86729
The value of the integral \(\int_{-1}^{1}\left\{\frac{x^{2015}}{e^{|x|}\left(x^{2}+\cos x\right)}+\frac{1}{e^{|x|}}\right\} d x\) is equal to
1 0
2 \(1-\mathrm{e}^{-1}\)
3 \(2 \mathrm{e}^{-1}\)
4 \(2\left(1-\mathrm{e}^{-1}\right)\)
Explanation:
(D) : Let, \(I=\int_{-1}^{1}\left\{\frac{x^{2015}}{e^{|x|\left(x^{2}+\cos x\right)}}+\frac{1}{e^{|x|}}\right\} d x\) \(=\int_{-1}^{1} \frac{x^{2015}}{e^{|x|}\left(x^{2}+\cos x\right)} d x+\int_{-1}^{1} \frac{1}{e^{|x|}} d x\) Let \(f(x)=\frac{x^{2015}}{e^{|x|}\left(x^{2}+\cos x\right)}\) and \(g(x)=\frac{1}{e^{|x|}}\) \(f(-x)=\frac{(-x)^{2015}}{e^{-|x|}\left[(-x)^{2}+\cos (-x)\right]}\) \(=\frac{-x^{2015}}{e^{|x|}\left(x^{2}+\cos x\right)}\) and \(g(-x)=\frac{1}{e^{|-x|}}=\frac{1}{e^{|x|}}=g(x)\) \(\because \mathrm{f}(\mathrm{x})\) is odd function and \(\mathrm{g}(\mathrm{x})\) is even function. \(\int_{-1}^{1} \mathrm{f}(\mathrm{x}) \mathrm{dx}=0 \text { and }\) \(\int_{-1}^{1} \mathrm{~g}(\mathrm{x}) \mathrm{dx}=2 \int_{0}^{1} \mathrm{~g}(\mathrm{x}) \mathrm{dx}\) \(\mathrm{I}=2 \int_{0}^{1} \frac{1}{\mathrm{e}^{\mid \mathrm{x|}}} \mathrm{dx}\) \(=2 \int_{0}^{1} \frac{1}{\mathrm{e}^{\mathrm{x}}} \mathrm{dx}=2 \int_{0}^{1} \mathrm{e}^{-\mathrm{x}} \mathrm{dx}\) \(= 2\left[\mathrm{e}^{-\mathrm{x}}\right]_{0}^{1}=2\left[\mathrm{e}^{-1}-\mathrm{e}^{0}\right]\) \(= 2\left[\mathrm{e}^{-1}-1\right]\) \(= 2\left[1-\mathrm{e}^{-1}\right]\)
WB JEE-2019
Integral Calculus
86730
The value of the integral \(\int_{-1 / 2}^{1 / 2}\left\{\left(\frac{x+1}{x-1}\right)^{2}+\left(\frac{x-1}{x+1}\right)^{2}-2\right\}^{1 / 2} d x\) is equal to
(C): Given \(\int_{-1 / 2}^{1 / 2}\left\{\left(\frac{x+1}{x-1}\right)^{2}+\left(\frac{x-1}{x+1}\right)^{2}-2\right\}^{1 / 2} d x\) \(=\int_{-1 / 2}^{1 / 2}\left[\left(\frac{x+1}{x-1}-\frac{x-1}{x+1}\right)^{2}\right]^{1 / 2} d x=\int_{-1 / 2}^{1 / 2}\left(\frac{4 x}{x^{2}-1}\right) d x\) \(=\int_{-1 / 2}^{0}\left(\frac{4 x}{x^{2}-1}\right) d x-\int_{0}^{1 / 2}\left(\frac{4 x}{x^{2}-1}\right) d x\) \(=-4 \int_{-1 / 2}^{0} \frac{x}{1-x^{2}} d x+4 \int_{0}^{1 / 2} \frac{x}{1-x^{2}} d x\) \(=2\left[\log \left(1-x^{2}\right)\right]_{-1 / 2}^{0}-2\left[\log \left(1-x^{2}\right)\right]_{0}^{1 / 2}\) \(=-2 \log \left(1-\frac{1}{4}\right)-2 \log \left(1-\frac{1}{4}\right)\) \(=-4 \log \frac{3}{4}=4 \log \frac{4}{3}\)
WB JEE-2021
Integral Calculus
86731
\(\int_{-\pi / 4}^{\pi / 4} \frac{d x}{1+\cos 2 x}\) is equal to
1 1
2 0
3 2
4 4
Explanation:
