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JEE MATHEMATICS MCQ: INDEFINITE INTEGRATION

1 Easy

1 of 1

Application +4 | -1

\int \frac{e^x(1+\sin x)}{1+\cos x} \mathrm{~d} x

is equal to

View answer and explanation

Correct option: B

e^x \tan x / 2+C

Option B (

e^x \tan x / 2+C

) is correct because the integral simplifies to the form

\int e^x (f(x) + f'(x)) dx = e^x f(x) + C

, where

f(x) = \tan(x/2)

.

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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

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

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

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

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

f(1)

is equal to (where

{C}

is an arbitrary constant)

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

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