Substitutes for a headache!

Calculus Level 5

$\large \int_{0}^{\frac{\pi}{2}} \prod_{r=1}^{2017} \cos rx \sum_{r=1}^{2017}r \tan rx \ dx = \ ?$

The answer is 1.

1 solution

Rohith M.Athreya
Apr 16, 2017

Let $\displaystyle \large y=\prod_{r=1}^{n}cosrx$

$\displaystyle \large logy=\sum_{r=1}^{n}log(cosrx)$

$\displaystyle \large \frac{y'}{y}=-\sum_{r=1}^{n}rtanrx$

$\displaystyle \large \int_{0}^{\frac{\pi}{2}} (\prod_{r=1}^{n} cosrx )(\sum_{r=1}^{n}rtanrx)dx=\int_{0}^{\frac{\pi}{2}}-y' dx = -y|_{0}^{\frac{\pi}{2}}=1$

Did the same!! Btw, better to ask the value of the integral for a particular value of n, otherwise it gives a hint that the value doesn't depend on n, substituting n =1 , gives the ans directly

Sumanth R Hegde - 4 years, 1 month ago

oops!I didnt see that!

have changed the form of the question.

thanks! :)

Rohith M.Athreya - 4 years, 1 month ago

Substitutes for a headache!

The answer is 1.

1 solution

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