Question

The value of integral $\int [\sin x-\cos x]dx$ will

• A. $0$ if $x\in (0,\frac{\pi }{2})$
• B. $[-\cos x-\sin x]\forall x\in (\frac{\pi }{2},\pi )$
• C. Some constant if $x\in [0,\frac{\pi }{2}]$
• D. $[-\cos x-\sin x]\forall x\in (0,\frac{\pi }{2})$

Answer 1

The correct option is C Some constant if $x\in [0,\frac{\pi }{2}]$

Given : $\int [\sin x-\cos x]dx$

For, $x\in [0,\frac{\pi }{2}]$

$\int [\sin x-\cos x]dx={\int }_0^{\frac{\pi }{4}}[\sin x-\cos x]dx+{\int }_{\frac{\pi }{4}}^{\frac{\pi }{2}}[\sin x-\cos x]dxx\in [0,\frac{\pi }{4}],\sin x-\cos x\in (-1,0)x\in [\frac{\pi }{4},\frac{\pi }{2}],\sin x-\cos x\in (0,1)\int [\sin x-\cos x]dx={\int }_0^{\frac{\pi }{4}}(-1)dx+{\int }_{\frac{\pi }{4}}^{\frac{\pi }{2}}0dx=-{\left[x\right]}_0^{\frac{\pi }{4}}-\frac{\pi }{4}$

Sum is constant if $x\in [0,\frac{\pi }{2}]$