1
$\begingroup$

In the folowing :

Let $h(x)=(f\circ g)(x)+K$ where $K$ is any constant. If $$\frac{d}{dx}\left(h(x)\right)=-\frac{\sin{x}}{\cos^2{(\cos{x})}}$$ then compute the value of $j(0)$ where $$j(x)=\int_{g(x)}^{f(x)}\frac{f(t)}{g(t)}dt$$ where $f$ and $g$ are trigonometric functions.

I tried it and got $f(x) = \tan {x}$ and $g(x) = \cos {x}$.

But how can we find $j(0)$?

  • 0
    Please, follow [this guide](http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference) for how to write mathematics on this site. While it may seem irrelevant, if people do it the right way it greatly increases the accuracy of both the search engine and the "related questions" list. In addition, it will fit better with the formatting of the front page. Finally, the math looks better than whatever you're using. – 2017-01-05

2 Answers 2

2

Hint: We know that $f (0)=\tan 0=0$ and $g (0)=\cos 0=1$ Basically we $$j (0)=\int_{1}^{0} \frac {\sin x}{\cos ^2 x} dx $$ Substitute for $u=\cos x $ and get the answer. Hope it helps.

1

\begin{align}j(0)=\int_{\cos(0)}^{\tan(0)}\frac{\tan{x}}{\cos{x}}dx=\int_{1}^0\sec{(x)}\tan{(x)}\;dx=-\int_0^1(\sec{x})'\;dx=1-\sec{1}\end{align}