JEE MATHEMATICS MCQ: INDEFINITE INTEGRATION

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Analyzing +4 | -1

\int \frac{\sin ^4 x}{\cos x} d x=\frac{1}{2} \log _e\left(\frac{1+\sin x}{1-\sin x}\right)-g(x)+c

where

g(x)

equals

View answer and explanation

Correct option: A

\frac{1}{3} \sin ^3 x+\sin x

Option A (

\left[-1,-\frac{3}{4}\right]

) is correct because the range of the expression

\frac{1}{4}(\sin 2 \theta)^2-1

falls within this interval.

More Questions from INDEFINITE INTEGRATION

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\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