NEET Test Series from KOTA - 10 Papers In MS WORD
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Integral Calculus
86363
If \(\frac{1}{x(x+1)(x+2) \ldots(x+n)}=\frac{A_{0}}{x}+\frac{A_{1}}{x+1}\) \(+\frac{A_{2}}{x+1}+\ldots . \frac{A_{n}}{x+n}\) then \(A_{r}\) is equal to
(B) : We have- \(\int \frac{d x}{x^{2}\left(x^{n}+1\right)^{\frac{(n-1)}{n}}}=\int \frac{d x}{x^{2} \cdot x^{n-1}\left(1+\frac{1}{x^{n}}\right)^{\left(\frac{n-1}{n}\right)}}\) \(=\int \frac{d x}{x^{n+1}\left(1+x^{-n}\right)^{\left.\frac{n-1}{n}\right)}}\) Put, \(\quad 1+\mathrm{x}^{-\mathrm{n}}=\mathrm{t}\) \(-n x^{-n-1} d x=d t\) \(-\mathrm{nx}^{-(\mathrm{n}+1)} \mathrm{dx}=\mathrm{dt}\) \(\frac{1}{\mathrm{x}^{\mathrm{n}+1}} \mathrm{dx}=\frac{-\mathrm{dt}}{\mathrm{n}}\) So, \(I =\int \frac{-d t / n}{t^{1-1 / n}}=-\frac{1}{n}\left[\int t^{\frac{1}{n}-1} d t\right]\) \(=-\frac{1}{n} \frac{t^{\frac{1}{n}-1+1}}{\frac{1}{n}-1+1}+C=-\frac{1}{n} \cdot n\left(1+x^{-n}\right)^{1 / n}+C\) \(=-\left(1+x^{-n}\right)^{1 / n}+C=\left[1+x^{-n}\right]^{1 / n}+C^{\prime}\) Then from, \([f(x)]^{1 / n}+C\) So, \(\quad f(x)=\left(1+x^{-n}\right)\)
Manipal-2013
Integral Calculus
86260
The value of \(\int \frac{1}{\sin ^{2} x \cos ^{2} x} d x\) is-
1 \(\tan x-\cot x+C\)
2 \(\tan x+\cot x+C\)
3 \(-\tan x-\cot x+C\)
4 None of these
Explanation:
(A) : \(\int \frac{1}{\sin ^{2} x \cos ^{2} x} d x\) \(=\int \frac{\sin ^{2} x+\cos ^{2} x}{\sin ^{2} x \cos ^{2} x} d x=\int \frac{1}{\cos ^{2} x}+\frac{1}{\sin ^{2} x} d x\) \(=\int \sec ^{2} x d x+\int \operatorname{cosec}^{2} x d x=\tan x-\cot x+C\)
86363
If \(\frac{1}{x(x+1)(x+2) \ldots(x+n)}=\frac{A_{0}}{x}+\frac{A_{1}}{x+1}\) \(+\frac{A_{2}}{x+1}+\ldots . \frac{A_{n}}{x+n}\) then \(A_{r}\) is equal to
(B) : We have- \(\int \frac{d x}{x^{2}\left(x^{n}+1\right)^{\frac{(n-1)}{n}}}=\int \frac{d x}{x^{2} \cdot x^{n-1}\left(1+\frac{1}{x^{n}}\right)^{\left(\frac{n-1}{n}\right)}}\) \(=\int \frac{d x}{x^{n+1}\left(1+x^{-n}\right)^{\left.\frac{n-1}{n}\right)}}\) Put, \(\quad 1+\mathrm{x}^{-\mathrm{n}}=\mathrm{t}\) \(-n x^{-n-1} d x=d t\) \(-\mathrm{nx}^{-(\mathrm{n}+1)} \mathrm{dx}=\mathrm{dt}\) \(\frac{1}{\mathrm{x}^{\mathrm{n}+1}} \mathrm{dx}=\frac{-\mathrm{dt}}{\mathrm{n}}\) So, \(I =\int \frac{-d t / n}{t^{1-1 / n}}=-\frac{1}{n}\left[\int t^{\frac{1}{n}-1} d t\right]\) \(=-\frac{1}{n} \frac{t^{\frac{1}{n}-1+1}}{\frac{1}{n}-1+1}+C=-\frac{1}{n} \cdot n\left(1+x^{-n}\right)^{1 / n}+C\) \(=-\left(1+x^{-n}\right)^{1 / n}+C=\left[1+x^{-n}\right]^{1 / n}+C^{\prime}\) Then from, \([f(x)]^{1 / n}+C\) So, \(\quad f(x)=\left(1+x^{-n}\right)\)
Manipal-2013
Integral Calculus
86260
The value of \(\int \frac{1}{\sin ^{2} x \cos ^{2} x} d x\) is-
1 \(\tan x-\cot x+C\)
2 \(\tan x+\cot x+C\)
3 \(-\tan x-\cot x+C\)
4 None of these
Explanation:
(A) : \(\int \frac{1}{\sin ^{2} x \cos ^{2} x} d x\) \(=\int \frac{\sin ^{2} x+\cos ^{2} x}{\sin ^{2} x \cos ^{2} x} d x=\int \frac{1}{\cos ^{2} x}+\frac{1}{\sin ^{2} x} d x\) \(=\int \sec ^{2} x d x+\int \operatorname{cosec}^{2} x d x=\tan x-\cot x+C\)