(A) : Given, \(\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{\mathrm{dx}}{1+\cos 2 \mathrm{x}}\) Let, \(\quad \mathrm{I}=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{d \mathrm{x}}{1+2 \cos ^{2} \mathrm{x}-1}\) \(=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{\mathrm{dx}}{2 \cos ^{2} \mathrm{x}}\) \(\because \quad \mathrm{f}(\mathrm{x})=\frac{1}{\cos ^{2} \mathrm{x}}\) is an even function \(\therefore \quad\) By using property \(\int_{-a}^{a} f(x) d x=2 \int_{0}^{a} f(x) d x\) We get, \(I=2 x \frac{1}{2} \int_{0}^{\pi / 4} \sec ^{2} x d x\) \(=|\tan x|_{0}^{\pi / 4}=\tan \frac{\pi}{4}-\tan 0\) \(=1\)
86729
The value of the integral \(\int_{-1}^{1}\left\{\frac{x^{2015}}{e^{|x|}\left(x^{2}+\cos x\right)}+\frac{1}{e^{|x|}}\right\} d x\) is equal to
1 0
2 \(1-\mathrm{e}^{-1}\)
3 \(2 \mathrm{e}^{-1}\)
4 \(2\left(1-\mathrm{e}^{-1}\right)\)
Explanation:
(D) : Let, \(I=\int_{-1}^{1}\left\{\frac{x^{2015}}{e^{|x|\left(x^{2}+\cos x\right)}}+\frac{1}{e^{|x|}}\right\} d x\) \(=\int_{-1}^{1} \frac{x^{2015}}{e^{|x|}\left(x^{2}+\cos x\right)} d x+\int_{-1}^{1} \frac{1}{e^{|x|}} d x\) Let \(f(x)=\frac{x^{2015}}{e^{|x|}\left(x^{2}+\cos x\right)}\) and \(g(x)=\frac{1}{e^{|x|}}\) \(f(-x)=\frac{(-x)^{2015}}{e^{-|x|}\left[(-x)^{2}+\cos (-x)\right]}\) \(=\frac{-x^{2015}}{e^{|x|}\left(x^{2}+\cos x\right)}\) and \(g(-x)=\frac{1}{e^{|-x|}}=\frac{1}{e^{|x|}}=g(x)\) \(\because \mathrm{f}(\mathrm{x})\) is odd function and \(\mathrm{g}(\mathrm{x})\) is even function. \(\int_{-1}^{1} \mathrm{f}(\mathrm{x}) \mathrm{dx}=0 \text { and }\) \(\int_{-1}^{1} \mathrm{~g}(\mathrm{x}) \mathrm{dx}=2 \int_{0}^{1} \mathrm{~g}(\mathrm{x}) \mathrm{dx}\) \(\mathrm{I}=2 \int_{0}^{1} \frac{1}{\mathrm{e}^{\mid \mathrm{x|}}} \mathrm{dx}\) \(=2 \int_{0}^{1} \frac{1}{\mathrm{e}^{\mathrm{x}}} \mathrm{dx}=2 \int_{0}^{1} \mathrm{e}^{-\mathrm{x}} \mathrm{dx}\) \(= 2\left[\mathrm{e}^{-\mathrm{x}}\right]_{0}^{1}=2\left[\mathrm{e}^{-1}-\mathrm{e}^{0}\right]\) \(= 2\left[\mathrm{e}^{-1}-1\right]\) \(= 2\left[1-\mathrm{e}^{-1}\right]\)
WB JEE-2019
Integral Calculus
86730
The value of the integral \(\int_{-1 / 2}^{1 / 2}\left\{\left(\frac{x+1}{x-1}\right)^{2}+\left(\frac{x-1}{x+1}\right)^{2}-2\right\}^{1 / 2} d x\) is equal to
