# Struggling to find a Closed Form for an Integral

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



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!

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

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! :)

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 $u=\cos\theta$.

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.

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