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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)

best_answer

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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