(C): Given \(\int_{-1 / 2}^{1 / 2}\left\{\left(\frac{x+1}{x-1}\right)^{2}+\left(\frac{x-1}{x+1}\right)^{2}-2\right\}^{1 / 2} d x\) \(=\int_{-1 / 2}^{1 / 2}\left[\left(\frac{x+1}{x-1}-\frac{x-1}{x+1}\right)^{2}\right]^{1 / 2} d x=\int_{-1 / 2}^{1 / 2}\left(\frac{4 x}{x^{2}-1}\right) d x\) \(=\int_{-1 / 2}^{0}\left(\frac{4 x}{x^{2}-1}\right) d x-\int_{0}^{1 / 2}\left(\frac{4 x}{x^{2}-1}\right) d x\) \(=-4 \int_{-1 / 2}^{0} \frac{x}{1-x^{2}} d x+4 \int_{0}^{1 / 2} \frac{x}{1-x^{2}} d x\) \(=2\left[\log \left(1-x^{2}\right)\right]_{-1 / 2}^{0}-2\left[\log \left(1-x^{2}\right)\right]_{0}^{1 / 2}\) \(=-2 \log \left(1-\frac{1}{4}\right)-2 \log \left(1-\frac{1}{4}\right)\) \(=-4 \log \frac{3}{4}=4 \log \frac{4}{3}\)
WB JEE-2021
Integral Calculus
86731
\(\int_{-\pi / 4}^{\pi / 4} \frac{d x}{1+\cos 2 x}\) is equal to
1 1
2 0
3 2
4 4
Explanation:
(A) : Given, \(\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{\mathrm{dx}}{1+\cos 2 \mathrm{x}}\) Let, \(\quad \mathrm{I}=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{d \mathrm{x}}{1+2 \cos ^{2} \mathrm{x}-1}\) \(=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{\mathrm{dx}}{2 \cos ^{2} \mathrm{x}}\) \(\because \quad \mathrm{f}(\mathrm{x})=\frac{1}{\cos ^{2} \mathrm{x}}\) is an even function \(\therefore \quad\) By using property \(\int_{-a}^{a} f(x) d x=2 \int_{0}^{a} f(x) d x\) We get, \(I=2 x \frac{1}{2} \int_{0}^{\pi / 4} \sec ^{2} x d x\) \(=|\tan x|_{0}^{\pi / 4}=\tan \frac{\pi}{4}-\tan 0\) \(=1\)
86729
The value of the integral \(\int_{-1}^{1}\left\{\frac{x^{2015}}{e^{|x|}\left(x^{2}+\cos x\right)}+\frac{1}{e^{|x|}}\right\} d x\) is equal to
1 0
2 \(1-\mathrm{e}^{-1}\)
3 \(2 \mathrm{e}^{-1}\)
4 \(2\left(1-\mathrm{e}^{-1}\right)\)
Explanation:
(D) : Let, \(I=\int_{-1}^{1}\left\{\frac{x^{2015}}{e^{|x|\left(x^{2}+\cos x\right)}}+\frac{1}{e^{|x|}}\right\} d x\) \(=\int_{-1}^{1} \frac{x^{2015}}{e^{|x|}\left(x^{2}+\cos x\right)} d x+\int_{-1}^{1} \frac{1}{e^{|x|}} d x\) Let \(f(x)=\frac{x^{2015}}{e^{|x|}\left(x^{2}+\cos x\right)}\) and \(g(x)=\frac{1}{e^{|x|}}\) \(f(-x)=\frac{(-x)^{2015}}{e^{-|x|}\left[(-x)^{2}+\cos (-x)\right]}\) \(=\frac{-x^{2015}}{e^{|x|}\left(x^{2}+\cos x\right)}\) and \(g(-x)=\frac{1}{e^{|-x|}}=\frac{1}{e^{|x|}}=g(x)\) \(\because \mathrm{f}(\mathrm{x})\) is odd function and \(\mathrm{g}(\mathrm{x})\) is even function. \(\int_{-1}^{1} \mathrm{f}(\mathrm{x}) \mathrm{dx}=0 \text { and }\) \(\int_{-1}^{1} \mathrm{~g}(\mathrm{x}) \mathrm{dx}=2 \int_{0}^{1} \mathrm{~g}(\mathrm{x}) \mathrm{dx}\) \(\mathrm{I}=2 \int_{0}^{1} \frac{1}{\mathrm{e}^{\mid \mathrm{x|}}} \mathrm{dx}\) \(=2 \int_{0}^{1} \frac{1}{\mathrm{e}^{\mathrm{x}}} \mathrm{dx}=2 \int_{0}^{1} \mathrm{e}^{-\mathrm{x}} \mathrm{dx}\) \(= 2\left[\mathrm{e}^{-\mathrm{x}}\right]_{0}^{1}=2\left[\mathrm{e}^{-1}-\mathrm{e}^{0}\right]\) \(= 2\left[\mathrm{e}^{-1}-1\right]\) \(= 2\left[1-\mathrm{e}^{-1}\right]\)