86363
If \(\frac{1}{x(x+1)(x+2) \ldots(x+n)}=\frac{A_{0}}{x}+\frac{A_{1}}{x+1}\) \(+\frac{A_{2}}{x+1}+\ldots . \frac{A_{n}}{x+n}\) then \(A_{r}\) is equal to
(B) : We have- \(\int \frac{d x}{x^{2}\left(x^{n}+1\right)^{\frac{(n-1)}{n}}}=\int \frac{d x}{x^{2} \cdot x^{n-1}\left(1+\frac{1}{x^{n}}\right)^{\left(\frac{n-1}{n}\right)}}\) \(=\int \frac{d x}{x^{n+1}\left(1+x^{-n}\right)^{\left.\frac{n-1}{n}\right)}}\) Put, \(\quad 1+\mathrm{x}^{-\mathrm{n}}=\mathrm{t}\) \(-n x^{-n-1} d x=d t\) \(-\mathrm{nx}^{-(\mathrm{n}+1)} \mathrm{dx}=\mathrm{dt}\) \(\frac{1}{\mathrm{x}^{\mathrm{n}+1}} \mathrm{dx}=\frac{-\mathrm{dt}}{\mathrm{n}}\) So, \(I =\int \frac{-d t / n}{t^{1-1 / n}}=-\frac{1}{n}\left[\int t^{\frac{1}{n}-1} d t\right]\) \(=-\frac{1}{n} \frac{t^{\frac{1}{n}-1+1}}{\frac{1}{n}-1+1}+C=-\frac{1}{n} \cdot n\left(1+x^{-n}\right)^{1 / n}+C\) \(=-\left(1+x^{-n}\right)^{1 / n}+C=\left[1+x^{-n}\right]^{1 / n}+C^{\prime}\) Then from, \([f(x)]^{1 / n}+C\) So, \(\quad f(x)=\left(1+x^{-n}\right)\)
Manipal-2013
Integral Calculus
86260
The value of \(\int \frac{1}{\sin ^{2} x \cos ^{2} x} d x\) is-
1 \(\tan x-\cot x+C\)
2 \(\tan x+\cot x+C\)
3 \(-\tan x-\cot x+C\)
4 None of these
Explanation:
(A) : \(\int \frac{1}{\sin ^{2} x \cos ^{2} x} d x\) \(=\int \frac{\sin ^{2} x+\cos ^{2} x}{\sin ^{2} x \cos ^{2} x} d x=\int \frac{1}{\cos ^{2} x}+\frac{1}{\sin ^{2} x} d x\) \(=\int \sec ^{2} x d x+\int \operatorname{cosec}^{2} x d x=\tan x-\cot x+C\)
86363
If \(\frac{1}{x(x+1)(x+2) \ldots(x+n)}=\frac{A_{0}}{x}+\frac{A_{1}}{x+1}\) \(+\frac{A_{2}}{x+1}+\ldots . \frac{A_{n}}{x+n}\) then \(A_{r}\) is equal to
(B) : We have- \(\int \frac{d x}{x^{2}\left(x^{n}+1\right)^{\frac{(n-1)}{n}}}=\int \frac{d x}{x^{2} \cdot x^{n-1}\left(1+\frac{1}{x^{n}}\right)^{\left(\frac{n-1}{n}\right)}}\) \(=\int \frac{d x}{x^{n+1}\left(1+x^{-n}\right)^{\left.\frac{n-1}{n}\right)}}\) Put, \(\quad 1+\mathrm{x}^{-\mathrm{n}}=\mathrm{t}\) \(-n x^{-n-1} d x=d t\) \(-\mathrm{nx}^{-(\mathrm{n}+1)} \mathrm{dx}=\mathrm{dt}\) \(\frac{1}{\mathrm{x}^{\mathrm{n}+1}} \mathrm{dx}=\frac{-\mathrm{dt}}{\mathrm{n}}\) So, \(I =\int \frac{-d t / n}{t^{1-1 / n}}=-\frac{1}{n}\left[\int t^{\frac{1}{n}-1} d t\right]\) \(=-\frac{1}{n} \frac{t^{\frac{1}{n}-1+1}}{\frac{1}{n}-1+1}+C=-\frac{1}{n} \cdot n\left(1+x^{-n}\right)^{1 / n}+C\) \(=-\left(1+x^{-n}\right)^{1 / n}+C=\left[1+x^{-n}\right]^{1 / n}+C^{\prime}\) Then from, \([f(x)]^{1 / n}+C\) So, \(\quad f(x)=\left(1+x^{-n}\right)\)
Manipal-2013
Integral Calculus
86260
The value of \(\int \frac{1}{\sin ^{2} x \cos ^{2} x} d x\) is-
1 \(\tan x-\cot x+C\)
2 \(\tan x+\cot x+C\)
3 \(-\tan x-\cot x+C\)
4 None of these
Explanation:
(A) : \(\int \frac{1}{\sin ^{2} x \cos ^{2} x} d x\) \(=\int \frac{\sin ^{2} x+\cos ^{2} x}{\sin ^{2} x \cos ^{2} x} d x=\int \frac{1}{\cos ^{2} x}+\frac{1}{\sin ^{2} x} d x\) \(=\int \sec ^{2} x d x+\int \operatorname{cosec}^{2} x d x=\tan x-\cot x+C\)
