0Math Nerd 1729Best Answer · Jul 29The answer is ½ ln(1+√2) + 1.5 √2I figured it out by rewriting the integrand as:cos(t) sqrt(1+16sin²(t))Then I substituted u=sin(t) and then did the resulting integral via trigonometric substitution.Hope that helps! :)

SoulDeath7d agoWoah... actually I forgot that this trigonometric identity was also usable for powers of functions. Thanks!

The answer is ½ ln(1+√2) + 1.5 √2

I figured it out by rewriting the integrand as:

cos(t) sqrt(1+16sin²(t))

Then I substituted u=sin(t) and then did the resulting integral via trigonometric substitution.

Hope that helps! :)

Woah... actually I forgot that this trigonometric identity was also usable for powers of functions. Thanks!