JEE MATHEMATICS MCQ: INDEFINITE INTEGRATION

1 Medium

1 of 1

Analyzing +4 | -1

\int \frac{\cos ^4 x}{\sin ^2 x} d x=A \cot x+B \sin 2 x+\frac{C}{2} x+D

, then

View answer and explanation

Correct option: B

B=-1 / 4, C=-3

Option B is correct because after evaluating the integral, we find that

B = -\frac{1}{4}

and

C = -3

More Questions from INDEFINITE INTEGRATION

Practice All Chapter MCQs →

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

f(x)

\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)

{C}

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

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