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. $None\, of\, the\, above$

Answer 1

Answer: (B)

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