Answer to Question #123908 in Calculus for Deepanshu

Question

Use integration by parts to integrate x sec x tan x dx ∫
(2)
b) Integrate 3( x )1 4x 12x 5 dx 2
∫
+ + + (3)
c) Integrate dx
x( x()4 )1
x x 5

Answer 1

$(1) \int xsecx tanx dx$

Using integration by parts we get

$\smallint udv=uv-\smallint vdu$

u=x dv= secxtanxdx

du= dx v= secx

$$\smallint xsecxtanxdx= xsecx-\smallint secxdx \implies xsecx- In |secx+ tanx|+C$$

(2)

$I=\int (3x+1)\sqrt{(2x+3)^2-4}dx=\frac{1}{2}\int(\frac{3(u-3)}{2}+1)\sqrt{u^2-4}du\, where \,u=2x+3 \,hence \,I=\frac{3}{4}\int u\sqrt{u^2-4} du-\frac{7}{4}\int\sqrt{u^2-4} du\,\, if\, u=\frac{2}{ \ cost},du=\frac{2\ sint}{ \cos^2t}dt\,\,then \,I=3\int\frac{ \sin^3t}{ \cos^4t} dt-\frac{7}{2}\int \frac{ \sin^2t}{\cos^3t} dt\, as H.B.Dwight \,Tables\, of \,integrals , \,N.Y.,\,The \,Mcmillan\, corp,\,1961,\,f.(452.23),\,f.(452.39) \,hence\,\, I=\frac{1-3\cos^2t}{ \cos^3t}-\frac{7}{2}(\frac{ \ sint}{2\ cos^2t}-\frac{1}{2}\ln\tan(\frac{\pi}{4}+\frac{t}{2}))\,\,where \,t=\arccos\frac{2}{2x+3}$

(3)

$I=\int(\frac{1}{x+1}+\frac{1}{x^2+4}) dx= \int ln |x+1 |+\frac{1}{2}\arctan\frac{x}{2}+C$