\pi}{8} \log 2\)
Explanation:
(D) : Let, \(I=\int_{0}^{\pi / 4} \log \left(\frac{\sin x+\cos x}{\cos x}\right) d x\) \(I=\int_{0}^{\pi / 4} \log [1+\tan x]\) \(I=\int_{0}^{\pi / 4} \log \left[1+\tan \left(\frac{\pi}{4}-x\right)\right]\) \(\quad\left[\because \int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x\right]\) \(=\int_{0}^{\pi / 4} \log \left[1+\frac{\tan \frac{\pi}{4}-\tan x}{1+\tan \frac{\pi}{4} \tan x}\right] d x\) \(=\int_{0}^{\pi / 4} \log \left[1+\frac{1-\tan x}{1+\tan x}\right] d x\) \(=\int_{0}^{\pi / 4} \log \left[\frac{1+\tan x+1-\tan x}{1+\tan x}\right] d x\) \(I=\int_{0}^{\pi / 4} \log \left(\frac{2}{1+\tan x}\right) d x \tag{ii}\) Adding equation (i) and (ii) \(2 I=\int_{0}^{\pi / 4} \log \left[\frac{2}{(1+\tan x)} \times(1+\tan x)\right] d x\) \(2 I=\log 2 \int_{0}^{\pi / 4} 1 \mathrm{dx}\) \(2 I=\log 2[x]_{0}^{\pi / 4}\) \(2 I=\log 2 \times \frac{\pi}{4}\) \(I=\frac{\pi}{8} \log 2\)
(C) : Let,\(I=\int_{0}^{1} \sqrt{\frac{1+x}{1-x}} d x=\int_{0}^{1} \sqrt{\frac{(1+x)^{2}}{\left(1^{2}-x^{2}\right)}} d x\) \(=\int_{0}^{1} \frac{1+\mathrm{x}}{\sqrt{1-\mathrm{x}^{2}}} \mathrm{dx}=\int_{0}^{1} \frac{1}{\sqrt{1-\mathrm{x}^{2}}} \mathrm{dx}+\int_{0}^{1} \frac{\mathrm{x}}{\sqrt{1-\mathrm{x}^{2}}} \mathrm{dx}\) Put, \(1-\mathrm{x}^{2}=\mathrm{t}\) \(-2 \mathrm{xdx}=\mathrm{dt}\) Now, when \(\mathrm{x}=0, \mathrm{t}=1\) and \(\mathrm{x}=1, \mathrm{t}=0\) \(=\left[\sin ^{-1} \mathrm{x}\right]_{0}^{1}-\frac{1}{2} \int_{1}^{0} \frac{\mathrm{dt}}{\sqrt{\mathrm{t}}}=\left(\sin ^{-1} 1-\sin ^{-1} 0\right)+\frac{1}{2} \int_{0}^{1} \frac{\mathrm{dt}}{\sqrt{\mathrm{t}}}\) \(=\frac{\pi}{2}+[\sqrt{\mathrm{t}}]_{0}^{1}=\frac{\pi}{2}+1\)
Karnataka CET-2019
Integral Calculus
86692
The value of \(\int_{2}^{8} \frac{\sqrt{10-x}}{\sqrt{x}+\sqrt{10-x}} d x\) is
1 10
2 0
3 8
4 3
Explanation:
(D) : Let, \(I=\int_{2}^{8} \frac{\sqrt{10-x}}{\sqrt{x}+\sqrt{10-x}} d x\) \(=\int_{2}^{8} \frac{\sqrt{10-10+x}}{\sqrt{10-x}+\sqrt{10-10+x}} d x\) \(I=\int_{2}^{8} \frac{\sqrt{x}}{\sqrt{10-x}+\sqrt{x}} d x\) Adding equation (i) and (ii) \(2 \mathrm{I}=\int_{2}^{8} \mathrm{dx}\) \(2 \mathrm{I}=[\mathrm{x}]_{2}^{8}\) \(2 \mathrm{I}=[8-2]=6\) \(\therefore \mathrm{I}=3\)
Karnataka CET-2016
Integral Calculus
86693
\(\int_{0}^{\pi / 4} \log \left(\frac{\sin x+\cos x}{\cos x}\right) d x\) is equal to
1 \(\frac{\pi}{2} \log 2\)
2 \(\log 2\)
3 \(\frac{\pi}{4} \log 2\)
4 \(\frac{\pi}{8} \log 2\)
Explanation:
