It looks hard

Calculus Level 4

$\large \int_0^{\pi /4} (\tan x)^{\sec x} \left [\ln ( ( \tan x)^{\sec x \tan x} ) + \dfrac{\sec^3 x}{\tan x} \right ] \, dx = \, ?$

\infty

1

0

\frac{\sqrt{2}}{2}

1 solution

Hana Wehbi
Jun 3, 2016

Let $\large y= (tanx)^{secx}$ , then

$\large ln y = secx \times ln(tanx)$ ,

$\large\frac{y'}{y}$ = $\large secxtanx \times ln(tanx)$ $+$ $\large \frac{sec^{3}x}{tanx}$ ,

$\large y'=$ $\large (secxtanx \times ln(tanx)$ $+$ $\large \frac{sec^{3}x}{tanx})$ $(\large tanx)^{secx}$ ,

Thus, $\large\int y'dy= y= (tanx)^{secx}] _ {0}^{\pi/4}$ = $\large1-0=1$

The integral doesn't have a value because there was no dx

J D - 5 years ago

Oh, i am sorry, i will correct it right now. It took me time to deal with latex, went on forgetting the dx. I appreciate that you didn't hit me with a report. Thanks.

Hana Wehbi - 5 years ago

It looks hard

1 solution

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