(C) : \(I=\int_{0}^{\pi / 2} \frac{\sin x+\cos x}{\sqrt{1+\sin 2 x}} d x\) \(I=\int_{0}^{\pi / 2} \frac{\sin x+\cos x}{\sqrt{\sin ^{2} x+\cos ^{2} x+2 \sin x \cos x}} d x\) \(I=\int_{0}^{\pi / 2} \frac{\sin x+\cos x}{\sqrt{(\sin x+\cos x)^{2}}} d x\) \(I=\int_{0}^{\pi / 2} d x=[x]_{0}^{\pi / 2}=\frac{\pi}{2}-0\) So, \(\quad \mathrm{I}=\frac{\pi}{2}\)
MHT CET-2005
Integral Calculus
86517
\(\int_{0}^{\pi / 2} \frac{1}{1+\sqrt{\tan x}} d x=\)
1 \(\frac{\pi}{4}\)
2 \(\frac{\pi}{2}\)
3 \(\pi\)
4 \(2 \pi\)
Explanation:
(A) : Given, \(\mathrm{I}=\int_{0}^{\pi / 2} \frac{1}{1+\sqrt{\tan x}} \mathrm{dx} \tag{i}\) \(I=\int_{0}^{\pi / 2} \frac{1}{1+\sqrt{\tan \left(\frac{\pi}{2}-x\right)}} d x\) \(\mathrm{I}=\int_{0}^{\pi / 2} \frac{\mathrm{dx}}{1+\sqrt{\cot \mathrm{x}}}=\int_{0}^{\pi / 2} \frac{1}{1+\sqrt{\frac{1}{\tan x}}} \mathrm{dx}\) \(\mathrm{I}=\int_{0}^{\pi / 2} \frac{1}{1+\frac{1}{\sqrt{\tan \mathrm{x}}}} \mathrm{dx}\) \(I=\int_{0}^{\pi / 2} \frac{\sqrt{\tan x}}{1+\sqrt{\tan x}} d x \tag{ii}\) On adding equation (i) and (ii), we get- \(2 I=\int_{0}^{\pi / 2} \frac{1+\sqrt{\tan x}}{1+\sqrt{\tan x}} d x=\int_{0}^{\pi / 2} 1 d x\) \(2 I=[x]_{0}^{\pi / 2} \Rightarrow I=\frac{\pi}{4}\)
MHT CET-2006
Integral Calculus
86518
The value of \(\int_{0}^{\pi / 2} \log (\tan x) d x\) is
1 2
2 1
3 0
4 3
Explanation:
(C) : Let, \(I=\int_{0}^{\pi / 2} \log (\tan x) d x\) Apply rule- \(I=\int_{0}^{\pi / 2} \log \tan \left(\frac{\pi}{2}-x\right) d x=\int_{0}^{\pi / 2} \log \cot x d x\) On adding equation (i) and (ii), we get - \(2 I=\int_{0}^{\pi / 2} \log (\tan x \times \cot x) d x=\int_{0}^{\pi / 2} \log 1 d x=0\)
(C) : \(I=\int_{0}^{\pi / 2} \frac{\sin x+\cos x}{\sqrt{1+\sin 2 x}} d x\) \(I=\int_{0}^{\pi / 2} \frac{\sin x+\cos x}{\sqrt{\sin ^{2} x+\cos ^{2} x+2 \sin x \cos x}} d x\) \(I=\int_{0}^{\pi / 2} \frac{\sin x+\cos x}{\sqrt{(\sin x+\cos x)^{2}}} d x\) \(I=\int_{0}^{\pi / 2} d x=[x]_{0}^{\pi / 2}=\frac{\pi}{2}-0\) So, \(\quad \mathrm{I}=\frac{\pi}{2}\)
MHT CET-2005
Integral Calculus
86517
\(\int_{0}^{\pi / 2} \frac{1}{1+\sqrt{\tan x}} d x=\)
1 \(\frac{\pi}{4}\)
2 \(\frac{\pi}{2}\)
3 \(\pi\)
4 \(2 \pi\)
Explanation:
(A) : Given, \(\mathrm{I}=\int_{0}^{\pi / 2} \frac{1}{1+\sqrt{\tan x}} \mathrm{dx} \tag{i}\) \(I=\int_{0}^{\pi / 2} \frac{1}{1+\sqrt{\tan \left(\frac{\pi}{2}-x\right)}} d x\) \(\mathrm{I}=\int_{0}^{\pi / 2} \frac{\mathrm{dx}}{1+\sqrt{\cot \mathrm{x}}}=\int_{0}^{\pi / 2} \frac{1}{1+\sqrt{\frac{1}{\tan x}}} \mathrm{dx}\) \(\mathrm{I}=\int_{0}^{\pi / 2} \frac{1}{1+\frac{1}{\sqrt{\tan \mathrm{x}}}} \mathrm{dx}\) \(I=\int_{0}^{\pi / 2} \frac{\sqrt{\tan x}}{1+\sqrt{\tan x}} d x \tag{ii}\) On adding equation (i) and (ii), we get- \(2 I=\int_{0}^{\pi / 2} \frac{1+\sqrt{\tan x}}{1+\sqrt{\tan x}} d x=\int_{0}^{\pi / 2} 1 d x\) \(2 I=[x]_{0}^{\pi / 2} \Rightarrow I=\frac{\pi}{4}\)
MHT CET-2006
Integral Calculus
86518
The value of \(\int_{0}^{\pi / 2} \log (\tan x) d x\) is
1 2
2 1
3 0
4 3
Explanation:
(C) : Let, \(I=\int_{0}^{\pi / 2} \log (\tan x) d x\) Apply rule- \(I=\int_{0}^{\pi / 2} \log \tan \left(\frac{\pi}{2}-x\right) d x=\int_{0}^{\pi / 2} \log \cot x d x\) On adding equation (i) and (ii), we get - \(2 I=\int_{0}^{\pi / 2} \log (\tan x \times \cot x) d x=\int_{0}^{\pi / 2} \log 1 d x=0\)
(C) : \(I=\int_{0}^{\pi / 2} \frac{\sin x+\cos x}{\sqrt{1+\sin 2 x}} d x\) \(I=\int_{0}^{\pi / 2} \frac{\sin x+\cos x}{\sqrt{\sin ^{2} x+\cos ^{2} x+2 \sin x \cos x}} d x\) \(I=\int_{0}^{\pi / 2} \frac{\sin x+\cos x}{\sqrt{(\sin x+\cos x)^{2}}} d x\) \(I=\int_{0}^{\pi / 2} d x=[x]_{0}^{\pi / 2}=\frac{\pi}{2}-0\) So, \(\quad \mathrm{I}=\frac{\pi}{2}\)
MHT CET-2005
Integral Calculus
86517
\(\int_{0}^{\pi / 2} \frac{1}{1+\sqrt{\tan x}} d x=\)
1 \(\frac{\pi}{4}\)
2 \(\frac{\pi}{2}\)
3 \(\pi\)
4 \(2 \pi\)
Explanation:
(A) : Given, \(\mathrm{I}=\int_{0}^{\pi / 2} \frac{1}{1+\sqrt{\tan x}} \mathrm{dx} \tag{i}\) \(I=\int_{0}^{\pi / 2} \frac{1}{1+\sqrt{\tan \left(\frac{\pi}{2}-x\right)}} d x\) \(\mathrm{I}=\int_{0}^{\pi / 2} \frac{\mathrm{dx}}{1+\sqrt{\cot \mathrm{x}}}=\int_{0}^{\pi / 2} \frac{1}{1+\sqrt{\frac{1}{\tan x}}} \mathrm{dx}\) \(\mathrm{I}=\int_{0}^{\pi / 2} \frac{1}{1+\frac{1}{\sqrt{\tan \mathrm{x}}}} \mathrm{dx}\) \(I=\int_{0}^{\pi / 2} \frac{\sqrt{\tan x}}{1+\sqrt{\tan x}} d x \tag{ii}\) On adding equation (i) and (ii), we get- \(2 I=\int_{0}^{\pi / 2} \frac{1+\sqrt{\tan x}}{1+\sqrt{\tan x}} d x=\int_{0}^{\pi / 2} 1 d x\) \(2 I=[x]_{0}^{\pi / 2} \Rightarrow I=\frac{\pi}{4}\)
