1
$\begingroup$

$$\large{\int_0^\theta (\tan\theta-\sin\theta)\sec^2\theta d\theta}$$

$$$$

The following Integral came up while computing the value of Work performed by a Spring force. I tried to search for the Closed Form on wolfram Alpha too, but could not get any Closed Form. $$$$I would truly appreciate i if somebody would kindly show me how to compute the Integral. Many thanks!

$\endgroup$
  • $\begingroup$ It works $\endgroup$ – user99914 Oct 21 '15 at 12:33
  • $\begingroup$ a primitive function is this here $$ \frac{\sec ^2(x)}{2}-\sec (x)$$ $\endgroup$ – Dr. Sonnhard Graubner Oct 21 '15 at 12:37
0
$\begingroup$

First of all, you might notice that the limit of integration and the variable of integration cannot be the same! So the right way of writing this down is

$$\int_0^\theta {\left( {\tan x - \sin x} \right){{\sec }^2}xdx} $$

As I just memorize the derivatives for ${\sin x }$ and ${\cos x }$, I do it this way

$$\eqalign{ & \int_0^\theta {\left( {\tan x - \sin x} \right){{\sec }^2}xdx} = \int_0^\theta {\tan x{{\sec }^2}xdx} - \int_0^\theta {\sin x{{\sec }^2}xdx} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \,\int_0^\theta {{{\sin x} \over {{{\cos }^3}x}}dx - \int_0^\theta {{{\sin x} \over {{{\cos }^2}x}}dx} } \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left. {{{{{\cos }^{ - 2}}x} \over 2}} \right|_0^\theta - \left. {{{{{\cos }^{ - 1}}x} \over 1}} \right|_0^\theta \cr} $$

and I leave the rest of computation for your! :)

$\endgroup$
1
$\begingroup$

Hint: Converting everything to sines and cosines sometimes helps. In this case

$$\int(\tan\theta-\sin\theta)\sec^2\theta\,d\theta=\int\left({1\over\cos^3\theta}-{1\over\cos^2\theta}\right)\sin\theta\,d\theta$$

Now let $u=\cos\theta$.

$\endgroup$
1
$\begingroup$

Split the integral as,

$$\displaystyle\int_{0}^\theta\tan\theta\sec^2\theta\ d\theta-\displaystyle\int_{0}^\theta\sec\theta\tan\theta\ d\theta$$

For the first integral substitute $\sec^2\theta=u$ or $\tan\theta=u$. I hope you can do the second one.

$\endgroup$
0
$\begingroup$

$\textbf{hint}$ $$ (\tan \theta - \sin \theta)\sec^2\theta = \left(\sec \theta -1\right)\sin \theta \sec^2\theta = \left(\sec \theta -1\right)\tan\theta \sec\theta $$ what is the derivative of $\sec \theta -1$?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.