Question

$\int_{}^{}{\tan^{n}xdx,}$equals –

• A. $\frac{\tan^{n–1}x}{(n - 1)}$ – $\int_{}^{}{\tan^{n–2}x}$dx
• B. $\frac{\tan^{n–1}x}{(n - 2)}$ – $\int_{}^{}\tan^{n–3}$x dx
• C. $\frac{\tan^{n–2}x}{(n - 1)}$ – $\int_{}^{}\tan^{n}$x dx
• D. None of these

Answer 1

Answer: (A)

$\frac{\tan^{n–1}x}{(n - 1)}$ – $\int_{}^{}{\tan^{n–2}x}$dx

Answer 2

Let I_n = $\int_{}^{}\tan^{n}$x dx

= $\int_{}^{}\tan^{n - 2}$x . tan² x dx**=** $\int_{}^{}\tan^{n - 2}$x . (sec² x – 1) dx

= $\int_{}^{}\tan^{n - 2}$x . sec²x dx – $\int_{}^{}\tan^{n - 2}$x dx I_n = $\frac{\tan^{n - 1}x}{(n - 1)}$– I_{n – 2}

Hence $\int_{}^{}\tan^{n}$x dx = $\frac{\tan^{n - 1}x}{(n - 1)}$– $\int_{}^{}{\tan^{n - 2}xdx}$