Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSModel Paper 63 Marks
Question
Evaluate $\int_{-\pi}^\pi(\cos a x-\sin b x)^2 d x$.
✓
Answer
$\text { Given, } I=\int_{-\pi}^\pi(\cos a x-\sin b x)^2 d x$
$=\int_{-\pi}^\pi\left(\cos ^2 a x+\sin ^2 b x-2 \cos a x \sin b x\right) d x$
$=\int_{-\pi}^\pi\left(\cos ^2 a x+\sin ^2 b x\right) d x-2 \int_{-\pi}^\pi \cos a x \sin b x d x$
$=I_1-I_2$
Let,
$I_1=\int_{-\pi}^\pi\left(\cos ^2 a x+\sin ^2 b x\right) d x \text { ( be an even function) }$
$=2 \int_0^\pi\left(\cos ^2 a x+\sin ^2 b x\right) d x$
$=2 \int_0^\pi\left(\frac{1+\cos 2 a x}{2}+\frac{1-\cos 2 b x}{2}\right) d x$
$=\int_0^\pi(1+\cos 2 a x+1-\cos 2 b x) d x$
$=\int_0^\pi(2+\cos 2 a x-\cos 2 b x) d x$
$=\left(2 x+\frac{\sin 2 a x}{2 a}-\frac{\sin 2 b x}{2 b}\right)_0^\pi$
$=\left(2 \pi+\frac{\sin 2 a \pi}{2 a}-\frac{\sin 2 b \pi}{2 b}\right)-0$
$=2 \pi+\frac{\sin 2 a \pi}{2 a}-\frac{\sin 2 b \pi}{2 b}$
$I_2=2 \int_{-\pi}^\pi(\cos a x \sin b x) d x \text { ( be a odd function) }$
$=0\binom{\because \int_{-a}^a f(x)=2 \int_0^a f(x) d x, \text { if } f(x) \text { is even }}{0, \text { if } f(x) \text { is odd }}$
$\therefore I=I_1-I_2=2 \pi+\frac{\sin 2 a \pi}{2 a}-\frac{\sin 2 b \pi}{2 b}$
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