Question

lf $f\left(x\right)=\cos x-{\cos }^{2}x+{\cos }^{3}x-\dots \dots \dots$ to $\infty$, then $\int f\left(x\right)dx$ is equal to

• A. $\tan \frac{x}{2}+c$
• B. $x-\tan \frac{x}{2}+c$
• C. $x-\frac{1}{2}\tan \frac{x}{2}+c$
• D. $\frac{x-\tan \frac{x}{2}}{3}+c$

Answer 1

Answer: (B)

$x-\tan \frac{x}{2}+c$

$f\left(x\right)=\cos x-{\cos }^{2}x+{\cos }^{3}x-...\infty$

The function is a G.P with a common ratio of $-\cos x$.

Hence

$f\left(x\right)=\frac{a}{1-r}$ where $a$ is the first term and $r$ is the common ratio.

$\therefore f\left(x\right)=\frac{\cos x}{1-(-\cos x)}$

$=\frac{\cos x}{1+\cos x}$

$=\frac{1+\cos x-1}{1+\cos x}$

$=1-\frac{1}{1+\cos x}$

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

$=1-\frac{{\sec }^2\frac{x}{2}}{2}$

$=1-\frac{1}{2}\left({\sec }^2\frac{x}{2}\right)$

Hence

$\int f\left(x\right).dx$

$=\int (1-\frac{1}{2}\left({\sec }^2\frac{x}{2}\right))dx$

$=x-\frac{1}{2}\left(\frac{\tan \frac{x}{2}}{\frac{1}{2}}\right)+c$

$=x-\tan \left(\frac{x}{2}\right)+c$.