Evaluate $\int_{-\pi}^{\pi}(\cos a x-\sin b x)^{2} d x$

$2 \pi+\frac{\sin 2 a \pi}{2 a}-\frac{\sin 2 b \pi}{2 b}$

CBSE Class 12 · 2015 · Definite IntegralsLong Answer

Evaluate

\int_{- π}^{π} (cos a x - sin b x)^{2} d x

Answer: $2 π + \frac{s i n 2 aπ}{2 a} - \frac{s i n 2 bπ}{2 b}$

Expand:

(cos a x - sin b x)^{2} = cos^{2} a x + sin^{2} b x - 2 cos a x sin b x

.
On the symmetric interval

[- π, π]

cos^{2} a x

and

sin^{2} b x

are even, while

cos a x sin b x

is odd (its integral is

0

I = 2 \int_{0}^{π} cos^{2} a x d x + 2 \int_{0}^{π} sin^{2} b x d x

Use

cos^{2} a x = \frac{1 + c o s 2 a x}{2}

and

sin^{2} b x = \frac{1 - c o s 2 b x}{2}

I = \int_{0}^{π} (2 + cos 2 a x - cos 2 b x) d x = 2 π + \frac{sin 2 aπ}{2 a} - \frac{sin 2 bπ}{2 b}

CBSE Class 12 Board · 2015 · Subjective