Answer to Question #104840 in Calculus for SHIVAM KUMAR

Question

Use integration by parts to integrate ∫x secx tanx dx

Answer 1

Solution:

$\int {x\sec x \tan x}~dx~~~\small{=>} ~~~|\sec x = \frac{1}{\small\cos x}|~~~ \small{=>}~~~\large\int \frac{\small{x\cdot\sin x}} {\small{\cos^2x}}\small{dx} ~~~~~~~~~~~~~~(1)$

After getting expression (1) we can use method integrate by parts: $\int v~du=u\cdot v - \int u~dv$

$\large\int \frac{\small{x\cdot\sin x}} {\small{\cos^2x}}\small{dx} =\begin{bmatrix} v = x, ~dv = dx~~~~~~~~~~~~~~~~~~\\ du=\frac{\small{\sin x}} {\small{\cos^2x}}dx, u^* = \frac{\small{1}} {\small{\cos x}} \end{bmatrix} = \frac{\small{x}} {\small{\cos x}}-\int\frac{\small{dx}} {\small{\cos x}} = \frac{\small{ x}} {\small{\cos x}}-(**) + C;$

$(**)~~~\int\frac{\small{d x}} {\small{\cos x}} = \int \frac{\small{\cos x}} {\small{\cos^2x}}dx = \int \frac{\small{d(\sin x})} {\small{\cos^2x}} = \int \frac{\small{d(\sin x)}} {\small{1-\sin^2x}}~~\small{=>}~~|\sin x =t~|~~\small{=>}~~\int \frac{\small{dt}} {\small{1-t^2}}$

So, we get a tabular integrale: $\int \frac{\small{dt}} {\small{1-t^2}}=-\frac{1}{2}\ln|\frac{t-1}{t+1}|+C;$

$(**)~~~\int\frac{\small{d x}} {\small{\cos x}} =-\frac{1}{2}\ln\large{|}\frac{\small{\sin x -1}}{\small{\sin x + 1}}\large{|};$

$\large\int \frac{\small{x\cdot\sin x}} {\small{\cos^2x}}\small{dx} =\frac{\small{ x}} {\small{\cos x}}+\frac{1}{2}\ln\large{|}\frac{\small{\sin x -1}}{\small{\sin x + 1}}\large{|} + C;$

$u^*=\int \frac{\small{\sin x}} {\small{\cos^2x}}dx = -\int \frac{\small{d(\cos x)}} {\small{\cos^2x}}=\frac{\small{1}} {\small{\cos x}};$

Answer: $\large\int \frac{\small{x\cdot\sin x}} {\small{\cos^2x}}\small{dx} =\frac{\small{ x}} {\small{\cos x}}+\frac{1}{2}\ln\large{|}\frac{\small{\sin x -1}}{\small{\sin x + 1}}\large{|} + C;$