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)

\frac {-1}{(secx+tan \, x)^{11/2}} \bigg \\{ \frac {1}{11}+ \frac {1}{7}(secx+tan \, x)^2 \bigg \\}+K

Answer 2

PLAN Integration by Substitution \hspace5mm i.e. \, \, \, \, I= \int \limits f \\{g(x) \\}.. g^1(x)dx \hspace5mm put \, \, \, \, g(x)=t \Rightarrow g^1(x)dx=dt \hspace5mm \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 \hspace20mm sec^2x-tan^2x=1 $\Rightarrow \, \, \, \, \, \, (secx-tanx)(secx+tanx)=1$ $\Rightarrow \, \, \, \, \, \, \, \, \, \, \, \, secx-tanx= \frac {1}{t}$ Here, \hspace15mm I= \int \limits \frac {sec^2dx}{(secx+tanx)^{9/2}} Put secx+tanx=t $\Rightarrow (secxtanx+sec^2x)dx=dt$ \Rightarrow \hspace10mm secx.tdx=dt \Rightarrow \hspace10mm secxdx=\frac {dt}{t} \therefore \hspace8mm secx-tanx= \frac {1}{t} \Rightarrow secx= \frac {1}{2} \bigg (t+ \frac {1}{t} \bigg ) \therefore \hspace12mm I=\int \limits \frac {secx.secxdx}{(secx+tanx)^ {9/2}} $\Rightarrow I= \int \limits \frac {\frac {1}{2}\bigg (t+\frac {1}{t}\bigg ).\frac {dt}{t}}{t^{9/2}}= \frac {1}{2}\int \limits \bigg (\frac {1}{t^ {9/2}}+ \frac {1}{t^ {13/2}}\bigg ) dt$ \hspace5mm =- \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