Question

${\int (\sin x-\cos x)}^4(\sin x+\cos x)dx$ =

• A. $\frac{\sin x-\cos x}{5}+c$
• B. $\frac{{(\sin x-\cos x)}^5}{5}+c$
• C. $\frac{{(\sin x-\cos x)}^4}{4}+c$
• D. $\frac{{(\sin x+\cos x)}^5}{5}+c$
• E. $Noneoftheabove$

Answer 1

Answer: (B)

Let $I=\int {(\sin x-\cos x)}^4(\sin x+\cos x)dx$
Put $sinx-cosx=t$
$\Rightarrow$ $(\cos x+\sin x)dx=dt$
$\therefore$ $I=\int t^{4}dt=\frac{t^5}{5}+c$
$=\frac{{(\sin x-\cos x)}^5}{5}+c$