Question

The intergral $\int \frac{se{c}^{2}x}{(secx+tanx{)}^{\frac{9}{2}}}$dx equal (for some arbitrary constant k)

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

Answer 1

The correct option is C $-\frac{1}{(secx+tanx{)}^{\frac{11}{2}}}\{\frac{1}{11}+\frac{1}{7}(secx+tanx{)}^2\}+k$

$I=\int \frac{se{c}^{2}x}{(secx+tanx{)}^{9/2}}dx\text{Let}secx+tanx=t\Rightarrow secx-tanx=\frac{1}{t}\Rightarrow secx=\frac{1}{2}(t+\frac{1}{t})\text{Also}secx(secx+tanx)dx=dt\Rightarrow secxdx=\frac{dt}{t}\therefore I=\frac{1}{2}\int \frac{(t+\frac{1}{t})dt}{t^{\frac{9}{2}}.t}=\frac{1}{2}\int (t^{-9/2}+t^{-13/2})dt=\frac{1}{2}[\frac{t^{-9/2+1}}{\frac{-9}{2}+1}+\frac{t^{-13/2+1}}{\frac{-13}{2}+1}]+K=\frac{-1}{7}{t}^{-7/2}-\frac{1}{11}{t}^{\frac{-11}{2}}+k=\frac{-1}{7t^{7/2}}-\frac{1}{11t^{11/2}}+K=-\frac{1}{t^{11/2}}(\frac{1}{11}+\frac{t^2}{7})+K=\frac{-1}{(secx+tanx{)}^{11/2}}\{\frac{1}{11}+\frac{1}{7}(secx+tanx{)}^2\}+K$