Question

$\int \frac{se{c}^{2}x}{(secx+tanx{)}^{9/2}}dx$ equals to (for some arbitrary constant K)

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

Answer 1

Answer: (C)

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