(D) : Let, \(I=\int_{0}^{\pi / 4} \log \left(\frac{\sin x+\cos x}{\cos x}\right) d x\) \(I=\int_{0}^{\pi / 4} \log [1+\tan x]\) \(I=\int_{0}^{\pi / 4} \log \left[1+\tan \left(\frac{\pi}{4}-x\right)\right]\) \(\quad\left[\because \int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x\right]\) \(=\int_{0}^{\pi / 4} \log \left[1+\frac{\tan \frac{\pi}{4}-\tan x}{1+\tan \frac{\pi}{4} \tan x}\right] d x\) \(=\int_{0}^{\pi / 4} \log \left[1+\frac{1-\tan x}{1+\tan x}\right] d x\) \(=\int_{0}^{\pi / 4} \log \left[\frac{1+\tan x+1-\tan x}{1+\tan x}\right] d x\) \(I=\int_{0}^{\pi / 4} \log \left(\frac{2}{1+\tan x}\right) d x \tag{ii}\) Adding equation (i) and (ii) \(2 I=\int_{0}^{\pi / 4} \log \left[\frac{2}{(1+\tan x)} \times(1+\tan x)\right] d x\) \(2 I=\log 2 \int_{0}^{\pi / 4} 1 \mathrm{dx}\) \(2 I=\log 2[x]_{0}^{\pi / 4}\) \(2 I=\log 2 \times \frac{\pi}{4}\) \(I=\frac{\pi}{8} \log 2\)
(C) : Let,\(I=\int_{0}^{1} \sqrt{\frac{1+x}{1-x}} d x=\int_{0}^{1} \sqrt{\frac{(1+x)^{2}}{\left(1^{2}-x^{2}\right)}} d x\) \(=\int_{0}^{1} \frac{1+\mathrm{x}}{\sqrt{1-\mathrm{x}^{2}}} \mathrm{dx}=\int_{0}^{1} \frac{1}{\sqrt{1-\mathrm{x}^{2}}} \mathrm{dx}+\int_{0}^{1} \frac{\mathrm{x}}{\sqrt{1-\mathrm{x}^{2}}} \mathrm{dx}\) Put, \(1-\mathrm{x}^{2}=\mathrm{t}\) \(-2 \mathrm{xdx}=\mathrm{dt}\) Now, when \(\mathrm{x}=0, \mathrm{t}=1\) and \(\mathrm{x}=1, \mathrm{t}=0\) \(=\left[\sin ^{-1} \mathrm{x}\right]_{0}^{1}-\frac{1}{2} \int_{1}^{0} \frac{\mathrm{dt}}{\sqrt{\mathrm{t}}}=\left(\sin ^{-1} 1-\sin ^{-1} 0\right)+\frac{1}{2} \int_{0}^{1} \frac{\mathrm{dt}}{\sqrt{\mathrm{t}}}\) \(=\frac{\pi}{2}+[\sqrt{\mathrm{t}}]_{0}^{1}=\frac{\pi}{2}+1\)
Karnataka CET-2019
Integral Calculus
86692
The value of \(\int_{2}^{8} \frac{\sqrt{10-x}}{\sqrt{x}+\sqrt{10-x}} d x\) is
1 10
2 0
3 8
4 3
Explanation:
(D) : Let, \(I=\int_{2}^{8} \frac{\sqrt{10-x}}{\sqrt{x}+\sqrt{10-x}} d x\) \(=\int_{2}^{8} \frac{\sqrt{10-10+x}}{\sqrt{10-x}+\sqrt{10-10+x}} d x\) \(I=\int_{2}^{8} \frac{\sqrt{x}}{\sqrt{10-x}+\sqrt{x}} d x\) Adding equation (i) and (ii) \(2 \mathrm{I}=\int_{2}^{8} \mathrm{dx}\) \(2 \mathrm{I}=[\mathrm{x}]_{2}^{8}\) \(2 \mathrm{I}=[8-2]=6\) \(\therefore \mathrm{I}=3\)
Karnataka CET-2016
Integral Calculus
86693
\(\int_{0}^{\pi / 4} \log \left(\frac{\sin x+\cos x}{\cos x}\right) d x\) is equal to
1 \(\frac{\pi}{2} \log 2\)
2 \(\log 2\)
3 \(\frac{\pi}{4} \log 2\)
4 \(\frac{\pi}{8} \log 2\)
Explanation:
(D) : Let, \(I=\int_{0}^{\pi / 4} \log \left(\frac{\sin x+\cos x}{\cos x}\right) d x\) \(I=\int_{0}^{\pi / 4} \log [1+\tan x]\) \(I=\int_{0}^{\pi / 4} \log \left[1+\tan \left(\frac{\pi}{4}-x\right)\right]\) \(\quad\left[\because \int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x\right]\) \(=\int_{0}^{\pi / 4} \log \left[1+\frac{\tan \frac{\pi}{4}-\tan x}{1+\tan \frac{\pi}{4} \tan x}\right] d x\) \(=\int_{0}^{\pi / 4} \log \left[1+\frac{1-\tan x}{1+\tan x}\right] d x\) \(=\int_{0}^{\pi / 4} \log \left[\frac{1+\tan x+1-\tan x}{1+\tan x}\right] d x\) \(I=\int_{0}^{\pi / 4} \log \left(\frac{2}{1+\tan x}\right) d x \tag{ii}\) Adding equation (i) and (ii) \(2 I=\int_{0}^{\pi / 4} \log \left[\frac{2}{(1+\tan x)} \times(1+\tan x)\right] d x\) \(2 I=\log 2 \int_{0}^{\pi / 4} 1 \mathrm{dx}\) \(2 I=\log 2[x]_{0}^{\pi / 4}\) \(2 I=\log 2 \times \frac{\pi}{4}\) \(I=\frac{\pi}{8} \log 2\)
