Question #104840
Use integration by parts to integrate ∫x secx tanx dx
1
Expert's answer
2020-03-09T10:56:24-0400

Solution:


xsecxtanx dx   =>   secx=1cosx   =>   xsinxcos2xdx              (1)\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: v du=uvu dv\int v~du=u\cdot v - \int u~dv

xsinxcos2xdx=[v=x, dv=dx                  du=sinxcos2xdx,u=1cosx]=xcosxdxcosx=xcosx()+C;\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;


()   dxcosx=cosxcos2xdx=d(sinx)cos2x=d(sinx)1sin2x  =>  sinx=t   =>  dt1t2(**)~~~\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: dt1t2=12lnt1t+1+C;\int \frac{\small{dt}} {\small{1-t^2}}=-\frac{1}{2}\ln|\frac{t-1}{t+1}|+C;


()   dxcosx=12lnsinx1sinx+1;(**)~~~\int\frac{\small{d x}} {\small{\cos x}} =-\frac{1}{2}\ln\large{|}\frac{\small{\sin x -1}}{\small{\sin x + 1}}\large{|};


xsinxcos2xdx=xcosx+12lnsinx1sinx+1+C;\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=sinxcos2xdx=d(cosx)cos2x=1cosx;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: xsinxcos2xdx=xcosx+12lnsinx1sinx+1+C;\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;




Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS