Question

If $\int {\sin }^{13}x{\cos }^{3}xdx=A{\sin }^{14}x+B{\sin }^{16}x+C$, then $A+B=$_______.

• A. $\frac{1}{110}$
• B. $\frac{17}{112}$
• C. $\frac{15}{112}$
• D. $\frac{1}{112}$

Answer 1

The correct option is D $\frac{1}{112}$

$\int {\sin }^{13}x{\cos }^{3}xdx=A{\sin }^{14}x+B{\sin }^{16}x+C$

Let $\sin x=t$
Then $\cos xdx=dt$
$\Rightarrow \int {\sin }^{13}x{\cos }^{3}xdx=\int t^{13}(1-t^2)dt$
$=\int (t^{13}-t^{15})dt$
$=\frac{t^{14}}{14}-\frac{t^{16}}{16}+C$
$=\frac{{\sin }^{14}x}{14}-\frac{{\sin }^{16}x}{16}+C$
$\Rightarrow A=\frac{1}{14},B=-\frac{1}{16}$
$\therefore A+B=\frac{1}{14}-\frac{1}{16}=\frac{1}{112}$