Answer to Question #108788 in Calculus for Nimra

Question

∫cos⁻¹ax dx

Answer 1

$\int \cfrac {dx}{\cos ax}= ...$

multipy all to $a\cos ax$

... = $\int \cfrac {a\cos ax dx}{a\cos^2 ax}= \int \cfrac { d\sin ax}{a\cos^2 ax} = ...$

(use formula $\cos^2 ax = 1 - \sin^2 ax$ )

$... = \int \cfrac { d\sin ax}{a(1-\sin^2 ax)} = ...$

take the constant a out of brackets and make a substitution $y = \sin ax$

$... =\frac 1a \int \cfrac {dy}{(1-y^2)} = ...$

Using a decomposition into fractions

$\cfrac {1}{(1-y^2)} = \cfrac {1}{(1-y)(1+y)} = \cfrac {A}{(1-y)}+\cfrac {B}{(1+y)} = \cfrac {A(1+y) +B(1-y) }{(1-y)(1+y)}=\cfrac {A+Ay +B-By }{(1-y)(1+y)} = \cfrac {A(1+y) +B(1-y) }{(1-y)(1+y)}=\cfrac {A+B+y(A -B) }{(1-y)(1+y)}.$

$\begin{cases} A+B=1 \\ A-B = 0 \end{cases}$ , $\begin{cases} A=1/2 \\ B=1/2 \end{cases}$ .

$... =\cfrac {1}{2a} (\int \cfrac {dy}{(1-y)}+ \int \cfrac {dy}{(1+y)})= \cfrac {1}{2a} (-\int \cfrac {d(1-y)}{(1-y)}+ \int \cfrac {d(1+y)}{(1+y)})=\cfrac {1}{2a} (-\ln (1-y) + const_1+\ln (1+y)+const_2) = \cfrac {1}{2a} (-\ln (1-\sin ax) + const_1+\ln (1+\sin ax)+const_2) = \cfrac {1}{2a} (\ln (1+\sin ax)-\ln (1-\sin ax)) + Const.$

Answer: $\int \cfrac {dx}{\cos ax}= \cfrac {1}{2a} (\ln (1+\sin ax)-\ln (1-\sin ax)) + Const.$