Question

$\int \limits \frac {sec^2x}{(secx+tan \, x)^{9/2}}dx $ equals to (for some arbitrary constant K)

• A. $\frac {-1}{(secx+tan \, x)^{11/2}} \bigg \{ \frac {1}{11}- \frac {1}{7}(secx+tan \, x)^2 \bigg \}+K$
• B. $\frac {1}{(secx+tan \, x)^{11/2}} \bigg \{ \frac {1}{11}- \frac {1}{7}(secx+tan \, x)^2 \bigg \}+K$
• C. $\frac {-1}{(secx+tan \, x)^{11/2}} \bigg \{ \frac {1}{11}+ \frac {1}{7}(secx+tan \, x)^2 \bigg \}+K$
• D. $\frac {1}{(secx+tan \, x)^{11/2}} \bigg \{ \frac {1}{11}+ \frac {1}{7}(secx+tan \, x)^2 \bigg \}+K$

Answer 1

Answer: (C)

PLAN Integration by Substitution
$ i.e. \, \, \, \, I= \int \limits f \{g(x) \}.. g^1(x)dx$
$ put \, \, \, \, g(x)=t \Rightarrow g^1(x)dx=dt$
$ \therefore \, \, \, \, I= \int \limits f(t)dt$
Description of Situation Generally, students gets
confused after substitution, i.e. secx+tanx=t.
Now, for secx, we should use
$ sec^2x-tan^2x=1$
$\Rightarrow \, \, \, \, \, \, (secx-tanx)(secx+tanx)=1$
$\Rightarrow \, \, \, \, \, \, \, \, \, \, \, \, secx-tanx= \frac {1}{t}$
Here, $ I= \int \limits \frac {sec^2dx}{(secx+tanx)^{9/2}}$
Put secx+tanx=t $\Rightarrow (secxtanx+sec^2x)dx=dt$
$\Rightarrow secx.tdx=dt$
$\Rightarrow secxdx=\frac {dt}{t}$
$\therefore secx-tanx= \frac {1}{t} \Rightarrow secx= \frac {1}{2} \bigg (t+ \frac {1}{t} \bigg )$
$\therefore I=\int \limits \frac {secx.secxdx}{(secx+tanx)^ {9/2}}$
$\Rightarrow I= \int \limits \frac {\frac {1}{2}\bigg (t+\frac {1}{t}\bigg ).\frac {dt}{t}}{t^{9/2}}= \frac {1}{2}\int \limits \bigg (\frac {1}{t^ {9/2}}+ \frac {1}{t^ {13/2}}\bigg ) dt$
$ =- \frac {1}{2} \bigg \{\frac {2}{7t^{7/2}}+ \frac {2}{11t ^{11/2}}\bigg \}+K$
$= -\bigg [ \frac {1}{7(secx+tanx)^{7/2}} + \frac {1}{11(secx+tanx)^{11/2}}\bigg ]+K$
$= \frac {-1}{(secx+tanx)^{11/2}}\bigg \{\frac {1}{11}+ \frac {1}{7}(secx+tan x)^2 \bigg \}+K$