Question
What is $$\int {{{\sec }^n}x\,\tan \,x\,dx} $$ equal to ?
Where $$'c’$$ is a constant of integration.
A.
$$\frac{{{{\sec }^n}x}}{n} + c$$
B.
$$\frac{{{{\sec }^{n - 1}}x}}{{n - 1}} + c$$
C.
$$\frac{{{{\tan }^n}x}}{n} + c$$
D.
$$\frac{{{{\tan }^{n - 1}}x}}{{n - 1}} + c$$
Answer :
$$\frac{{{{\sec }^n}x}}{n} + c$$
Solution :
$$\eqalign{
& {\text{Let }}I = \int {{{\sec }^n}x\,\tan \,x\,dx} \cr
& {\text{Put, }}\sec \,x = t \Rightarrow \sec \,x\,\tan \,x\,dx = dt \cr
& \therefore \,I = \int {{t^n}.\frac{{dt}}{t}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\, = \int {{t^{n - 1}}dt} \cr
& \,\,\,\,\,\,\,\,\,\,\,\, = \frac{{{t^n}}}{n} + c \cr
& \,\,\,\,\,\,\,\,\,\,\,\, = \frac{{{{\sec }^n}x}}{n} + c \cr} $$
where $$'c’$$ is a constant of integration.