MHT CET-2006
Integral Calculus
86518
The value of \(\int_{0}^{\pi / 2} \log (\tan x) d x\) is
1 2
2 1
3 0
4 3
Explanation:
(C) : Let, \(I=\int_{0}^{\pi / 2} \log (\tan x) d x\) Apply rule- \(I=\int_{0}^{\pi / 2} \log \tan \left(\frac{\pi}{2}-x\right) d x=\int_{0}^{\pi / 2} \log \cot x d x\) On adding equation (i) and (ii), we get - \(2 I=\int_{0}^{\pi / 2} \log (\tan x \times \cot x) d x=\int_{0}^{\pi / 2} \log 1 d x=0\)
(C) : \(I=\int_{0}^{\pi / 2} \frac{\sin x+\cos x}{\sqrt{1+\sin 2 x}} d x\) \(I=\int_{0}^{\pi / 2} \frac{\sin x+\cos x}{\sqrt{\sin ^{2} x+\cos ^{2} x+2 \sin x \cos x}} d x\) \(I=\int_{0}^{\pi / 2} \frac{\sin x+\cos x}{\sqrt{(\sin x+\cos x)^{2}}} d x\) \(I=\int_{0}^{\pi / 2} d x=[x]_{0}^{\pi / 2}=\frac{\pi}{2}-0\) So, \(\quad \mathrm{I}=\frac{\pi}{2}\)
MHT CET-2005
Integral Calculus
86517
\(\int_{0}^{\pi / 2} \frac{1}{1+\sqrt{\tan x}} d x=\)
1 \(\frac{\pi}{4}\)
2 \(\frac{\pi}{2}\)
3 \(\pi\)
4 \(2 \pi\)
Explanation:
(A) : Given, \(\mathrm{I}=\int_{0}^{\pi / 2} \frac{1}{1+\sqrt{\tan x}} \mathrm{dx} \tag{i}\) \(I=\int_{0}^{\pi / 2} \frac{1}{1+\sqrt{\tan \left(\frac{\pi}{2}-x\right)}} d x\) \(\mathrm{I}=\int_{0}^{\pi / 2} \frac{\mathrm{dx}}{1+\sqrt{\cot \mathrm{x}}}=\int_{0}^{\pi / 2} \frac{1}{1+\sqrt{\frac{1}{\tan x}}} \mathrm{dx}\) \(\mathrm{I}=\int_{0}^{\pi / 2} \frac{1}{1+\frac{1}{\sqrt{\tan \mathrm{x}}}} \mathrm{dx}\) \(I=\int_{0}^{\pi / 2} \frac{\sqrt{\tan x}}{1+\sqrt{\tan x}} d x \tag{ii}\) On adding equation (i) and (ii), we get- \(2 I=\int_{0}^{\pi / 2} \frac{1+\sqrt{\tan x}}{1+\sqrt{\tan x}} d x=\int_{0}^{\pi / 2} 1 d x\) \(2 I=[x]_{0}^{\pi / 2} \Rightarrow I=\frac{\pi}{4}\)
MHT CET-2006
Integral Calculus
86518
The value of \(\int_{0}^{\pi / 2} \log (\tan x) d x\) is
1 2
2 1
3 0
4 3
Explanation:
(C) : Let, \(I=\int_{0}^{\pi / 2} \log (\tan x) d x\) Apply rule- \(I=\int_{0}^{\pi / 2} \log \tan \left(\frac{\pi}{2}-x\right) d x=\int_{0}^{\pi / 2} \log \cot x d x\) On adding equation (i) and (ii), we get - \(2 I=\int_{0}^{\pi / 2} \log (\tan x \times \cot x) d x=\int_{0}^{\pi / 2} \log 1 d x=0\)