(C) : Let,\(I=\int_{0}^{1} \sqrt{\frac{1+x}{1-x}} d x=\int_{0}^{1} \sqrt{\frac{(1+x)^{2}}{\left(1^{2}-x^{2}\right)}} d x\) \(=\int_{0}^{1} \frac{1+\mathrm{x}}{\sqrt{1-\mathrm{x}^{2}}} \mathrm{dx}=\int_{0}^{1} \frac{1}{\sqrt{1-\mathrm{x}^{2}}} \mathrm{dx}+\int_{0}^{1} \frac{\mathrm{x}}{\sqrt{1-\mathrm{x}^{2}}} \mathrm{dx}\) Put, \(1-\mathrm{x}^{2}=\mathrm{t}\) \(-2 \mathrm{xdx}=\mathrm{dt}\) Now, when \(\mathrm{x}=0, \mathrm{t}=1\) and \(\mathrm{x}=1, \mathrm{t}=0\) \(=\left[\sin ^{-1} \mathrm{x}\right]_{0}^{1}-\frac{1}{2} \int_{1}^{0} \frac{\mathrm{dt}}{\sqrt{\mathrm{t}}}=\left(\sin ^{-1} 1-\sin ^{-1} 0\right)+\frac{1}{2} \int_{0}^{1} \frac{\mathrm{dt}}{\sqrt{\mathrm{t}}}\) \(=\frac{\pi}{2}+[\sqrt{\mathrm{t}}]_{0}^{1}=\frac{\pi}{2}+1\)
Karnataka CET-2019
Integral Calculus
86692
The value of \(\int_{2}^{8} \frac{\sqrt{10-x}}{\sqrt{x}+\sqrt{10-x}} d x\) is
1 10
2 0
3 8
4 3
Explanation:
(D) : Let, \(I=\int_{2}^{8} \frac{\sqrt{10-x}}{\sqrt{x}+\sqrt{10-x}} d x\) \(=\int_{2}^{8} \frac{\sqrt{10-10+x}}{\sqrt{10-x}+\sqrt{10-10+x}} d x\) \(I=\int_{2}^{8} \frac{\sqrt{x}}{\sqrt{10-x}+\sqrt{x}} d x\) Adding equation (i) and (ii) \(2 \mathrm{I}=\int_{2}^{8} \mathrm{dx}\) \(2 \mathrm{I}=[\mathrm{x}]_{2}^{8}\) \(2 \mathrm{I}=[8-2]=6\) \(\therefore \mathrm{I}=3\)
Karnataka CET-2016
Integral Calculus
86693
\(\int_{0}^{\pi / 4} \log \left(\frac{\sin x+\cos x}{\cos x}\right) d x\) is equal to
1 \(\frac{\pi}{2} \log 2\)
2 \(\log 2\)
3 \(\frac{\pi}{4} \log 2\)
4 \(\frac{\pi}{8} \log 2\)
Explanation:
(D) : Let, \(I=\int_{0}^{\pi / 4} \log \left(\frac{\sin x+\cos x}{\cos x}\right) d x\) \(I=\int_{0}^{\pi / 4} \log [1+\tan x]\) \(I=\int_{0}^{\pi / 4} \log \left[1+\tan \left(\frac{\pi}{4}-x\right)\right]\) \(\quad\left[\because \int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x\right]\) \(=\int_{0}^{\pi / 4} \log \left[1+\frac{\tan \frac{\pi}{4}-\tan x}{1+\tan \frac{\pi}{4} \tan x}\right] d x\) \(=\int_{0}^{\pi / 4} \log \left[1+\frac{1-\tan x}{1+\tan x}\right] d x\) \(=\int_{0}^{\pi / 4} \log \left[\frac{1+\tan x+1-\tan x}{1+\tan x}\right] d x\) \(I=\int_{0}^{\pi / 4} \log \left(\frac{2}{1+\tan x}\right) d x \tag{ii}\) Adding equation (i) and (ii) \(2 I=\int_{0}^{\pi / 4} \log \left[\frac{2}{(1+\tan x)} \times(1+\tan x)\right] d x\) \(2 I=\log 2 \int_{0}^{\pi / 4} 1 \mathrm{dx}\) \(2 I=\log 2[x]_{0}^{\pi / 4}\) \(2 I=\log 2 \times \frac{\pi}{4}\) \(I=\frac{\pi}{8} \log 2\)