Question

If $I=\int {\sin }^{7}xdx$, then $I$ equals

• A. $-\cos x+{\cos }^{3}x-{\cos }^{5}x+\frac{1}{7}{\cos }^{7}x+C$
• B. $-\cos x+{\cos }^{3}x-\frac{3}{5}{\cos }^{5}x+\frac{1}{7}{\cos }^{7}x+C$
• C. $\tan x-{\tan }^{3}x+\sin x-\frac{1}{3}{\cos }^{3}x+C$
• D. $-\cos x+{\cos }^{5}x-\frac{1}{3}{\cos }^{3}x+\frac{1}{7}{\cos }^{7}x+C$

Answer 1

The correct option is B $-\cos x+{\cos }^{3}x-\frac{3}{5}{\cos }^{5}x+\frac{1}{7}{\cos }^{7}x+C$

Let $I=\int {\sin }^{7}xdx$

We know that, ${\sin }^{2}x=1-{\cos }^{2}x$

$=\int {(1-{\cos }^{2}x)}^3\sin xdx$

Substitute $\cos x=t\Rightarrow -\sin xdx=dt$

$\therefore I=-\int {(1-t^2)}^{3}dt$

$I=-\int (1-3t^2+3t^4-t^6)dt$

$I=-\cos x+{\cos }^{3}x-\frac{3}{5}{\cos }^{5}x+\frac{1}{7}{\cos }^{7}x+C$