I know the derivative of -secx is -sinx/cos^2 x. the first trouble is that the problem has e^(-x) instead of e^(x). the second is that -sin/cos^2 is the derivative of -secx, not secx. I'm sure these two things tie together somehow through a basic simplification but I can't figure this basic simplification outBvU said:So you want a primitive of ##\displaystyle { \sin x\over \cos^2 x}## .
It is almost handed over on a silver platter: if ##\ \ -\displaystyle {\sin x\over \cos^2 x}\ ## is ##f'(x)##, what do you have left over for ##f(x)## ?
AHHH! how INCREDIBLY stupid of me. I just had to put -x=t (sorry you had to read this question). I'm never going to be a scientist this way...BvU said:So you want a primitive of ##\displaystyle { \sin x\over \cos^2 x}## .
It is almost handed over on a silver platter: if ##\ \ -\displaystyle {\sin x\over \cos^2 x}\ ## is ##f'(x)##, what do you have left over for ##f(x)## ?
Integration by parts is a mathematical technique used to evaluate integrals of products of functions. It is based on the product rule of differentiation, where the derivative of a product of two functions is equal to the first function multiplied by the derivative of the second function plus the second function multiplied by the derivative of the first function.
Integration by parts is used when an integral involves a product of two functions, and it is difficult to evaluate the integral using other methods such as substitution or trigonometric identities. It is particularly useful for integrals involving logarithmic, exponential, or inverse trigonometric functions.
The formula for integration by parts is ∫u dv = uv - ∫v du, where u and v are functions and dv and du are their respective differentials. This formula is derived from the product rule of differentiation.
When choosing u and dv, it is important to follow the acronym "LIATE": logarithmic, inverse trigonometric, algebraic, trigonometric, exponential. In general, u should be the function that becomes simpler when differentiated, and dv should be the function that becomes easier to integrate when differentiated. It may take some trial and error to determine the best choice for u and dv.
One common mistake is to choose u and dv in a way that leads to a more complicated integral than the original one. It is important to choose u and dv strategically to simplify the integral. Additionally, forgetting to include the constant of integration when integrating dv can lead to incorrect results. It is also important to check for any symmetry in the integral, as this may allow for a simpler solution using other integration techniques.