If \int \frac{\sin^{4}x}{\cos x}dx=\frac{1}{2} \log_{e}(\frac{1+\sin x}{1-\sin x})- g(x)+c

where g(x) =

  • Option 1)

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

  • Option 2)

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

  • Option 3)

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

  • Option 4)

    \frac{1}{3}\ cos^{3}x - \cos x

 

Answers (1)

 

Integration of trigonometric function of power m -

\int sin^{m}xdx    and 

 \int cos^{m}xdx

- wherein

for m=2,

sin^{2}x=\frac{1-cos 2x}{2}

for m=3,

sin3x=3sinx-4sin^{3}x

 

 \int \frac{\sin ^{4}x}{\cos x}dx=\frac{1}{2}log_{e}\left ( \frac{1+\sin x}{1-\sin x} \right )-g(x)+C

=\int \frac{\sin ^{4}x.\cos x}{\cos ^{2}x}dx

=\int \frac{\sin ^{4}x}{1-\sin ^{2}x}\cos xdx

=\int \frac{t^{4}}{1-t^{2}}dt

\left [ \because let \:Sinx=t \: \cos xdx=dt \right ]

=>\int \frac{t^{4}-1+1}{1-t^{2}}dt

=\int \frac{(t^{2}-1)(t^{2}+1)}{1-t^{2}}dt+\int \frac{1}{1-t^{2}}dt

=\int-(t^{2}+1)dt+\int \frac{1}{1-t^{2}}dt

-\left ( \frac{t^{3}}{3}+t \right )+\frac{1}{2}log\frac{1+t}{1-t}+C

=>\frac{1}{2}log\left ( \frac{1+\sin x}{1-\sin x} \right )-\left ( \sin x+\frac{\sin ^{3}x}{3} \right )

\therefore g(x)=\frac{\sin ^{3}x}{3}+\sin x


Option 1)

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

Option is Correct

Option 2)

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

Option is incorrect

Option 3)

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

Option is incorrect

Option 4)

\frac{1}{3}\ cos^{3}x - \cos x

Option is incorrect

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