Show that the integral of tan (x) is ln|sec (x)| + C where C is a ...
If the integral ∫ tan x + tan α tan x − tan α d x = A (x) cos 2 α + B (x) sin 2 α + C, where C is a constant of integration, then the functions A (x) and B (x) are respectively.
Let alpha epsilon (0, pi/2) be fixed. If the integral int dfrac ... - Toppr
If ∫ 5tanx tanx−2dx = x+aln|sinx −2cosx|+C for arbitrary constant of integration C, then the value of a is
If the integral displaystyle int { dfrac { 5tan { x } } { tan { x } -2 ...
Integrate sec^2 (x)tan (X)dx This can be done with integration by substitution. If we let u=tanx then du/dx=sec^2 (X). If we substitute U into the integrand we get it being u (sec^2 (X))dx. rearranging the du/dx equation to make dx the subject and we get dx=1/ (sec^2 (x)) du and so subbing this into the equation we see the sec^2 (x) cancel.
∫π/4 0 log(1+tanx)dx is equal to π 8loge2 π 4logee π 4loge2 π 8loge(1 2)
The integral which you describe has no closed form which is to say that it cannot be expressed in elementary functions. For example, you can express $\int x^2 \mathrm {d}x$ in elementary functions such as $\frac {x^3} {3} +C$. However, the indefinite integral from $ (-\infty,\infty)$ does exist and it is $\sqrt {\pi}$ so explicitly: $$\int^ {\infty}_ {-\infty} e^ {-x^2} = \sqrt {\pi}$$ Note ...