Question

Evaluate the integral ${\int }_0^{\pi /2}sin|2x-\alpha |dxwhere\alpha ϵ[0,\pi ]$.

• A. - 1
• B. 1
• C. 2

Answer 1

The correct option is C 2

Observe the integral first. This is a very fundamental function with a linear expression as its argument. In such case we can follow the substitution method.
So, $t=2x-\alpha x=\frac{(t+\alpha )}{2}dx=\left(\frac{1}{2}\right).\left[dt\right]Lowerlimit=2.0-\alpha =-\alpha Upperlimit=2.\frac{\pi }{2}-\alpha =\pi -\alpha \therefore {\int }_0^{\pi /2}sin|2x-\alpha |dx={\int }_{-\alpha }^{\pi -\alpha }sin|t|dt$
Observe the limits. Since $\alpha$ is positive $\left(\alpha ϵ\right[0,\pi \left]\right)$, the lower limit is negative and upper limit is greater than or equal to zero. Also there is a modulus inside the sin function and this is a critical point since the function is changing signs. So we will split the interval into two. A negative interval from $-\alpha to0$ and a positive interval from $0to\pi -\alpha$.
$\therefore {\int }_{-\alpha }^{\pi -\alpha }sin|t|dt={\int }_{-\alpha }^0-sin\left(t\right)dt+{\int }_0^{\pi -\alpha }sin\left(t\right)dt=+cos\left(x\right){|}_{-\alpha }^0-cos\left(x\right){|}_0^{\pi -\alpha }=1-cos(-\alpha )-cos(\pi -\alpha )+1=1-cos\left(\alpha \right)+cos\left(\alpha \right)+1=2$
Hence the correct option is (d)