Question

Expand log(1 + sin² x) in powers of x as far as the term in x⁶

Answer 1

Expand f (x) = log (1 + sin2
x) in powers of x upto x
6
.

Answer 2

log (1 + sin²x)

= sin²x - \(\frac{sin^4x}{2}\) + \(\frac{sin^6x}{3}\) - ...

(∵ log ( 1 + x) = x - \(\frac{x^2}{2}\)+ \(\frac{x^3}{3}\) ... if 1 ≤ x ≤ 1 and here 0 ≤ sin²x ≤ 1)

= (x - \(\frac{x^3}{3!}\)+ \(\frac{x^5}{5!}\)- ...)² - \(\frac{1}{2}\)(x - \(\frac{x^3}{3!}\)+ \(\frac{x^5}{5!}\)- ...)⁴ + \(\frac{1}{3}\)(x - \(\frac{x^3}{3!}\)+ \(\frac{x^5}{5!}\)- ...)⁶ - .....

(∵ sinx = x - \(\frac{x^3}{3!}\)+ \(\frac{x^5}{5!}\)- ...)

Since,

We consider power of x at the terms x⁶.

Therefore,

Remove other terms which gives power more than 6.

= (x - \(\frac{x^3}{3!}\)+ \(\frac{x^5}{5!}\)+ ...)² - \(\frac{1}{2}\)(x - \(\frac{x^3}{3!}\)+ \(\frac{x^5}{5!}\)+ ...)⁴ + (\(\frac{x^6}{3}\) + ....)

= (x² + \(\frac{x^6}{(3!)^2}\) - \(\frac{2x^4}{3!}\) + \(\frac{2x^6}{5!}\) + ....) - \(\frac{1}{2}\)(x⁴ - \(\frac{4x^6}{3!}\)+...) + (\(\frac{x^6}{3}\) + ...)

= x² - x⁴(\(\frac{2}{3!}\) + \(\frac{1}{2}\)) + x⁶(\(\frac{1}{(3!)^2}\)+\(\frac{2}{5!}\)+\(\frac{2}{3!}\)+\(\frac{1}{3}\)) + ....

= x² - x⁴(\(\frac{2}{6}\) + \(\frac{1}{2}\)) + x⁶(\(\frac{1}{36}\)+\(\frac{2}{120}\)+\(\frac{2}{6}\)+\(\frac{1}{3}\)) + ....

(∵ 3! = 6, 5! = 120)

= x² - x⁴\((\frac{2+3}{6})\) + x⁶ \(\times\)\(\frac{10+6+120+120}{360}\) + ....

= x² - \(\frac{5}{6}\)x⁴ + \(\frac{256}{360}\)x⁶ +....

=  x² - \(\frac{5}{6}\)x⁴ + \(\frac{32}{45}\)x⁶ +....

Hence,

Expansion of log(1 + sin²x) in power of x at terms of x⁶ in log(1+sin²x) = x² - \(\frac{5}{6}\)x⁴ + \(\frac{32}{45}\)x⁶ + ....