JEE MATHEMATICS MCQ: INDEFINITE INTEGRATION

1 Easy

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Application +4 | -1

\int \cos ^{-1}\left(\frac{1}{x}\right) d x

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Correct option: B

x \sec ^{-1} x-\cosh ^{-1} x+C

Option B is correct because the integral of arccos(1/x) dx is equivalent to the integral of arcsec(x) dx, which after integration by parts results in x arcsec(x) - arccosh(x) + C.

More Questions from INDEFINITE INTEGRATION

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\int \sin ^5 x \cos ^4 x d x=

f(x)

is differentibale and

\int\limits_{0}^{t^2} x f(x) d x=\frac{2}{5} t^5

then

f\left(\frac{4}{25}\right)

equals If

\int \frac{1-7 \cos ^2 x}{\sin ^7 x \cos ^2 x} d x=\frac{f(x)}{\sin ^7 x}+C

is equal to If

\int {e}^{{x}^2}\left(2-\frac{1}{{x}^2}\right) {de}={e}^{{x}^2} {f}({x})+{C}

and

{f}\left(\frac{1}{2}\right)=2

, then

f(1)

is equal to (where

{C}

is an arbitrary constant)

\int \frac{x^7}{\left(1-x^2\right)^5} d x

is equal to:

\int \,\frac{{a{x^3} + \,\,b{x^2} + c}}{{{x^4}}}\;\,dx

equals