WB JEE-2019
Integral Calculus
86730
The value of the integral \(\int_{-1 / 2}^{1 / 2}\left\{\left(\frac{x+1}{x-1}\right)^{2}+\left(\frac{x-1}{x+1}\right)^{2}-2\right\}^{1 / 2} d x\) is equal to
(C): Given \(\int_{-1 / 2}^{1 / 2}\left\{\left(\frac{x+1}{x-1}\right)^{2}+\left(\frac{x-1}{x+1}\right)^{2}-2\right\}^{1 / 2} d x\) \(=\int_{-1 / 2}^{1 / 2}\left[\left(\frac{x+1}{x-1}-\frac{x-1}{x+1}\right)^{2}\right]^{1 / 2} d x=\int_{-1 / 2}^{1 / 2}\left(\frac{4 x}{x^{2}-1}\right) d x\) \(=\int_{-1 / 2}^{0}\left(\frac{4 x}{x^{2}-1}\right) d x-\int_{0}^{1 / 2}\left(\frac{4 x}{x^{2}-1}\right) d x\) \(=-4 \int_{-1 / 2}^{0} \frac{x}{1-x^{2}} d x+4 \int_{0}^{1 / 2} \frac{x}{1-x^{2}} d x\) \(=2\left[\log \left(1-x^{2}\right)\right]_{-1 / 2}^{0}-2\left[\log \left(1-x^{2}\right)\right]_{0}^{1 / 2}\) \(=-2 \log \left(1-\frac{1}{4}\right)-2 \log \left(1-\frac{1}{4}\right)\) \(=-4 \log \frac{3}{4}=4 \log \frac{4}{3}\)
WB JEE-2021
Integral Calculus
86731
\(\int_{-\pi / 4}^{\pi / 4} \frac{d x}{1+\cos 2 x}\) is equal to
1 1
2 0
3 2
4 4
Explanation:
(A) : Given, \(\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{\mathrm{dx}}{1+\cos 2 \mathrm{x}}\) Let, \(\quad \mathrm{I}=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{d \mathrm{x}}{1+2 \cos ^{2} \mathrm{x}-1}\) \(=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{\mathrm{dx}}{2 \cos ^{2} \mathrm{x}}\) \(\because \quad \mathrm{f}(\mathrm{x})=\frac{1}{\cos ^{2} \mathrm{x}}\) is an even function \(\therefore \quad\) By using property \(\int_{-a}^{a} f(x) d x=2 \int_{0}^{a} f(x) d x\) We get, \(I=2 x \frac{1}{2} \int_{0}^{\pi / 4} \sec ^{2} x d x\) \(=|\tan x|_{0}^{\pi / 4}=\tan \frac{\pi}{4}-\tan 0\) \(=1\)
86729
The value of the integral \(\int_{-1}^{1}\left\{\frac{x^{2015}}{e^{|x|}\left(x^{2}+\cos x\right)}+\frac{1}{e^{|x|}}\right\} d x\) is equal to
1 0
2 \(1-\mathrm{e}^{-1}\)
3 \(2 \mathrm{e}^{-1}\)
4 \(2\left(1-\mathrm{e}^{-1}\right)\)
Explanation:
(D) : Let, \(I=\int_{-1}^{1}\left\{\frac{x^{2015}}{e^{|x|\left(x^{2}+\cos x\right)}}+\frac{1}{e^{|x|}}\right\} d x\) \(=\int_{-1}^{1} \frac{x^{2015}}{e^{|x|}\left(x^{2}+\cos x\right)} d x+\int_{-1}^{1} \frac{1}{e^{|x|}} d x\) Let \(f(x)=\frac{x^{2015}}{e^{|x|}\left(x^{2}+\cos x\right)}\) and \(g(x)=\frac{1}{e^{|x|}}\) \(f(-x)=\frac{(-x)^{2015}}{e^{-|x|}\left[(-x)^{2}+\cos (-x)\right]}\) \(=\frac{-x^{2015}}{e^{|x|}\left(x^{2}+\cos x\right)}\) and \(g(-x)=\frac{1}{e^{|-x|}}=\frac{1}{e^{|x|}}=g(x)\) \(\because \mathrm{f}(\mathrm{x})\) is odd function and \(\mathrm{g}(\mathrm{x})\) is even function. \(\int_{-1}^{1} \mathrm{f}(\mathrm{x}) \mathrm{dx}=0 \text { and }\) \(\int_{-1}^{1} \mathrm{~g}(\mathrm{x}) \mathrm{dx}=2 \int_{0}^{1} \mathrm{~g}(\mathrm{x}) \mathrm{dx}\) \(\mathrm{I}=2 \int_{0}^{1} \frac{1}{\mathrm{e}^{\mid \mathrm{x|}}} \mathrm{dx}\) \(=2 \int_{0}^{1} \frac{1}{\mathrm{e}^{\mathrm{x}}} \mathrm{dx}=2 \int_{0}^{1} \mathrm{e}^{-\mathrm{x}} \mathrm{dx}\) \(= 2\left[\mathrm{e}^{-\mathrm{x}}\right]_{0}^{1}=2\left[\mathrm{e}^{-1}-\mathrm{e}^{0}\right]\) \(= 2\left[\mathrm{e}^{-1}-1\right]\) \(= 2\left[1-\mathrm{e}^{-1}\right]\)
WB JEE-2019
Integral Calculus
86730
The value of the integral \(\int_{-1 / 2}^{1 / 2}\left\{\left(\frac{x+1}{x-1}\right)^{2}+\left(\frac{x-1}{x+1}\right)^{2}-2\right\}^{1 / 2} d x\) is equal to
(C): Given \(\int_{-1 / 2}^{1 / 2}\left\{\left(\frac{x+1}{x-1}\right)^{2}+\left(\frac{x-1}{x+1}\right)^{2}-2\right\}^{1 / 2} d x\) \(=\int_{-1 / 2}^{1 / 2}\left[\left(\frac{x+1}{x-1}-\frac{x-1}{x+1}\right)^{2}\right]^{1 / 2} d x=\int_{-1 / 2}^{1 / 2}\left(\frac{4 x}{x^{2}-1}\right) d x\) \(=\int_{-1 / 2}^{0}\left(\frac{4 x}{x^{2}-1}\right) d x-\int_{0}^{1 / 2}\left(\frac{4 x}{x^{2}-1}\right) d x\) \(=-4 \int_{-1 / 2}^{0} \frac{x}{1-x^{2}} d x+4 \int_{0}^{1 / 2} \frac{x}{1-x^{2}} d x\) \(=2\left[\log \left(1-x^{2}\right)\right]_{-1 / 2}^{0}-2\left[\log \left(1-x^{2}\right)\right]_{0}^{1 / 2}\) \(=-2 \log \left(1-\frac{1}{4}\right)-2 \log \left(1-\frac{1}{4}\right)\) \(=-4 \log \frac{3}{4}=4 \log \frac{4}{3}\)
WB JEE-2021
Integral Calculus
86731
\(\int_{-\pi / 4}^{\pi / 4} \frac{d x}{1+\cos 2 x}\) is equal to
1 1
2 0
3 2
4 4
Explanation:
(A) : Given, \(\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{\mathrm{dx}}{1+\cos 2 \mathrm{x}}\) Let, \(\quad \mathrm{I}=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{d \mathrm{x}}{1+2 \cos ^{2} \mathrm{x}-1}\) \(=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{\mathrm{dx}}{2 \cos ^{2} \mathrm{x}}\) \(\because \quad \mathrm{f}(\mathrm{x})=\frac{1}{\cos ^{2} \mathrm{x}}\) is an even function \(\therefore \quad\) By using property \(\int_{-a}^{a} f(x) d x=2 \int_{0}^{a} f(x) d x\) We get, \(I=2 x \frac{1}{2} \int_{0}^{\pi / 4} \sec ^{2} x d x\) \(=|\tan x|_{0}^{\pi / 4}=\tan \frac{\pi}{4}-\tan 0\) \(=1\)
